Create a dictionary based on a function and a list of strings

Question

I have a list of strings, lostrings, and a function, splitter, which splits a string.

lostrings = 
    ['308 921 q53 C13 0000000200',
     '300 920 q51 C13 000000199',
     '318 921 q53 C12 0000000199']

def slitter(s: str) -> list:
    value1 = s[:3]
    value2 = s[4:7]
    value3 = s[9:12]
    value4 = s[14:17]
    value5 = s[25:]
    return [value1, value2, value3, value4, value5]

Example: splitter(lostrings[0]) will output ['308', '921', 'q53', 'C13', '200'].

What I am trying to do is to create a key-value dictionary where keys are 'value1, 'value2', 'value3', 'value4', 'value5' and values are lists. Desired output is as follows:

{'value1': ['308', '300', '318'],
 'value2': ['912', '920', '921'],
 'value3': ['q53', 'q51', 'q53'],
 'value4': ['C13', 'C13', 'C12'],
 'value5': ['200', '199', '199']}

I tried the following:
1.

dict(zip(['value1', 'value2', 'value3', 'value4', 'value5'], [splitter[lostrings[row]] for row in range(len(lostrings))]))

This does not give the correct output. I am not sure how to create a dictionary of 'str':list out of a list of strings based on a function.

Asked By: Joe

||

Source

Answer 1

You can use lostrings[0].split(' ') instead of using your splitter function.

lostrings = ['308 921 q53 C13 0000000200', '300 920 q51 C13 0000000199', '318 921 q53 C12 0000000199']

mydict = {'value1': [], 'value2': [], 'value3': [], 'value4': [], 'value5': []}

for targetstr in lostrings:
    for idx, targetsubstr in enumerate(targetstr.split(' ')):
        mydict['value%d' % (idx+1)].append(targetsubstr)

mydict = 
{'value1': ['308', '300', '318'],
 'value2': ['921', '920', '921'],
 'value3': ['q53', 'q51', 'q53'],
 'value4': ['C13', 'C13', 'C12'],
 'value5': ['0000000200', '0000000199', '0000000199']}

Answered By: J. Choi

Answer 2

Here is a solution, using defaultdict to initialize & use enumerate to track index.

from collections import defaultdict

lo_strings = [
    '308 921 q53 C13 0000000200',
    '300 920 q51 C13 0000000199',
    '318 921 q53 C12 0000000199'
]

collect_ = defaultdict(list)

for i in lo_strings:
    for k, v in enumerate(i.split(), 1):
        collect_[f'value{k}'].append(v)

print(collect_)

defaultdict(<class 'list'>, {'value1': ['308', '300', '318'], 'value2': ['921', '920', '921'], 'value3': ['q53', 'q51', 'q53'], 'value4': ['C13', 'C13', 'C12'], 'value5': ['0000000200', '0000000199', '0000000199']})

Answered By: sushanth

Answer 3

We could split the lostring by space. This would give us all the substrings. Next, we could put them in a dictionary by their indices.

defaultdict is essentially a wrapper over the normal dict. Whenever a key doesn’t exist, it returns a default value. In this case, it will return a list.

from collections import defaultdict


lostrings = [
    '308 921 q53 C13 0000000200',
    '300 920 q51 C13 0000000199',
    '318 921 q53 C12 0000000199'
]

values = defaultdict(list)

for lostring in lostrings:
    substrings = lostring.split(" ")
    
    for index, substring in enumerate(substrings, start=1):
        values[f"value{index}"].append(substring)

print(values)

Answered By: Preet Mishra

Answer 4

You can map the list of strings to the splitter function, transpose the output so that the lists align with the keys, which you can then zip together to construct a dict:

dict(
    zip(
        ['value1', 'value2', 'value3', 'value4', 'value5'],
        map(
            list,
            zip(*map(splitter, lostrings))
        )
    )
)

Demo: https://replit.com/@blhsing/OutstandingWeakGnudebugger#main.py

Answered By: blhsing

Create a dictionary based on a function and a list of strings

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