找到k个连续数字后如何终止字符串？

Question

Say I have some list with files of the form *.1243.* , and I wish to obtain everything before these 4 digits. How do I do this efficiently?

An ugly, inefficient example of working code is:

names = []
for file in file_list:
    words = file.split('.')
    for i, word in enumerate(words):
        if word.isdigit():
            if int(word)>999 and int(word)<10000:
                names.append(' '.join(words[:i]))
                break
print(names)

Obviously though, this is far from ideal and I was wondering about better ways to do this.

Answer 1

You may want to use regular expressions for this.

import re

name = []
for file in file_list:
    m = re.match(r'^(.+?)\.\d{4}\.', file)
    if m:
        name.append(m.groups()[0])

Answer 2

Using a regular expression, this would become simpler

import re

names = ['hello.1235.sas','test.5678.hai']

for fn in names:
    myreg = r'(.*)\.(?:\d{4})\..*'
    output = re.findall(myreg,fn)
    print(output)

output:

['hello']
['test']

Answer 3

If you know that all entries has the same format, here is list comprehension approach:

[item[0] for item in filter(lambda start, digit, end: len(digit) == 4, (item.split('.') for item in file_list))]

To be fair I also like solution, provided by @James. Note, that downside of this list comprehension is three loops: 1. On all items to split 2. Filtering all items, that match 3. Returning result.

With regular for loop it could be be more sufficient:

output = []
for item in file_list:
    begging, digits, end = item.split('.')
    if len(digits) == 4:
        output.append(begging)

It does only one loop, which way better.

Answer 4

You can use Positive Lookahead (?=(\\.\\d{4}))

import re
pattern=r'(.*)(?=(\.\d{4}))'

text=['*hello.1243.*','*.1243.*','hello.1235.sas','test.5678.hai','a.9999']


print(list(map(lambda x:re.search(pattern,x).group(0),text)))

output:

['*hello', '*', 'hello', 'test', 'a']