Testing nested dictionaries for a user input and retrieving directly related information
up vote
0
down vote
favorite
I’m new to Python 3 and have been experimenting with Dictionaries but I’m having problems testing for a specific key and retrieving its related values from nested dictionaries.
I want a user defined input to be checked against nested dictionaries and if the user input is found I then want to collect the details for that item and add it to another dictionary.
E.g.
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
userQuery = input("What food do you want to check for? ")
userQuery = "Apple"
Desired result:
basketDict = {"Fruit": {"Apple": "2"}}
I've attempted to use dictionary comprehension to form a new dictionary of only food items (Apple, Banana, Lettuce etc.) but keep running into issues when trying to collect the related category ("Fruit"/"Vegetable" and quantity information from the nested dictionaries.
Here's my (broken) code:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
shopCheck =
userQuery = input("What food do you want to check for? ")
for category, food in shopDict.items():
for each in food:
shopCheck.append(each)
if userQuery not in shopCheck:
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[category] = [food]
print(basketDict)
python python-3.x dictionary nested dictionary-comprehension
add a comment |
up vote
0
down vote
favorite
I’m new to Python 3 and have been experimenting with Dictionaries but I’m having problems testing for a specific key and retrieving its related values from nested dictionaries.
I want a user defined input to be checked against nested dictionaries and if the user input is found I then want to collect the details for that item and add it to another dictionary.
E.g.
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
userQuery = input("What food do you want to check for? ")
userQuery = "Apple"
Desired result:
basketDict = {"Fruit": {"Apple": "2"}}
I've attempted to use dictionary comprehension to form a new dictionary of only food items (Apple, Banana, Lettuce etc.) but keep running into issues when trying to collect the related category ("Fruit"/"Vegetable" and quantity information from the nested dictionaries.
Here's my (broken) code:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
shopCheck =
userQuery = input("What food do you want to check for? ")
for category, food in shopDict.items():
for each in food:
shopCheck.append(each)
if userQuery not in shopCheck:
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[category] = [food]
print(basketDict)
python python-3.x dictionary nested dictionary-comprehension
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I’m new to Python 3 and have been experimenting with Dictionaries but I’m having problems testing for a specific key and retrieving its related values from nested dictionaries.
I want a user defined input to be checked against nested dictionaries and if the user input is found I then want to collect the details for that item and add it to another dictionary.
E.g.
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
userQuery = input("What food do you want to check for? ")
userQuery = "Apple"
Desired result:
basketDict = {"Fruit": {"Apple": "2"}}
I've attempted to use dictionary comprehension to form a new dictionary of only food items (Apple, Banana, Lettuce etc.) but keep running into issues when trying to collect the related category ("Fruit"/"Vegetable" and quantity information from the nested dictionaries.
Here's my (broken) code:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
shopCheck =
userQuery = input("What food do you want to check for? ")
for category, food in shopDict.items():
for each in food:
shopCheck.append(each)
if userQuery not in shopCheck:
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[category] = [food]
print(basketDict)
python python-3.x dictionary nested dictionary-comprehension
I’m new to Python 3 and have been experimenting with Dictionaries but I’m having problems testing for a specific key and retrieving its related values from nested dictionaries.
I want a user defined input to be checked against nested dictionaries and if the user input is found I then want to collect the details for that item and add it to another dictionary.
E.g.
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
userQuery = input("What food do you want to check for? ")
userQuery = "Apple"
Desired result:
basketDict = {"Fruit": {"Apple": "2"}}
I've attempted to use dictionary comprehension to form a new dictionary of only food items (Apple, Banana, Lettuce etc.) but keep running into issues when trying to collect the related category ("Fruit"/"Vegetable" and quantity information from the nested dictionaries.
Here's my (broken) code:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
shopCheck =
userQuery = input("What food do you want to check for? ")
for category, food in shopDict.items():
for each in food:
shopCheck.append(each)
if userQuery not in shopCheck:
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[category] = [food]
print(basketDict)
python python-3.x dictionary nested dictionary-comprehension
python python-3.x dictionary nested dictionary-comprehension
edited Nov 12 at 15:31
toti08
1,74721523
1,74721523
asked Nov 12 at 14:35
MHSalehi
32
32
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
0
down vote
accepted
This code is enough for your requirements:
for category, food in shopDict.items():
if userQuery in food:
basketDict[category] = {userQuery: food[userQuery]}
But if you want to add new items to the basket, you should update it properly:
for category, food in shopDict.items():
if userQuery in food:
basketDict.update(
{category: {**basketDict.get(category, {}),
**{userQuery: food[userQuery]}}}
)
You want to merge the actual dictionary of the category (the basket's actual content) with the selected item's key-value.
