CODEWITHSUNDEEP
pythonstringlist
Okenite
Okenite

Reputation: 189

How to separate numbers from letters in text of list

I have a list[] which contains phrases like list = ['123Abc','234Asx','456Aio','...'].I would like to separate numbers from letters. The recurrent model is 3 numbers followed by capital A. How can I do that? I tried many ways using list.replace but I don't know how to set it, or whether the existence or not of a better method to use.

Upvotes: 2

Views: 2030

Answers (5)

Rahul P
Rahul P

Reputation: 2663

Here is another way to do it to get the result you want. This is done using re.findall and is independent of the scenario where the separating character is 'A'.

import re

test_list = ['123Abc','234Asx','456Aio']

result = [' '.join(re.findall(r'[A-Za-z]+|\d+', x)) for x in test_list]

print(result)

Output

['123 Abc', '234 Asx', '456 Aio']

Time taken to run:

3.62 µs ± 13 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

This seems to run faster than the accepted solution. From my analysis, the accepted solution runs in approximately:

5.79 µs ± 132 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Upvotes: 0

kederrac
kederrac

Reputation: 17322

you can use list comprehension and add a space at position 3 in each string:

[e[:3] + ' ' + e[3:] for e in my_list]

output:

['123 Abc', '234 Asx', '456 Aio', '... ']

Upvotes: 1

Jongware
Jongware

Reputation: 22457

To get continuous groups of items, using a function to determine to what group each item belongs, use itertools.groupby. In your case, you (presumably) want continuous sets of digits and non-digits, so the key is isdigit(), applied to each of the characters in your input string.

The number of digits can be variable, the letter-part does not need to start with A, and you actually can have any series of digits and letters in any order: groupby will sort it out.

The result of groupby is an iterator (so you must convert it to something definitive such as a list), in the format "key result, group object". You can see the isdigit fired correctly for all characters by printing out the immediate result:

from itertools import groupby

l = ['123Abc','234Asx','456Aio']
print ([(i,list(j)) for i,j in groupby (l[0], key=lambda x:x[0].isdigit())])

which is

[(True, ['1', '2', '3']), (False, ['A', 'b', 'c'])]

Converting the list(j) back to a single string only needs a join. You don't want to know whether you have a list of digits or not, so you can discard the boolean result and store only the strings. This

print ([''.join(j) for i,j in groupby (l[0], key=lambda x:x[0].isdigit())])

yields

['123', 'Abc']

for the first item in your list, and if your desired result is a list for every item in your current list, add an iteration around this one:

print ([[''.join(j) for i,j in groupby (item, key=lambda x:x[0].isdigit())] for item in l])

returns

[['123', 'Abc'], ['234', 'Asx'], ['456', 'Aio']]

Upvotes: 1

Negar37
Negar37

Reputation: 362

try this :

text = ['A'+i.split('A')[1] for i in list]
numbers = [i.split('A')[0] for i in list]

output:

['Abc', 'Asx', 'Aio']
['123', '234', '456']

Upvotes: 0

Sam Chats
Sam Chats

Reputation: 2321

numbers = [int(val[:3]) for val in list]
text = [val[3:] for val in list]

Upvotes: 0

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