How can you split an input to go into 2 different lists using a ;
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Ive been given a problem where given two lists, returns the list of all the elements that occur multiple times in both lists. The returned list should be in ascending order, without duplicates. Your main program will allow the enter two lists of numbers on one line separated by a semicolon and end input with a blank line. For example
Lists: 1 3 4 2 1 2 1 3; 4 4 2 4 3 2 4 4 3 1 3
List1: 1 3 4 2 1 2 1 3
List 2: 4 4 2 4 3 2 4 4 3 1 3
The result for this example would be: [2, 3]
I have a majority of the programming working, its just that after implementing the first for loop in the calculator function, I get an error.
ValueError: invalid literal for int() with base 10: '3;'
It may be a possibility that due to my inexperience there is a syntax error somewhere. Any help or advice is greatly appreciated.
def calculator(userinput):
marker = ":"
for x in userinput:
if x == ";":
marker = ";"
elif marker != ";":
numlist1 = [int(x) for x in userinput]
elif marker == ";": numlist2
numlist2 = [int(x) for x in userinput]
else:
pass
for y in numlist1:
list1 = numlist1.count(y)
list2 = numlist2.count(y)
if list1 > 1 and list2 > 1:
if y not in multiples:
multiples.append(int(y))
else:
continue
multiples.sort()
print(multiples)
while True:
multiples =
userinput = input("Lists: ").split() # asks for first input from user
if len(userinput) == 0: # breaks if user inputs nothing
break
calculator(userinput) # calls the calculator function
python-3.x
add a comment |
up vote
0
down vote
favorite
Ive been given a problem where given two lists, returns the list of all the elements that occur multiple times in both lists. The returned list should be in ascending order, without duplicates. Your main program will allow the enter two lists of numbers on one line separated by a semicolon and end input with a blank line. For example
Lists: 1 3 4 2 1 2 1 3; 4 4 2 4 3 2 4 4 3 1 3
List1: 1 3 4 2 1 2 1 3
List 2: 4 4 2 4 3 2 4 4 3 1 3
The result for this example would be: [2, 3]
I have a majority of the programming working, its just that after implementing the first for loop in the calculator function, I get an error.
ValueError: invalid literal for int() with base 10: '3;'
It may be a possibility that due to my inexperience there is a syntax error somewhere. Any help or advice is greatly appreciated.
def calculator(userinput):
marker = ":"
for x in userinput:
if x == ";":
marker = ";"
elif marker != ";":
numlist1 = [int(x) for x in userinput]
elif marker == ";": numlist2
numlist2 = [int(x) for x in userinput]
else:
pass
for y in numlist1:
list1 = numlist1.count(y)
list2 = numlist2.count(y)
if list1 > 1 and list2 > 1:
if y not in multiples:
multiples.append(int(y))
else:
continue
multiples.sort()
print(multiples)
while True:
multiples =
userinput = input("Lists: ").split() # asks for first input from user
if len(userinput) == 0: # breaks if user inputs nothing
break
calculator(userinput) # calls the calculator function
python-3.x
1
What error do you get? Why not update your question with the output you see when you run your script?
– Red Cricket
Nov 11 at 8:44
What output would you expect from the input in your example?
– Red Cricket
Nov 11 at 9:19
@RedCricket hi sorry, I edited my original post. Didn't think to post it.
– Reese
Nov 11 at 9:22
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Ive been given a problem where given two lists, returns the list of all the elements that occur multiple times in both lists. The returned list should be in ascending order, without duplicates. Your main program will allow the enter two lists of numbers on one line separated by a semicolon and end input with a blank line. For example
Lists: 1 3 4 2 1 2 1 3; 4 4 2 4 3 2 4 4 3 1 3
List1: 1 3 4 2 1 2 1 3
List 2: 4 4 2 4 3 2 4 4 3 1 3
The result for this example would be: [2, 3]
I have a majority of the programming working, its just that after implementing the first for loop in the calculator function, I get an error.