Generally in python3 if you have 2 dictionaries this is the way to merge them into one:
d1 = {"one": 1, "two": 2}
d2 = {"three": 3}
merged = {**d1, **d2} # {'one': 1, 'two': 2, 'three': 3}
Because **
is used to unpack the dictionary.
Good way to understand it is if you have a function definition with some arguments, then you can call the function with a dictionary that has the argument names as keys. But you have to unpack it, otherwise you would pass the dictionary as the parameter, and not its values.
d = {"one": 1, "two": 2}
def example(one, two):
pass
example(d)
would throw a TypeError, because it's missing the second positional argument, but example(**d)
would work fine
This works wonderfully for my purpose. I am somewhat puzzled by the contents and structure of the basketDict.update() though - I'm not familiar with the '**' or structure of the contents within the (). Would you be able to provide an explanation or point me to resources where I can learn how it functions? Thanks for the solution.
– MHSalehi
Nov 12 at 20:29
I made some changes to be more clear. Also, you don't have to use the methodupdate
, simply set the key's value:basketDict[category] = {**basketDict.get(category, {}), **{userQuery: food[userQuery]}}
– M.Adam
Nov 13 at 9:18
add a comment |
up vote
0
down vote
Probably there's a simpler way to do this, but I just fixed your code.
When I run your code, everything inside shopDict
was going to basketDict
. This is because you wasn't cleaning your shopCheck
variable at each for
iteration.
And there was a second mistake when adding the elements to basketDict
. You need to add only the food
element that match the userQuery
. Here's the full fixed code:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
userQuery = input("What food do you want to check for? ")
for category, food in shopDict.items():
shopCheck =
for each in food:
shopCheck.append(each)
if userQuery not in shopCheck:
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[category] = {userQuery:food[userQuery]}
print(basketDict)
1
this too has indentation issues I guess
– Manoj Kengudelu
Nov 12 at 14:55
@ManojKengudelu You was right. Fixed!
– Hemerson Tacon
Nov 12 at 14:56
Thanks for the response. When I run this code and enter 'Apple', it erroneously produces, "That's not available." despite successfully adding it to the basketDict. Any ideas why?
– MHSalehi
Nov 12 at 20:41
@MHSalehi Actualy if you analyse the code carefully you will see that is not an error. Since theshopDict
has two pair of key-values (in this case category-food pairs), the outerfor
will be executed twice: one for the key "Fruit" and another for the key "Vegetables". That second iteration is generates that print becauseApple
indeed isn't present among the "Vegetable" values what is totaly right.
– Hemerson Tacon
Nov 13 at 14:50
add a comment |
up vote
0
down vote
I'd start by creating a dictionary containing all the items you can buy, whether they're fruit or vegetables. Then I'd simply check if the required item is on that dictionary:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
allVals = {}
for k, subDict in shopDict.items():
for k1, val in subDict.items():
allVals[k1] = val
userQuery = input("What food do you want to check for? ")
if userQuery not in allVals.keys():
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[userQuery] = allVals[userQuery]
print(basketDict)
AttributeError: 'dict' object has no attribute 'iteritems'
– Hemerson Tacon
Nov 12 at 14:52
Oops, sorry, I tested this with python2.7. I'll fix this!
– toti08
Nov 12 at 15:04
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53264397%2ftesting-nested-dictionaries-for-a-user-input-and-retrieving-directly-related-inf%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
This code is enough for your requirements:
for category, food in shopDict.items():
if userQuery in food:
basketDict[category] = {userQuery: food[userQuery]}
But if you want to add new items to the basket, you should update it properly:
for category, food in shopDict.items():
if userQuery in food:
basketDict.update(
{category: {**basketDict.get(category, {}),
**{userQuery: food[userQuery]}}}
)
You want to merge the actual dictionary of the category (the basket's actual content) with the selected item's key-value.
Generally in python3 if you have 2 dictionaries this is the way to merge them into one:
d1 = {"one": 1, "two": 2}
d2 = {"three": 3}
merged = {**d1, **d2} # {'one': 1, 'two': 2, 'three': 3}
Because **
is used to unpack the dictionary.
