从字符串中提取名称和数字

Question

Similar to this question I have a string of names and numbers separated by a colon:

s = 'Waz D: 5 l gu l: 5 GrinVe: 3 P LUK: 2 Cubbi: 1 2 nd dok: 1 maf 74: 1 abr12: 1 Waza D 5'

I'm trying to split this to get:

 ('Waz D', '5'),
 ('l gu l', '5'),
 ('GrinVe', '3'),
 ('P LUK', '2'),
 ('Cubbi', '1'),
 ('2 nd dok', '1')
 ('maf 74', '1')
 ('abr12', '1')

I have tried two regular expressions so far with mixed success:

re.findall(r"(.*?)[a-zA-Z0-9]+: (\d+)*", s)
[('Waz ', '5'),
 (' l gu ', '5'),
 (' ', '3'),
 (' P ', '2'),
 (' ', '1'),
 (' 2 nd ', '1'),
 (' maf ', '1'),
 (' ', '1')]

And:

re.findall(r"(.*?)([a-zA-Z0-9]+): (\d+)*", s)
[('Waz ', 'D', '5'),
 (' l gu ', 'l', '5'),
 (' ', 'GrinVe', '3'),
 (' P ', 'LUK', '2'),
 (' ', 'Cubbi', '1'),
 (' 2 nd ', 'dok', '1'),
 (' maf ', '74', '1'),
 (' ', 'abr12', '1')]

How can I adjust this to get the output I'm after?

Answer 1

Consume the whitespace greedily and don't put it into the matching groups.

>>> import re
>>> s = 'Waz D: 5 l gu l: 5 GrinVe: 3 P LUK: 2 Cubbi: 1 2 nd dok: 1 maf 74: 1 abr12: 1 Waza D 5'
>>> 
>>> re.findall('([^:]+?):\s*(\d+)\s*', s)
[('Waz D', '5'), ('l gu l', '5'), ('GrinVe', '3'), ('P LUK', '2'), ('Cubbi', '1'), ('2 nd dok', '1'), ('maf 74', '1'), ('abr12', '1')]

Answer 2

If we assume that the string is always followed by a semicolon-space-number-space sequence, you can do it like this:

re.findall(r"(.+?):\s(\d+)\s", s)

[('Waz D', '5'),
 ('l gu l', '5'),
 ('GrinVe', '3'),
 ('P LUK', '2'),
 ('Cubbi', '1'),
 ('2 nd dok', '1'),
 ('maf 74', '1'),
 ('abr12', '1')]

Answer 3

It comes down to splitting on the combination : \\d , nothing else (besides suppressing leading and following whitespace here and there). All it needs is a group of any length that does not contain a colon : , followed by that colon and then a single run of digits.

import re
s = 'Waz D: 5 l gu l: 5 GrinVe: 3 P LUK: 2 Cubbi: 1 2 nd dok: 1 maf 74: 1 abr12: 1 Waza D 5'

print (re.findall(r'([^:]+):\s*(\d+)\s+', s))

result:

[('Waz D', '5'),
 ('l gu l', '5'),
 ('GrinVe', '3'),
 ('P LUK', '2'),
 ('Cubbi', '1'),
 ('2 nd dok', '1'),
 ('maf 74', '1'),
 ('abr12', '1')]

Answer 4

You could match zero or more times a whitespace character followed by capturing in a group not a colon using a negated character class ([^:]+) .

Then match a colon, zero or more whitespace characters \\s* and capture in a group one or more digits (\\d+)

\\s*([^:]+):\\s*(\\d+)

Demo

Answer 5

In your sample the name starts generally from a letter, but in 1 case - from a digit.

So the first capturing group, for the name should:

• start with [az\\d] (remember of re.I flag at the end),
• then it should contain [^:]* - a sequence of chars other than : .

Your solution ( [a-zA-Z0-9]+ ) is wrong, because the name can contain spaces.

The second group, matching the number is simple - just \\d+ .

Between these 2 groups there should be :\\s* - a colon and a sequence of white chars.

The code contains a single call to re.findall , as follows:

re.findall(r"([a-z\d][^:]*):\s*(\d+)", s, flags=re.I)

But I am in doubt about Cubbi: 1 2 in your sample. Should the 2 really be a part of the next name?

If not, consider changing the regex to: ([az][^:]*):\\s*(\\d+(?: \\d+)?) . Differences:

• The name must start with a letter (not a digit),
• The number can contain the "second part", with a preceding single space - (?: \\d+)? .

Then 1 2 will be the "numer" for Cubbi and the next name will start from "nd".

And what about Waza D 5 at the end of your sample? Did you forget to put the colon before 5 ?

Answer 6

My solution

I've added a ':' after Waza D because I think there should be (I think it was a typo, because the rule should be name: number). The pattern , for me, is a name starting with a letter and followed by other letters/numbers and spaces until the : a space and a number.

s = 'Waz D: 5 l gu l: 5 GrinVe: 3 P LUK: 2 Cubbi: 1 2 nd dok: 1 maf 74: 1 abr12: 1 Waza D: 5'

import re

# \w find something starting with a letter
# [\w\s]+ followed by any number of letter and space
# : followed by a :
# \s[0-9] and a space and a number
x = re.findall(r"\w[\w\s]+:\s[0-9]", s)
print(*x, sep="\n")

output

Waz D: 5
l gu l: 5
GrinVe: 3
P LUK: 2
Cubbi: 1
2 nd dok: 1
maf 74: 1
abr12: 1
Waza D: 5