How do I fill my dictionary values with the values from another dictionary where their keys are the same?

I have one dictionary (dictDemCLass) with a key but the values are all 0 and I plan to fill them with the values from another dictionary (dictAvgGrade). I need to do so where the keys of the two dictionaries are the same.

dictAvgGrade = {k:sum(v)/4 for k,v in studentPerf.items()}



dictDemClass = {k:0 for k in classes}

When printed (dictAvgGrade is shortened):

print(dictAvgGrade)

{('Jeffery', 'male', 'junior'): 0.7749999999999999, ('Able', 'male', 'senior'): 0.8200000000000001, ('Don', 'male', 'junior'): 0.7974999999999999, ('Will', 'male', 'senior'): 0.7975000000000001}



print(dictDemClass)



{'junior': 0, 'senior': 0, 'sophomore': 0}

Ultimately I want to fill dictDemClass to show the average for each class. So that the output could look something like:

print(dictDemClass)



{'junior': 77.46, 'senior': 83.82, 'sophomore': 86.79}

edited Nov 19 '18 at 4:57

asked Nov 19 '18 at 4:17

Jacob Myer

496

So the final values in dictDemClass are the averages of all corresponding values in dictAvgGrade?

– slider
Nov 19 '18 at 5:11

yes that's what im hoping to achieve, obviously the ones there at the end of my post are just an example

– Jacob Myer
Nov 19 '18 at 5:28

add a comment |

dictAvgGrade = {k:sum(v)/4 for k,v in studentPerf.items()}



dictDemClass = {k:0 for k in classes}

When printed (dictAvgGrade is shortened):

print(dictAvgGrade)

{('Jeffery', 'male', 'junior'): 0.7749999999999999, ('Able', 'male', 'senior'): 0.8200000000000001, ('Don', 'male', 'junior'): 0.7974999999999999, ('Will', 'male', 'senior'): 0.7975000000000001}



print(dictDemClass)



{'junior': 0, 'senior': 0, 'sophomore': 0}

Ultimately I want to fill dictDemClass to show the average for each class. So that the output could look something like:

print(dictDemClass)



{'junior': 77.46, 'senior': 83.82, 'sophomore': 86.79}

edited Nov 19 '18 at 4:57

asked Nov 19 '18 at 4:17

Jacob Myer

496

So the final values in dictDemClass are the averages of all corresponding values in dictAvgGrade?

– slider
Nov 19 '18 at 5:11

yes that's what im hoping to achieve, obviously the ones there at the end of my post are just an example

– Jacob Myer
Nov 19 '18 at 5:28

add a comment |

dictAvgGrade = {k:sum(v)/4 for k,v in studentPerf.items()}



dictDemClass = {k:0 for k in classes}

When printed (dictAvgGrade is shortened):

print(dictAvgGrade)

{('Jeffery', 'male', 'junior'): 0.7749999999999999, ('Able', 'male', 'senior'): 0.8200000000000001, ('Don', 'male', 'junior'): 0.7974999999999999, ('Will', 'male', 'senior'): 0.7975000000000001}



print(dictDemClass)



{'junior': 0, 'senior': 0, 'sophomore': 0}

Ultimately I want to fill dictDemClass to show the average for each class. So that the output could look something like:

print(dictDemClass)



{'junior': 77.46, 'senior': 83.82, 'sophomore': 86.79}

edited Nov 19 '18 at 4:57

asked Nov 19 '18 at 4:17

Jacob Myer

496

dictAvgGrade = {k:sum(v)/4 for k,v in studentPerf.items()}



dictDemClass = {k:0 for k in classes}

When printed (dictAvgGrade is shortened):

print(dictAvgGrade)

{('Jeffery', 'male', 'junior'): 0.7749999999999999, ('Able', 'male', 'senior'): 0.8200000000000001, ('Don', 'male', 'junior'): 0.7974999999999999, ('Will', 'male', 'senior'): 0.7975000000000001}



print(dictDemClass)



{'junior': 0, 'senior': 0, 'sophomore': 0}

Ultimately I want to fill dictDemClass to show the average for each class. So that the output could look something like:

print(dictDemClass)



{'junior': 77.46, 'senior': 83.82, 'sophomore': 86.79}

python dictionary

edited Nov 19 '18 at 4:57

asked Nov 19 '18 at 4:17

Jacob Myer

496

edited Nov 19 '18 at 4:57

asked Nov 19 '18 at 4:17

Jacob Myer

496

edited Nov 19 '18 at 4:57

asked Nov 19 '18 at 4:17

Jacob Myer

496

asked Nov 19 '18 at 4:17

Jacob Myer

496

asked Nov 19 '18 at 4:17

Jacob Myer

496

So the final values in dictDemClass are the averages of all corresponding values in dictAvgGrade?