Good way to understand it is if you have a function definition with some arguments, then you can call the function with a dictionary that has the argument names as keys. But you have to unpack it, otherwise you would pass the dictionary as the parameter, and not its values.
d = {"one": 1, "two": 2}
def example(one, two):
pass
example(d)
would throw a TypeError, because it's missing the second positional argument, but example(**d)
would work fine
This works wonderfully for my purpose. I am somewhat puzzled by the contents and structure of the basketDict.update() though - I'm not familiar with the '**' or structure of the contents within the (). Would you be able to provide an explanation or point me to resources where I can learn how it functions? Thanks for the solution.
– MHSalehi
Nov 12 at 20:29
I made some changes to be more clear. Also, you don't have to use the methodupdate
, simply set the key's value:basketDict[category] = {**basketDict.get(category, {}), **{userQuery: food[userQuery]}}
– M.Adam
Nov 13 at 9:18
add a comment |
up vote
0
down vote
accepted
This code is enough for your requirements:
for category, food in shopDict.items():
if userQuery in food:
basketDict[category] = {userQuery: food[userQuery]}
But if you want to add new items to the basket, you should update it properly:
for category, food in shopDict.items():
if userQuery in food:
basketDict.update(
{category: {**basketDict.get(category, {}),
**{userQuery: food[userQuery]}}}
)
You want to merge the actual dictionary of the category (the basket's actual content) with the selected item's key-value.
Generally in python3 if you have 2 dictionaries this is the way to merge them into one:
d1 = {"one": 1, "two": 2}
d2 = {"three": 3}
merged = {**d1, **d2} # {'one': 1, 'two': 2, 'three': 3}
Because **
is used to unpack the dictionary.
Good way to understand it is if you have a function definition with some arguments, then you can call the function with a dictionary that has the argument names as keys. But you have to unpack it, otherwise you would pass the dictionary as the parameter, and not its values.
d = {"one": 1, "two": 2}
def example(one, two):
pass
example(d)
would throw a TypeError, because it's missing the second positional argument, but example(**d)
would work fine
This works wonderfully for my purpose. I am somewhat puzzled by the contents and structure of the basketDict.update() though - I'm not familiar with the '**' or structure of the contents within the (). Would you be able to provide an explanation or point me to resources where I can learn how it functions? Thanks for the solution.
– MHSalehi
Nov 12 at 20:29
I made some changes to be more clear. Also, you don't have to use the methodupdate
, simply set the key's value:basketDict[category] = {**basketDict.get(category, {}), **{userQuery: food[userQuery]}}
– M.Adam
Nov 13 at 9:18
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
This code is enough for your requirements:
for category, food in shopDict.items():
if userQuery in food:
basketDict[category] = {userQuery: food[userQuery]}
But if you want to add new items to the basket, you should update it properly:
for category, food in shopDict.items():
if userQuery in food:
basketDict.update(
{category: {**basketDict.get(category, {}),
**{userQuery: food[userQuery]}}}
)
You want to merge the actual dictionary of the category (the basket's actual content) with the selected item's key-value.
Generally in python3 if you have 2 dictionaries this is the way to merge them into one:
d1 = {"one": 1, "two": 2}
d2 = {"three": 3}
merged = {**d1, **d2} # {'one': 1, 'two': 2, 'three': 3}
Because **
is used to unpack the dictionary.
Good way to understand it is if you have a function definition with some arguments, then you can call the function with a dictionary that has the argument names as keys. But you have to unpack it, otherwise you would pass the dictionary as the parameter, and not its values.
d = {"one": 1, "two": 2}
def example(one, two):
pass
example(d)
would throw a TypeError, because it's missing the second positional argument, but example(**d)
would work fine
This code is enough for your requirements:
for category, food in shopDict.items():
if userQuery in food:
basketDict[category] = {userQuery: food[userQuery]}
But if you want to add new items to the basket, you should update it properly:
for category, food in shopDict.items():
if userQuery in food:
basketDict.update(
{category: {**basketDict.get(category, {}),
**{userQuery: food[userQuery]}}}
)
You want to merge the actual dictionary of the category (the basket's actual content) with the selected item's key-value.
Generally in python3 if you have 2 dictionaries this is the way to merge them into one:
d1 = {"one": 1, "two": 2}
d2 = {"three": 3}
merged = {**d1, **d2} # {'one': 1, 'two': 2, 'three': 3}
Because **
is used to unpack the dictionary.
