CODEWITHSUNDEEP
python
Verma
Verma

Reputation: 966

Fill missing data in python list

I have a dictionary of lists each with a different number of elements. I'd like to add default values to the beginning of each list to make them all the same size.

So, if I start with...

data = {'data1': [1,2,3], 'data2': [12,23,34,45,44], 'data3': [7,8,9,10], 'data4': [71, 72, 73, 74, 75, 76, 78], 'data5': []}

I would like to end up with:

data1: [0, 0, 0, 0, 1, 2, 3]
data2: [0, 0, 12, 23, 34, 45, 44]
data3: [0, 0, 0, 7, 8, 9, 10]
data4: [71, 72, 73, 74, 75, 76, 78]
data5: [0, 0, 0, 0, 0, 0, 0]

I have come up with the following code:

maxlen = 0
for d in data:
    if maxlen < len(data[d]):
        maxlen = len(data[d])

for d in data:
    if len(data[d]) < maxlen:
        data[d] = [0] * (maxlen - len(data[d])) + data[d]

This works, but is there a better way to do this?

Upvotes: 1

Views: 1924

Answers (1)

Cory Kramer
Cory Kramer

Reputation: 117866

You can figure out which list is the longest, then pad out each of the other lists with zeroes in a dict comprehension.

>>> longest = max(len(i) for i in data.values())
>>> data = {key: [0]*(longest-len(value)) + value for key,value in data.items()}
>>> data
{'data5': [0, 0, 0, 0, 0, 0, 0],
 'data1': [0, 0, 0, 0, 1, 2, 3],
 'data4': [71, 72, 73, 74, 75, 76, 78],
 'data3': [0, 0, 0, 7, 8, 9, 10],
 'data2': [0, 0, 12, 23, 34, 45, 44]}

Upvotes: 9

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