– slider
Nov 19 '18 at 5:11

yes that's what im hoping to achieve, obviously the ones there at the end of my post are just an example

– Jacob Myer
Nov 19 '18 at 5:28

add a comment |

So the final values in dictDemClass are the averages of all corresponding values in dictAvgGrade?

– slider
Nov 19 '18 at 5:11

yes that's what im hoping to achieve, obviously the ones there at the end of my post are just an example

– Jacob Myer
Nov 19 '18 at 5:28

So the final values in dictDemClass are the averages of all corresponding values in dictAvgGrade?

– slider
Nov 19 '18 at 5:11

yes that's what im hoping to achieve, obviously the ones there at the end of my post are just an example

– Jacob Myer
Nov 19 '18 at 5:28

add a comment |

2 Answers
2

active

oldest

votes

You can use itertools.groupby to groups items in dictAvgGrade by "class" (i.e. junior, senior, etc.). Then you can compute the average for each group and add it to dictDemClass.

So for the example your posted, it can be something like the following:

from itertools import groupby



dictAvgGrade = {('Jeffery', 'male', 'junior'): 0.7749999999999999, ('Able', 'male', 'senior'): 0.8200000000000001, ('Don', 'male', 'junior'): 0.7974999999999999, ('Will', 'male', 'senior'): 0.7975000000000001}

dictDemClass = {'junior': 0, 'senior': 0, 'sophomore': 0}



def get_class(x):

    return x[0][2]



for k, g in groupby(sorted(dictAvgGrade.items(), key=get_class), key=get_class):

    group = list(g)

    class_avg = sum(x[1] for x in group)/len(group)

    dictDemClass[k] = class_avg



print(dictDemClass)

Output

{'senior': 0.8087500000000001, 'junior': 0.7862499999999999, 'sophomore': 0}

edited Nov 19 '18 at 5:19

answered Nov 19 '18 at 4:24

slider

8,23811129

I might have been too vague, see my comment on the other response. Id appreciate if you could give this another look

– Jacob Myer
Nov 19 '18 at 4:44

@JacobMyer As alex pointed out, can you please add example data and the expected output?

– slider
Nov 19 '18 at 4:54

I added some more information, I think that should show you what my goal is

– Jacob Myer
Nov 19 '18 at 5:01

@JacobMyer I hope the updated answer is more helpful.

– slider
Nov 19 '18 at 5:22

@JacobMyer Great! If this solved your problem, please consider marking the answer as accepted (stackoverflow.com/help/someone-answers).

– slider
Nov 19 '18 at 5:37

add a comment |

if you know that dictAvgGrade is always a subset of dictDemClass you can do this

dictDemClass.update(dictAvgGrade)

otherwise

if you are using python3 you can do something like this

for key in (dictDemClass.keys() & dictAvgGrade.keys()):

    dictDemClass[key] = dictAvgGrade[key]

if you are using python2 you can do something like this

for key in (set(dictDemClass.keys()) & set(dictAvgGrade.keys())):

    dictDemClass[key] = dictAvgGrade[key]

answered Nov 19 '18 at 4:35

Alex

1581112

Maybe I wasn't clear enough. dictAvgGrade has all of the information that needs to go into dictDemClass, but its keys are in tuples and their is other information I hope to filter out by creating this new dictionary. I need the v values of dictAvgGrade to fill the 0s of dictDemClass in the places where the keys from both of those dictionaries are equivalent

– Jacob Myer
Nov 19 '18 at 4:40

2

sorry, I'm not sure I understand, can you add examples of the two dictionaries and the desired result?

– Alex
Nov 19 '18 at 4:46

no worries, I made some changes. Hopefully that's easier to understand

– Jacob Myer
Nov 19 '18 at 5:00

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

You can use itertools.groupby to groups items in dictAvgGrade by "class" (i.e. junior, senior, etc.). Then you can compute the average for each group and add it to dictDemClass.