Good way to understand it is if you have a function definition with some arguments, then you can call the function with a dictionary that has the argument names as keys. But you have to unpack it, otherwise you would pass the dictionary as the parameter, and not its values.
d = {"one": 1, "two": 2}
def example(one, two):
pass
example(d)
would throw a TypeError, because it's missing the second positional argument, but example(**d)
would work fine
edited Nov 13 at 9:20
answered Nov 12 at 15:21
M.Adam
8614
8614
This works wonderfully for my purpose. I am somewhat puzzled by the contents and structure of the basketDict.update() though - I'm not familiar with the '**' or structure of the contents within the (). Would you be able to provide an explanation or point me to resources where I can learn how it functions? Thanks for the solution.
– MHSalehi
Nov 12 at 20:29
I made some changes to be more clear. Also, you don't have to use the methodupdate
, simply set the key's value:basketDict[category] = {**basketDict.get(category, {}), **{userQuery: food[userQuery]}}
– M.Adam
Nov 13 at 9:18
add a comment |
This works wonderfully for my purpose. I am somewhat puzzled by the contents and structure of the basketDict.update() though - I'm not familiar with the '**' or structure of the contents within the (). Would you be able to provide an explanation or point me to resources where I can learn how it functions? Thanks for the solution.
– MHSalehi
Nov 12 at 20:29
I made some changes to be more clear. Also, you don't have to use the methodupdate
, simply set the key's value:basketDict[category] = {**basketDict.get(category, {}), **{userQuery: food[userQuery]}}
– M.Adam
Nov 13 at 9:18
This works wonderfully for my purpose. I am somewhat puzzled by the contents and structure of the basketDict.update() though - I'm not familiar with the '**' or structure of the contents within the (). Would you be able to provide an explanation or point me to resources where I can learn how it functions? Thanks for the solution.
– MHSalehi
Nov 12 at 20:29
This works wonderfully for my purpose. I am somewhat puzzled by the contents and structure of the basketDict.update() though - I'm not familiar with the '**' or structure of the contents within the (). Would you be able to provide an explanation or point me to resources where I can learn how it functions? Thanks for the solution.
– MHSalehi
Nov 12 at 20:29
I made some changes to be more clear. Also, you don't have to use the method
update
, simply set the key's value: basketDict[category] = {**basketDict.get(category, {}), **{userQuery: food[userQuery]}}
– M.Adam
Nov 13 at 9:18
I made some changes to be more clear. Also, you don't have to use the method
update
, simply set the key's value: basketDict[category] = {**basketDict.get(category, {}), **{userQuery: food[userQuery]}}
– M.Adam
Nov 13 at 9:18
add a comment |
up vote
0
down vote
Probably there's a simpler way to do this, but I just fixed your code.
When I run your code, everything inside shopDict
was going to basketDict
. This is because you wasn't cleaning your shopCheck
variable at each for
iteration.
And there was a second mistake when adding the elements to basketDict
. You need to add only the food
element that match the userQuery
. Here's the full fixed code:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
userQuery = input("What food do you want to check for? ")
for category, food in shopDict.items():
shopCheck =
for each in food:
shopCheck.append(each)
if userQuery not in shopCheck:
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[category] = {userQuery:food[userQuery]}
print(basketDict)
1
this too has indentation issues I guess
– Manoj Kengudelu
Nov 12 at 14:55
@ManojKengudelu You was right. Fixed!
– Hemerson Tacon
Nov 12 at 14:56
Thanks for the response. When I run this code and enter 'Apple', it erroneously produces, "That's not available." despite successfully adding it to the basketDict. Any ideas why?
– MHSalehi
Nov 12 at 20:41
@MHSalehi Actualy if you analyse the code carefully you will see that is not an error. Since theshopDict
has two pair of key-values (in this case category-food pairs), the outerfor
will be executed twice: one for the key "Fruit" and another for the key "Vegetables". That second iteration is generates that print becauseApple
indeed isn't present among the "Vegetable" values what is totaly right.
– Hemerson Tacon
Nov 13 at 14:50
add a comment |
up vote
0
down vote
Probably there's a simpler way to do this, but I just fixed your code.
When I run your code, everything inside shopDict
was going to basketDict
. This is because you wasn't cleaning your shopCheck
variable at each for
iteration.
And there was a second mistake when adding the elements to basketDict
. You need to add only the food
element that match the userQuery
. Here's the full fixed code:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
userQuery = input("What food do you want to check for? ")
for category, food in shopDict.items():
shopCheck =
for each in food:
shopCheck.append(each)
if userQuery not in shopCheck:
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[category] = {userQuery:food[userQuery]}
print(basketDict)
1
this too has indentation issues I guess
– Manoj Kengudelu
Nov 12 at 14:55
@ManojKengudelu You was right. Fixed!
