python using for loop to update nested dictionary values with comprehension

Question

i am struggling with this for loop. I want to update the nested dictionary's value with the list. the following works, but I would like to use for loop to condense the code. "status" is a nested dictionary that has outer key "A" and "B", and binary_list[] just all 1's.

Trucks = ["A", "B"]
Days= ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday",    
"Saturday", "Sunday"]

requests={"A": {"Monday":0, "Tuesday":0, "Wednesday":0, "Thursday":0     
                "Friday":0, "Saturday":0, "Sunday":0},
          "B": {"Monday":0, "Tuesday":0, "Wednesday":0, "Thursday":0,    
                "Friday":0, "Saturday":0, "Sunday":0}}


binary_list=[1,1,1,1,1,1,1,1,1,1,1,1,1,1]



status["A"]["Monday"]=binary_list[0]
status["A"]["Tuesday"]=binary_list[1]
status["A"]["Wednesday"]=binary_list[2]
status["A"]["Thursday"]=binary_list[3]
status["A"]["Friday"]=binary_list[4]
status["A"]["Saturday"]=binary_list[5]
status["A"]["Sunday"]=binary_list[6]

status["B"]["Monday"]=binary_list[7]
status["B"]["Tuesday"]=binary_list[8]
status["B"]["Wednesday"]=binary_list[9]
status["B"]["Thursday"]=binary_list[10]
status["B"]["Friday"]=binary_list[11]
status["B"]["Saturday"]=binary_list[12]
status["B"]["Sunday"]=binary_list[13]

what I tried:

   for truck in Trucks:
        for day in Days:
            requests[truck][day]=[j for j in binary_list]

this will produce 14 1's in a list as the value of each of the day. I understand why this happens. it's because of the 2 for loops, so I figured I need a nested dictionary comprehenssions to make everything into one line. I am not sure that is correct, but with my limited knowledge that seems to be the right direction. my questions is how do I do that? would someone please help. thanks again for your time.

blhsing · Accepted Answer · 2018-08-29 06:48:11Z

2

You can use an iterator to iterate through the values of binary_list instead:

iter_binary_list = iter(binary_list)
for truck in Trucks:
    for day in Days:
        requests[truck][day] = next(iter_binary_list)

answered Aug 29, 2018 at 6:48

blhsing

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1 Comment

fishtang Over a year ago

thank you blhsing, imtinan Azhar, sinwoobang moe A. I was able to read more on iter to learn more about your solution. I really appreciated.

Mohd · Accepted Answer · 2018-08-29 06:53:36Z

2

You can use dictionary comprehension to do it in a one liner and more "pythonic" way as the following:

it = iter(binary_list)
status = {x: {d: next(it) for d in Days} for x in Trucks}
# outputs: {'A': {'Monday': 1, 'Tuesday': 1, 'Wednesday': 1, 'Thursday': 1, 'Friday': 1, 'Saturday': 1, 'Sunday': 1}, 'B': {'Monday': 1, 'Tuesday': 1, 'Wednesday': 1, 'Thursday': 1, 'Friday': 1, 'Saturday': 1, 'Sunday': 1}}

answered Aug 29, 2018 at 6:53

Mohd

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Comments

Imtinan Azhar · Accepted Answer · 2018-08-29 06:48:18Z

1

what you are doing is wrong, j for j in binary_list] will create a list for each requests[truck][day] and this is not what you want, what you want is something like this

binary_list_iterator=0

for truck in Trucks:
    for day in Days:
        requests[truck][day]=binary_list[binary_list_iterator]
        binary_list_iterator+=1

answered Aug 29, 2018 at 6:48

Imtinan Azhar

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Comments

sinwoobang · Accepted Answer · 2018-08-29 06:50:14Z

1

Is this what you want?

requests = {t: {d: 1 for d in Days} for t in Trucks}

Or if you want to iterate everything

requests = {t: {d: binary_list[i*len(Days)+j] for j, d in enumerate(Days)} for i, t in enumerate(Trucks)}

answered Aug 29, 2018 at 6:50

sinwoobang

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