ValueError: invalid literal for int() with base 10: '3;'
It may be a possibility that due to my inexperience there is a syntax error somewhere. Any help or advice is greatly appreciated.
def calculator(userinput):
marker = ":"
for x in userinput:
if x == ";":
marker = ";"
elif marker != ";":
numlist1 = [int(x) for x in userinput]
elif marker == ";": numlist2
numlist2 = [int(x) for x in userinput]
else:
pass
for y in numlist1:
list1 = numlist1.count(y)
list2 = numlist2.count(y)
if list1 > 1 and list2 > 1:
if y not in multiples:
multiples.append(int(y))
else:
continue
multiples.sort()
print(multiples)
while True:
multiples =
userinput = input("Lists: ").split() # asks for first input from user
if len(userinput) == 0: # breaks if user inputs nothing
break
calculator(userinput) # calls the calculator function
python-3.x
Ive been given a problem where given two lists, returns the list of all the elements that occur multiple times in both lists. The returned list should be in ascending order, without duplicates. Your main program will allow the enter two lists of numbers on one line separated by a semicolon and end input with a blank line. For example
Lists: 1 3 4 2 1 2 1 3; 4 4 2 4 3 2 4 4 3 1 3
List1: 1 3 4 2 1 2 1 3
List 2: 4 4 2 4 3 2 4 4 3 1 3
The result for this example would be: [2, 3]
I have a majority of the programming working, its just that after implementing the first for loop in the calculator function, I get an error.
ValueError: invalid literal for int() with base 10: '3;'
It may be a possibility that due to my inexperience there is a syntax error somewhere. Any help or advice is greatly appreciated.
def calculator(userinput):
marker = ":"
for x in userinput:
if x == ";":
marker = ";"
elif marker != ";":
numlist1 = [int(x) for x in userinput]
elif marker == ";": numlist2
numlist2 = [int(x) for x in userinput]
else:
pass
for y in numlist1:
list1 = numlist1.count(y)
list2 = numlist2.count(y)
if list1 > 1 and list2 > 1:
if y not in multiples:
multiples.append(int(y))
else:
continue
multiples.sort()
print(multiples)
while True:
multiples =
userinput = input("Lists: ").split() # asks for first input from user
if len(userinput) == 0: # breaks if user inputs nothing
break
calculator(userinput) # calls the calculator function
python-3.x
python-3.x
edited Nov 11 at 9:29
asked Nov 11 at 8:30
Reese
11
11
1
What error do you get? Why not update your question with the output you see when you run your script?
– Red Cricket
Nov 11 at 8:44
What output would you expect from the input in your example?
– Red Cricket
Nov 11 at 9:19
@RedCricket hi sorry, I edited my original post. Didn't think to post it.
– Reese
Nov 11 at 9:22
add a comment |
1
What error do you get? Why not update your question with the output you see when you run your script?
– Red Cricket
Nov 11 at 8:44
What output would you expect from the input in your example?
– Red Cricket
Nov 11 at 9:19
@RedCricket hi sorry, I edited my original post. Didn't think to post it.
– Reese
Nov 11 at 9:22
1
1
What error do you get? Why not update your question with the output you see when you run your script?
– Red Cricket
Nov 11 at 8:44
What error do you get? Why not update your question with the output you see when you run your script?
– Red Cricket
Nov 11 at 8:44
What output would you expect from the input in your example?
– Red Cricket
Nov 11 at 9:19
What output would you expect from the input in your example?
– Red Cricket
Nov 11 at 9:19
@RedCricket hi sorry, I edited my original post. Didn't think to post it.
– Reese
Nov 11 at 9:22
@RedCricket hi sorry, I edited my original post. Didn't think to post it.
– Reese
Nov 11 at 9:22
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
I believe this does what you want …
$ cat intersec_two_list.py
#!/bin/env python
import sys
import collections
def intersect_list(lst1, lst2):
return sorted(list(set(lst1) & set(lst2)))
def find_mult(lst1, lst2):
c1 = collections.Counter(lst1)
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
c2 = collections.Counter(lst2)
mc2 = [mc[0] for mc in c2.items() if mc[1] > 1]
return intersect_list(mc1, mc2)
def main(l1,l2):
print find_mult(l1,l2)
if __name__ == "__main__":
user_input = sys.argv[1:][0]
[s1, s2] = user_input.split(';')
s1 = s1.strip()
s2 = s2.strip()
l1 = s1.split(' ')
l2 = s2.split(' ')
main(l1,l2)
You would execute like so ...
$ ./intersec_two_list.py "1 3 4 2 1 2 1 3; 4 4 2 4 3 2 4 4 3 1 3"
['2', '3']
Here is some explanation of the code above:
def intersect_list(lst1, lst2):
return sorted(list(set(lst1) & set(lst2)))
In this method I set()
to remove duplications from the lists.
>>> list = [ 1,2,3,3,3,3,4,5,6,6,6]
>>> set(list)
{1, 2, 3, 4, 5, 6}
Then we use &
like so set(lst1) & set(lst2)
to get a intersection of the two sets then convert that back to a list with list(set(lst1) & set(lst2))
and then we return a sorted
list.
In this method …
def find_mult(lst1, lst2):
c1 = collections.Counter(lst1)
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
c2 = collections.Counter(lst2)
mc2 = [mc[0] for mc in c2.items() if mc[1] > 1]
return intersect_list(mc1, mc2)
… use make use of Counter
's. What a counter does is this:
>>> from collections import Counter
>>> list1 = [1,2,3,3,3,3,4,5,6,6,6]
>>> Counter(list1)
>>> Counter({3: 4, 6: 3, 1: 1, 2: 1, 4: 1, 5: 1})
It counts the elements of a list. This way we can get rid of all the elements in out list that did not occur more than once like so:
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
The above is list comprehension which a short had for this python code:
mc1 =
for mc in c1.items():
if mc[1] > 1:
mc1.append(mc[0])
Then key to above code is c1.items()
. c1
is a collection and collections have a method items()
which return a list of (elem, cnt)
pairs from the collections. So if mc[1] > 1
means the element occurred more than once and mc[0]
is the element. We save these elements in a list called mc1
.
We do the same process on lst2
. We create collection, c2
and create a list mc2
. Now we get the intersection of mc1
and mc2
and return that and BAM! your done.
I understand "elements that occur multiple times in both lists" as "elements that occur multiple times in the first list and multiple times in the second list".
– Mr. T
Nov 11 at 9:12
Yeah "occur multiple times in each list" well does that mean "occur more than once in both lists" or is once enough since 1 can me considered a multiple :)
– Red Cricket
Nov 11 at 9:18
Yeah @Mr.T that is how the question is meant to be interpreted it. I should've been more specific when I wrote down the problem.
– Reese
Nov 11 at 9:29
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I believe this does what you want …
$ cat intersec_two_list.py
#!/bin/env python
import sys
import collections
def intersect_list(lst1, lst2):
return sorted(list(set(lst1) & set(lst2)))
def find_mult(lst1, lst2):
c1 = collections.Counter(lst1)
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
c2 = collections.Counter(lst2)
mc2 = [mc[0] for mc in c2.items() if mc[1] > 1]
return intersect_list(mc1, mc2)
def main(l1,l2):
print find_mult(l1,l2)
if __name__ == "__main__":
user_input = sys.argv[1:][0]
[s1, s2] = user_input.split(';')
s1 = s1.strip()
s2 = s2.strip()
l1 = s1.split(' ')
l2 = s2.split(' ')
main(l1,l2)
You would execute like so ...
$ ./intersec_two_list.py "1 3 4 2 1 2 1 3; 4 4 2 4 3 2 4 4 3 1 3"
['2', '3']
Here is some explanation of the code above:
def intersect_list(lst1, lst2):
return sorted(list(set(lst1) & set(lst2)))
In this method I set()
to remove duplications from the lists.
>>> list = [ 1,2,3,3,3,3,4,5,6,6,6]
>>> set(list)
{1, 2, 3, 4, 5, 6}
Then we use &
like so set(lst1) & set(lst2)
to get a intersection of the two sets then convert that back to a list with list(set(lst1) & set(lst2))
and then we return a sorted
list.
In this method …
def find_mult(lst1, lst2):
c1 = collections.Counter(lst1)
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
c2 = collections.Counter(lst2)
mc2 = [mc[0] for mc in c2.items() if mc[1] > 1]
return intersect_list(mc1, mc2)
… use make use of Counter
's. What a counter does is this:
>>> from collections import Counter
>>> list1 = [1,2,3,3,3,3,4,5,6,6,6]
>>> Counter(list1)
>>> Counter({3: 4, 6: 3, 1: 1, 2: 1, 4: 1, 5: 1})
It counts the elements of a list. This way we can get rid of all the elements in out list that did not occur more than once like so:
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
The above is list comprehension which a short had for this python code:
mc1 =
for mc in c1.items():
if mc[1] > 1:
mc1.append(mc[0])
Then key to above code is c1.items()
. c1
is a collection and collections have a method items()
which return a list of (elem, cnt)
pairs from the collections. So if mc[1] > 1
means the element occurred more than once and mc[0]
is the element. We save these elements in a list called mc1
.
We do the same process on lst2
. We create collection, c2
and create a list mc2
. Now we get the intersection of mc1
and mc2
and return that and BAM! your done.
I understand "elements that occur multiple times in both lists" as "elements that occur multiple times in the first list and multiple times in the second list".
– Mr. T
Nov 11 at 9:12
Yeah "occur multiple times in each list" well does that mean "occur more than once in both lists" or is once enough since 1 can me considered a multiple :)
– Red Cricket
Nov 11 at 9:18
Yeah @Mr.T that is how the question is meant to be interpreted it. I should've been more specific when I wrote down the problem.
– Reese
Nov 11 at 9:29
add a comment |
up vote
0
down vote
I believe this does what you want …
$ cat intersec_two_list.py
#!/bin/env python
import sys
import collections
def intersect_list(lst1, lst2):
return sorted(list(set(lst1) & set(lst2)))
def find_mult(lst1, lst2):
c1 = collections.Counter(lst1)
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
c2 = collections.Counter(lst2)
mc2 = [mc[0] for mc in c2.items() if mc[1] > 1]
return intersect_list(mc1, mc2)
def main(l1,l2):
print find_mult(l1,l2)
if __name__ == "__main__":
user_input = sys.argv[1:][0]
[s1, s2] = user_input.split(';')
s1 = s1.strip()
s2 = s2.strip()
l1 = s1.split(' ')
l2 = s2.split(' ')
main(l1,l2)
You would execute like so ...
$ ./intersec_two_list.py "1 3 4 2 1 2 1 3; 4 4 2 4 3 2 4 4 3 1 3"
['2', '3']
Here is some explanation of the code above:
def intersect_list(lst1, lst2):
return sorted(list(set(lst1) & set(lst2)))
In this method I set()
to remove duplications from the lists.
>>> list = [ 1,2,3,3,3,3,4,5,6,6,6]
>>> set(list)
{1, 2, 3, 4, 5, 6}
Then we use &
like so set(lst1) & set(lst2)
to get a intersection of the two sets then convert that back to a list with list(set(lst1) & set(lst2))
and then we return a sorted
list.
In this method …
def find_mult(lst1, lst2):
c1 = collections.Counter(lst1)
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
c2 = collections.Counter(lst2)
mc2 = [mc[0] for mc in c2.items() if mc[1] > 1]
return intersect_list(mc1, mc2)
… use make use of Counter
's. What a counter does is this:
>>> from collections import Counter
>>> list1 = [1,2,3,3,3,3,4,5,6,6,6]
>>> Counter(list1)
>>> Counter({3: 4, 6: 3, 1: 1, 2: 1, 4: 1, 5: 1})
It counts the elements of a list. This way we can get rid of all the elements in out list that did not occur more than once like so:
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
The above is list comprehension which a short had for this python code:
mc1 =
for mc in c1.items():
if mc[1] > 1:
mc1.append(mc[0])
Then key to above code is c1.items()
. c1
is a collection and collections have a method items()
which return a list of (elem, cnt)
pairs from the collections. So if mc[1] > 1
means the element occurred more than once and mc[0]
is the element. We save these elements in a list called mc1
.
We do the same process on lst2
. We create collection, c2
and create a list mc2
. Now we get the intersection of mc1
and mc2
and return that and BAM! your done.
I understand "elements that occur multiple times in both lists" as "elements that occur multiple times in the first list and multiple times in the second list".
– Mr. T
Nov 11 at 9:12
Yeah "occur multiple times in each list" well does that mean "occur more than once in both lists" or is once enough since 1 can me considered a multiple :)
– Red Cricket
Nov 11 at 9:18
Yeah @Mr.T that is how the question is meant to be interpreted it. I should've been more specific when I wrote down the problem.
– Reese
Nov 11 at 9:29
add a comment |
up vote
0
down vote
up vote
0
down vote
I believe this does what you want …
$ cat intersec_two_list.py
#!/bin/env python
import sys
import collections
def intersect_list(lst1, lst2):
return sorted(list(set(lst1) & set(lst2)))
def find_mult(lst1, lst2):
c1 = collections.Counter(lst1)
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
c2 = collections.Counter(lst2)
mc2 = [mc[0] for mc in c2.items() if mc[1] > 1]
return intersect_list(mc1, mc2)
def main(l1,l2):
print find_mult(l1,l2)
if __name__ == "__main__":
user_input = sys.argv[1:][0]
[s1, s2] = user_input.split(';')
s1 = s1.strip()
s2 = s2.strip()
l1 = s1.split(' ')
l2 = s2.split(' ')
main(l1,l2)
You would execute like so ...
$ ./intersec_two_list.py "1 3 4 2 1 2 1 3; 4 4 2 4 3 2 4 4 3 1 3"
['2', '3']
Here is some explanation of the code above:
def intersect_list(lst1, lst2):
return sorted(list(set(lst1) & set(lst2)))
In this method I set()
to remove duplications from the lists.
>>> list = [ 1,2,3,3,3,3,4,5,6,6,6]
>>> set(list)
{1, 2, 3, 4, 5, 6}
Then we use &
like so set(lst1) & set(lst2)
to get a intersection of the two sets then convert that back to a list with list(set(lst1) & set(lst2))
and then we return a sorted
list.
In this method …
def find_mult(lst1, lst2):
c1 = collections.Counter(lst1)
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
c2 = collections.Counter(lst2)
mc2 = [mc[0] for mc in c2.items() if mc[1] > 1]
return intersect_list(mc1, mc2)
… use make use of Counter
's. What a counter does is this:
>>> from collections import Counter
>>> list1 = [1,2,3,3,3,3,4,5,6,6,6]
>>> Counter(list1)
>>> Counter({3: 4, 6: 3, 1: 1, 2: 1, 4: 1, 5: 1})
It counts the elements of a list. This way we can get rid of all the elements in out list that did not occur more than once like so:
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
The above is list comprehension which a short had for this python code:
mc1 =
for mc in c1.items():
if mc[1] > 1:
mc1.append(mc[0])
Then key to above code is c1.items()
. c1
is a collection and collections have a method items()
which return a list of (elem, cnt)
pairs from the collections. So if mc[1] > 1
means the element occurred more than once and mc[0]
is the element. We save these elements in a list called mc1
.
We do the same process on lst2
. We create collection, c2
and create a list mc2
. Now we get the intersection of mc1
and mc2
and return that and BAM! your done.
I believe this does what you want …
$ cat intersec_two_list.py
#!/bin/env python
import sys
import collections
def intersect_list(lst1, lst2):
return sorted(list(set(lst1) & set(lst2)))
def find_mult(lst1, lst2):
c1 = collections.Counter(lst1)
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
c2 = collections.Counter(lst2)
mc2 = [mc[0] for mc in c2.items() if mc[1] > 1]
return intersect_list(mc1, mc2)
def main(l1,l2):
print find_mult(l1,l2)
if __name__ == "__main__":
user_input = sys.argv[1:][0]
[s1, s2] = user_input.split(';')
s1 = s1.strip()
s2 = s2.strip()
l1 = s1.split(' ')
l2 = s2.split(' ')
main(l1,l2)
You would execute like so ...
$ ./intersec_two_list.py "1 3 4 2 1 2 1 3; 4 4 2 4 3 2 4 4 3 1 3"
['2', '3']
Here is some explanation of the code above:
def intersect_list(lst1, lst2):
return sorted(list(set(lst1) & set(lst2)))
In this method I set()
to remove duplications from the lists.