– Hemerson Tacon
Nov 12 at 14:56
Thanks for the response. When I run this code and enter 'Apple', it erroneously produces, "That's not available." despite successfully adding it to the basketDict. Any ideas why?
– MHSalehi
Nov 12 at 20:41
@MHSalehi Actualy if you analyse the code carefully you will see that is not an error. Since theshopDict
has two pair of key-values (in this case category-food pairs), the outerfor
will be executed twice: one for the key "Fruit" and another for the key "Vegetables". That second iteration is generates that print becauseApple
indeed isn't present among the "Vegetable" values what is totaly right.
– Hemerson Tacon
Nov 13 at 14:50
add a comment |
up vote
0
down vote
up vote
0
down vote
Probably there's a simpler way to do this, but I just fixed your code.
When I run your code, everything inside shopDict
was going to basketDict
. This is because you wasn't cleaning your shopCheck
variable at each for
iteration.
And there was a second mistake when adding the elements to basketDict
. You need to add only the food
element that match the userQuery
. Here's the full fixed code:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
userQuery = input("What food do you want to check for? ")
for category, food in shopDict.items():
shopCheck =
for each in food:
shopCheck.append(each)
if userQuery not in shopCheck:
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[category] = {userQuery:food[userQuery]}
print(basketDict)
Probably there's a simpler way to do this, but I just fixed your code.
When I run your code, everything inside shopDict
was going to basketDict
. This is because you wasn't cleaning your shopCheck
variable at each for
iteration.
And there was a second mistake when adding the elements to basketDict
. You need to add only the food
element that match the userQuery
. Here's the full fixed code:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
userQuery = input("What food do you want to check for? ")
for category, food in shopDict.items():
shopCheck =
for each in food:
shopCheck.append(each)
if userQuery not in shopCheck:
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[category] = {userQuery:food[userQuery]}
print(basketDict)
edited Nov 12 at 15:01
answered Nov 12 at 14:51
Hemerson Tacon
887315
887315
1
this too has indentation issues I guess
– Manoj Kengudelu
Nov 12 at 14:55
@ManojKengudelu You was right. Fixed!
– Hemerson Tacon
Nov 12 at 14:56
Thanks for the response. When I run this code and enter 'Apple', it erroneously produces, "That's not available." despite successfully adding it to the basketDict. Any ideas why?
– MHSalehi
Nov 12 at 20:41
@MHSalehi Actualy if you analyse the code carefully you will see that is not an error. Since theshopDict
has two pair of key-values (in this case category-food pairs), the outerfor
will be executed twice: one for the key "Fruit" and another for the key "Vegetables". That second iteration is generates that print becauseApple
indeed isn't present among the "Vegetable" values what is totaly right.
– Hemerson Tacon
Nov 13 at 14:50
add a comment |
1
this too has indentation issues I guess
– Manoj Kengudelu
Nov 12 at 14:55
@ManojKengudelu You was right. Fixed!
– Hemerson Tacon
Nov 12 at 14:56
Thanks for the response. When I run this code and enter 'Apple', it erroneously produces, "That's not available." despite successfully adding it to the basketDict. Any ideas why?
– MHSalehi
Nov 12 at 20:41
@MHSalehi Actualy if you analyse the code carefully you will see that is not an error. Since theshopDict
has two pair of key-values (in this case category-food pairs), the outerfor
will be executed twice: one for the key "Fruit" and another for the key "Vegetables". That second iteration is generates that print becauseApple
indeed isn't present among the "Vegetable" values what is totaly right.
– Hemerson Tacon
Nov 13 at 14:50
1
1
this too has indentation issues I guess
– Manoj Kengudelu
Nov 12 at 14:55
this too has indentation issues I guess
– Manoj Kengudelu
Nov 12 at 14:55
@ManojKengudelu You was right. Fixed!
– Hemerson Tacon
Nov 12 at 14:56
@ManojKengudelu You was right. Fixed!
– Hemerson Tacon
Nov 12 at 14:56
Thanks for the response. When I run this code and enter 'Apple', it erroneously produces, "That's not available." despite successfully adding it to the basketDict. Any ideas why?
– MHSalehi
Nov 12 at 20:41
Thanks for the response. When I run this code and enter 'Apple', it erroneously produces, "That's not available." despite successfully adding it to the basketDict. Any ideas why?