So for the example your posted, it can be something like the following:

from itertools import groupby



dictAvgGrade = {('Jeffery', 'male', 'junior'): 0.7749999999999999, ('Able', 'male', 'senior'): 0.8200000000000001, ('Don', 'male', 'junior'): 0.7974999999999999, ('Will', 'male', 'senior'): 0.7975000000000001}

dictDemClass = {'junior': 0, 'senior': 0, 'sophomore': 0}



def get_class(x):

    return x[0][2]



for k, g in groupby(sorted(dictAvgGrade.items(), key=get_class), key=get_class):

    group = list(g)

    class_avg = sum(x[1] for x in group)/len(group)

    dictDemClass[k] = class_avg



print(dictDemClass)

Output

{'senior': 0.8087500000000001, 'junior': 0.7862499999999999, 'sophomore': 0}

edited Nov 19 '18 at 5:19

answered Nov 19 '18 at 4:24

slider

8,23811129

I might have been too vague, see my comment on the other response. Id appreciate if you could give this another look

– Jacob Myer
Nov 19 '18 at 4:44

@JacobMyer As alex pointed out, can you please add example data and the expected output?

– slider
Nov 19 '18 at 4:54

I added some more information, I think that should show you what my goal is

– Jacob Myer
Nov 19 '18 at 5:01

@JacobMyer I hope the updated answer is more helpful.

– slider
Nov 19 '18 at 5:22

@JacobMyer Great! If this solved your problem, please consider marking the answer as accepted (stackoverflow.com/help/someone-answers).

– slider
Nov 19 '18 at 5:37

add a comment |

You can use itertools.groupby to groups items in dictAvgGrade by "class" (i.e. junior, senior, etc.). Then you can compute the average for each group and add it to dictDemClass.

So for the example your posted, it can be something like the following:

from itertools import groupby



dictAvgGrade = {('Jeffery', 'male', 'junior'): 0.7749999999999999, ('Able', 'male', 'senior'): 0.8200000000000001, ('Don', 'male', 'junior'): 0.7974999999999999, ('Will', 'male', 'senior'): 0.7975000000000001}

dictDemClass = {'junior': 0, 'senior': 0, 'sophomore': 0}



def get_class(x):

    return x[0][2]



for k, g in groupby(sorted(dictAvgGrade.items(), key=get_class), key=get_class):

    group = list(g)

    class_avg = sum(x[1] for x in group)/len(group)

    dictDemClass[k] = class_avg



print(dictDemClass)

Output

{'senior': 0.8087500000000001, 'junior': 0.7862499999999999, 'sophomore': 0}

edited Nov 19 '18 at 5:19

answered Nov 19 '18 at 4:24

slider

8,23811129

I might have been too vague, see my comment on the other response. Id appreciate if you could give this another look

– Jacob Myer
Nov 19 '18 at 4:44

@JacobMyer As alex pointed out, can you please add example data and the expected output?

– slider
Nov 19 '18 at 4:54

I added some more information, I think that should show you what my goal is

– Jacob Myer
Nov 19 '18 at 5:01

@JacobMyer I hope the updated answer is more helpful.

– slider
Nov 19 '18 at 5:22

@JacobMyer Great! If this solved your problem, please consider marking the answer as accepted (stackoverflow.com/help/someone-answers).

– slider
Nov 19 '18 at 5:37

add a comment |

You can use itertools.groupby to groups items in dictAvgGrade by "class" (i.e. junior, senior, etc.). Then you can compute the average for each group and add it to dictDemClass.

So for the example your posted, it can be something like the following:

from itertools import groupby



dictAvgGrade = {('Jeffery', 'male', 'junior'): 0.7749999999999999, ('Able', 'male', 'senior'): 0.8200000000000001, ('Don', 'male', 'junior'): 0.7974999999999999, ('Will', 'male', 'senior'): 0.7975000000000001}

dictDemClass = {'junior': 0, 'senior': 0, 'sophomore': 0}



def get_class(x):

    return x[0][2]



for k, g in groupby(sorted(dictAvgGrade.items(), key=get_class), key=get_class):

    group = list(g)

    class_avg = sum(x[1] for x in group)/len(group)

    dictDemClass[k] = class_avg



print(dictDemClass)

Output

{'senior': 0.8087500000000001, 'junior': 0.7862499999999999, 'sophomore': 0}

edited Nov 19 '18 at 5:19

answered Nov 19 '18 at 4:24

slider

8,23811129

You can use itertools.groupby to groups items in dictAvgGrade by "class" (i.e. junior, senior, etc.). Then you can compute the average for each group and add it to dictDemClass.

So for the example your posted, it can be something like the following:

from itertools import groupby



dictAvgGrade = {('Jeffery', 'male', 'junior'): 0.7749999999999999, ('Able', 'male', 'senior'): 0.8200000000000001, ('Don', 'male', 'junior'): 0.7974999999999999, ('Will', 'male', 'senior'): 0.7975000000000001}

dictDemClass = {'junior': 0, 'senior': 0, 'sophomore': 0}



def get_class(x):

    return x[0][2]



for k, g in groupby(sorted(dictAvgGrade.items(), key=get_class), key=get_class):

    group = list(g)

    class_avg = sum(x[1] for x in group)/len(group)

    dictDemClass[k] = class_avg



print(dictDemClass)

Output

{'senior': 0.8087500000000001, 'junior': 0.7862499999999999, 'sophomore': 0}

edited Nov 19 '18 at 5:19

answered Nov 19 '18 at 4:24

slider

8,23811129

edited Nov 19 '18 at 5:19

answered Nov 19 '18 at 4:24

slider

8,23811129

answered Nov 19 '18 at 4:24

slider

8,23811129

answered Nov 19 '18 at 4:24

slider

8,23811129

I might have been too vague, see my comment on the other response. Id appreciate if you could give this another look

– Jacob Myer
Nov 19 '18 at 4:44

@JacobMyer As alex pointed out, can you please add example data and the expected output?

– slider
Nov 19 '18 at 4:54

I added some more information, I think that should show you what my goal is

– Jacob Myer
Nov 19 '18 at 5:01

@JacobMyer I hope the updated answer is more helpful.

– slider
Nov 19 '18 at 5:22

@JacobMyer Great! If this solved your problem, please consider marking the answer as accepted (stackoverflow.com/help/someone-answers).

– slider
Nov 19 '18 at 5:37

add a comment |

I might have been too vague, see my comment on the other response. Id appreciate if you could give this another look

– Jacob Myer
Nov 19 '18 at 4:44

@JacobMyer As alex pointed out, can you please add example data and the expected output?

– slider
Nov 19 '18 at 4:54

I added some more information, I think that should show you what my goal is

– Jacob Myer
Nov 19 '18 at 5:01

@JacobMyer I hope the updated answer is more helpful.

– slider
Nov 19 '18 at 5:22

@JacobMyer Great! If this solved your problem, please consider marking the answer as accepted (stackoverflow.com/help/someone-answers).

– slider
Nov 19 '18 at 5:37

I might have been too vague, see my comment on the other response. Id appreciate if you could give this another look

– Jacob Myer
Nov 19 '18 at 4:44

@JacobMyer As alex pointed out, can you please add example data and the expected output?

– slider
Nov 19 '18 at 4:54

I added some more information, I think that should show you what my goal is

– Jacob Myer
Nov 19 '18 at 5:01

@JacobMyer I hope the updated answer is more helpful.

– slider
Nov 19 '18 at 5:22

@JacobMyer Great! If this solved your problem, please consider marking the answer as accepted (stackoverflow.com/help/someone-answers).

– slider
Nov 19 '18 at 5:37

add a comment |

if you know that dictAvgGrade is always a subset of dictDemClass you can do this

dictDemClass.update(dictAvgGrade)

otherwise

if you are using python3 you can do something like this

for key in (dictDemClass.keys() & dictAvgGrade.keys()):

    dictDemClass[key] = dictAvgGrade[key]

if you are using python2 you can do something like this

for key in (set(dictDemClass.keys()) & set(dictAvgGrade.keys())):

    dictDemClass[key] = dictAvgGrade[key]

answered Nov 19 '18 at 4:35

Alex

1581112

Maybe I wasn't clear enough. dictAvgGrade has all of the information that needs to go into dictDemClass, but its keys are in tuples and their is other information I hope to filter out by creating this new dictionary. I need the v values of dictAvgGrade to fill the 0s of dictDemClass in the places where the keys from both of those dictionaries are equivalent

– Jacob Myer
Nov 19 '18 at 4:40

2

sorry, I'm not sure I understand, can you add examples of the two dictionaries and the desired result?

– Alex
Nov 19 '18 at 4:46

no worries, I made some changes. Hopefully that's easier to understand

– Jacob Myer
Nov 19 '18 at 5:00

add a comment |

if you know that dictAvgGrade is always a subset of dictDemClass you can do this

dictDemClass.update(dictAvgGrade)

otherwise

if you are using python3 you can do something like this

for key in (dictDemClass.keys() & dictAvgGrade.keys()):

    dictDemClass[key] = dictAvgGrade[key]

if you are using python2 you can do something like this

for key in (set(dictDemClass.keys()) & set(dictAvgGrade.keys())):

    dictDemClass[key] = dictAvgGrade[key]

answered Nov 19 '18 at 4:35

Alex

1581112

Maybe I wasn't clear enough. dictAvgGrade has all of the information that needs to go into dictDemClass, but its keys are in tuples and their is other information I hope to filter out by creating this new dictionary. I need the v values of dictAvgGrade to fill the 0s of dictDemClass in the places where the keys from both of those dictionaries are equivalent

– Jacob Myer
Nov 19 '18 at 4:40

2

sorry, I'm not sure I understand, can you add examples of the two dictionaries and the desired result?

– Alex
Nov 19 '18 at 4:46

no worries, I made some changes. Hopefully that's easier to understand

– Jacob Myer
Nov 19 '18 at 5:00

add a comment |

if you know that dictAvgGrade is always a subset of dictDemClass you can do this

dictDemClass.update(dictAvgGrade)

otherwise

if you are using python3 you can do something like this

for key in (dictDemClass.keys() & dictAvgGrade.keys()):

    dictDemClass[key] = dictAvgGrade[key]

if you are using python2 you can do something like this

for key in (set(dictDemClass.keys()) & set(dictAvgGrade.keys())):

    dictDemClass[key] = dictAvgGrade[key]

answered Nov 19 '18 at 4:35

Alex

1581112

if you know that dictAvgGrade is always a subset of dictDemClass you can do this

dictDemClass.update(dictAvgGrade)

otherwise

if you are using python3 you can do something like this

for key in (dictDemClass.keys() & dictAvgGrade.keys()):

    dictDemClass[key] = dictAvgGrade[key]

if you are using python2 you can do something like this

for key in (set(dictDemClass.keys()) & set(dictAvgGrade.keys())):

    dictDemClass[key] = dictAvgGrade[key]

answered Nov 19 '18 at 4:35

Alex

1581112

answered Nov 19 '18 at 4:35

Alex

1581112

answered Nov 19 '18 at 4:35

Alex

1581112

answered Nov 19 '18 at 4:35

Alex

1581112

Maybe I wasn't clear enough. dictAvgGrade has all of the information that needs to go into dictDemClass, but its keys are in tuples and their is other information I hope to filter out by creating this new dictionary. I need the v values of dictAvgGrade to fill the 0s of dictDemClass in the places where the keys from both of those dictionaries are equivalent

– Jacob Myer
Nov 19 '18 at 4:40

2

sorry, I'm not sure I understand, can you add examples of the two dictionaries and the desired result?

– Alex
Nov 19 '18 at 4:46

no worries, I made some changes. Hopefully that's easier to understand

– Jacob Myer
Nov 19 '18 at 5:00

add a comment |

Maybe I wasn't clear enough. dictAvgGrade has all of the information that needs to go into dictDemClass, but its keys are in tuples and their is other information I hope to filter out by creating this new dictionary. I need the v values of dictAvgGrade to fill the 0s of dictDemClass in the places where the keys from both of those dictionaries are equivalent

– Jacob Myer
Nov 19 '18 at 4:40

2

sorry, I'm not sure I understand, can you add examples of the two dictionaries and the desired result?

– Alex
Nov 19 '18 at 4:46

no worries, I made some changes. Hopefully that's easier to understand

– Jacob Myer
Nov 19 '18 at 5:00

Maybe I wasn't clear enough. dictAvgGrade has all of the information that needs to go into dictDemClass, but its keys are in tuples and their is other information I hope to filter out by creating this new dictionary. I need the v values of dictAvgGrade to fill the 0s of dictDemClass in the places where the keys from both of those dictionaries are equivalent

– Jacob Myer
Nov 19 '18 at 4:40

sorry, I'm not sure I understand, can you add examples of the two dictionaries and the desired result?

– Alex
Nov 19 '18 at 4:46

no worries, I made some changes. Hopefully that's easier to understand

– Jacob Myer
Nov 19 '18 at 5:00

add a comment |

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