>>> list = [ 1,2,3,3,3,3,4,5,6,6,6]
>>> set(list)
{1, 2, 3, 4, 5, 6}
Then we use &
like so set(lst1) & set(lst2)
to get a intersection of the two sets then convert that back to a list with list(set(lst1) & set(lst2))
and then we return a sorted
list.
In this method …
def find_mult(lst1, lst2):
c1 = collections.Counter(lst1)
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
c2 = collections.Counter(lst2)
mc2 = [mc[0] for mc in c2.items() if mc[1] > 1]
return intersect_list(mc1, mc2)
… use make use of Counter
's. What a counter does is this:
>>> from collections import Counter
>>> list1 = [1,2,3,3,3,3,4,5,6,6,6]
>>> Counter(list1)
>>> Counter({3: 4, 6: 3, 1: 1, 2: 1, 4: 1, 5: 1})
It counts the elements of a list. This way we can get rid of all the elements in out list that did not occur more than once like so:
mc1 = [mc[0] for mc in c1.items() if mc[1] > 1]
The above is list comprehension which a short had for this python code:
mc1 =
for mc in c1.items():
if mc[1] > 1:
mc1.append(mc[0])
Then key to above code is c1.items()
. c1
is a collection and collections have a method items()
which return a list of (elem, cnt)
pairs from the collections. So if mc[1] > 1
means the element occurred more than once and mc[0]
is the element. We save these elements in a list called mc1
.
We do the same process on lst2
. We create collection, c2
and create a list mc2
. Now we get the intersection of mc1
and mc2
and return that and BAM! your done.
edited Nov 11 at 10:18
answered Nov 11 at 9:09
Red Cricket
4,16883381
4,16883381
I understand "elements that occur multiple times in both lists" as "elements that occur multiple times in the first list and multiple times in the second list".
– Mr. T
Nov 11 at 9:12
Yeah "occur multiple times in each list" well does that mean "occur more than once in both lists" or is once enough since 1 can me considered a multiple :)
– Red Cricket
Nov 11 at 9:18
Yeah @Mr.T that is how the question is meant to be interpreted it. I should've been more specific when I wrote down the problem.
– Reese
Nov 11 at 9:29
add a comment |
I understand "elements that occur multiple times in both lists" as "elements that occur multiple times in the first list and multiple times in the second list".
– Mr. T
Nov 11 at 9:12
Yeah "occur multiple times in each list" well does that mean "occur more than once in both lists" or is once enough since 1 can me considered a multiple :)
– Red Cricket
Nov 11 at 9:18
Yeah @Mr.T that is how the question is meant to be interpreted it. I should've been more specific when I wrote down the problem.
– Reese
Nov 11 at 9:29
I understand "elements that occur multiple times in both lists" as "elements that occur multiple times in the first list and multiple times in the second list".
– Mr. T
Nov 11 at 9:12
I understand "elements that occur multiple times in both lists" as "elements that occur multiple times in the first list and multiple times in the second list".
– Mr. T
Nov 11 at 9:12
Yeah "occur multiple times in each list" well does that mean "occur more than once in both lists" or is once enough since 1 can me considered a multiple :)
– Red Cricket
Nov 11 at 9:18
Yeah "occur multiple times in each list" well does that mean "occur more than once in both lists" or is once enough since 1 can me considered a multiple :)
– Red Cricket
Nov 11 at 9:18
Yeah @Mr.T that is how the question is meant to be interpreted it. I should've been more specific when I wrote down the problem.
– Reese
Nov 11 at 9:29
Yeah @Mr.T that is how the question is meant to be interpreted it. I should've been more specific when I wrote down the problem.
– Reese
Nov 11 at 9:29
add a comment |
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What error do you get? Why not update your question with the output you see when you run your script?
– Red Cricket
Nov 11 at 8:44
What output would you expect from the input in your example?
– Red Cricket
Nov 11 at 9:19
@RedCricket hi sorry, I edited my original post. Didn't think to post it.
– Reese
Nov 11 at 9:22