– MHSalehi
Nov 12 at 20:41
@MHSalehi Actualy if you analyse the code carefully you will see that is not an error. Since the
shopDict
has two pair of key-values (in this case category-food pairs), the outer for
will be executed twice: one for the key "Fruit" and another for the key "Vegetables". That second iteration is generates that print because Apple
indeed isn't present among the "Vegetable" values what is totaly right.– Hemerson Tacon
Nov 13 at 14:50
@MHSalehi Actualy if you analyse the code carefully you will see that is not an error. Since the
shopDict
has two pair of key-values (in this case category-food pairs), the outer for
will be executed twice: one for the key "Fruit" and another for the key "Vegetables". That second iteration is generates that print because Apple
indeed isn't present among the "Vegetable" values what is totaly right.– Hemerson Tacon
Nov 13 at 14:50
add a comment |
up vote
0
down vote
I'd start by creating a dictionary containing all the items you can buy, whether they're fruit or vegetables. Then I'd simply check if the required item is on that dictionary:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
allVals = {}
for k, subDict in shopDict.items():
for k1, val in subDict.items():
allVals[k1] = val
userQuery = input("What food do you want to check for? ")
if userQuery not in allVals.keys():
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[userQuery] = allVals[userQuery]
print(basketDict)
AttributeError: 'dict' object has no attribute 'iteritems'
– Hemerson Tacon
Nov 12 at 14:52
Oops, sorry, I tested this with python2.7. I'll fix this!
– toti08
Nov 12 at 15:04
add a comment |
up vote
0
down vote
I'd start by creating a dictionary containing all the items you can buy, whether they're fruit or vegetables. Then I'd simply check if the required item is on that dictionary:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
allVals = {}
for k, subDict in shopDict.items():
for k1, val in subDict.items():
allVals[k1] = val
userQuery = input("What food do you want to check for? ")
if userQuery not in allVals.keys():
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[userQuery] = allVals[userQuery]
print(basketDict)
AttributeError: 'dict' object has no attribute 'iteritems'
– Hemerson Tacon
Nov 12 at 14:52
Oops, sorry, I tested this with python2.7. I'll fix this!
– toti08
Nov 12 at 15:04
add a comment |
up vote
0
down vote
up vote
0
down vote
I'd start by creating a dictionary containing all the items you can buy, whether they're fruit or vegetables. Then I'd simply check if the required item is on that dictionary:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
allVals = {}
for k, subDict in shopDict.items():
for k1, val in subDict.items():
allVals[k1] = val
userQuery = input("What food do you want to check for? ")
if userQuery not in allVals.keys():
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[userQuery] = allVals[userQuery]
print(basketDict)
I'd start by creating a dictionary containing all the items you can buy, whether they're fruit or vegetables. Then I'd simply check if the required item is on that dictionary:
basketDict = {}
shopDict = {"Fruit": {"Apple": "2", "Banana": "3"},
"Vegetables": {"Lettuce": "5", "Potato": "7"}}
allVals = {}
for k, subDict in shopDict.items():
for k1, val in subDict.items():
allVals[k1] = val
userQuery = input("What food do you want to check for? ")
if userQuery not in allVals.keys():
print("That's not available.")
else:
print(userQuery + " added to basket. ")
basketDict[userQuery] = allVals[userQuery]
print(basketDict)
edited Nov 12 at 15:04
answered Nov 12 at 14:49
toti08
1,74721523
1,74721523
AttributeError: 'dict' object has no attribute 'iteritems'
– Hemerson Tacon
Nov 12 at 14:52
Oops, sorry, I tested this with python2.7. I'll fix this!
– toti08
Nov 12 at 15:04
add a comment |
AttributeError: 'dict' object has no attribute 'iteritems'
– Hemerson Tacon
Nov 12 at 14:52
Oops, sorry, I tested this with python2.7. I'll fix this!
– toti08
Nov 12 at 15:04
AttributeError: 'dict' object has no attribute 'iteritems'
– Hemerson Tacon
Nov 12 at 14:52
AttributeError: 'dict' object has no attribute 'iteritems'
– Hemerson Tacon
Nov 12 at 14:52
Oops, sorry, I tested this with python2.7. I'll fix this!
– toti08
Nov 12 at 15:04
Oops, sorry, I tested this with python2.7. I'll fix this!
– toti08
Nov 12 at 15:04
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53264397%2ftesting-nested-dictionaries-for-a-user-input-and-retrieving-directly-related-inf%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown