正则表达式查找不包含某些字符的数字

Question

I currently have a text field that contains information about times that are to be used for scheduling purposes. As it is a text field the data is unstructured and is in many different formats. Examples of data include:

• Mon-Wed 6-7:30pm
• Tuesday/Thurs 5:00 - 6:30
• M/T/W 3:30 -7
• F 4-5

As such I am trying to write a parser to turn these into usable data points. I am working on the time components at the moment. In order to structure the data and have the ability to pass it into the dateutil parser I want to "fill out" all times. 6 would become 6:00, 7 would become 7:00 etc. To do so I am trying to use the regex expression:

reg = re.compile('[\d]([^:]|$)')

The idea is to get any digit that either does not have a : after it, or is at the end of the line. However, I realized that this will get too many data points as in the first example it would get the '3' of 7:30 and the 0 of 7:30.

What would be a better way to convert this data to a usable format?

Answer 1

I would do it in two-stage manner, harnessing one interesting feature of re.split , sample data:

line1 = 'Mon-Wed 6-7:30pm'
line2 = 'Tuesday/Thurs 5:00 - 6:30'
line3 = 'M/T/W 3:30 -7'
line4 = 'F 4-5'

Function:

def add_zeros(line):
    parts = re.split(r'(\d{1,2}:\d{1,2})',line)
    parts[::2] = [re.sub(r'(\d{1,2})',r'\1:00',p) for p in parts[::2]]
    return ''.join(parts)

Usage:

print(add_zeros(line1)) # Mon-Wed 6:00-7:30pm
print(add_zeros(line2)) # Tuesday/Thurs 5:00 - 6:30
print(add_zeros(line3)) # M/T/W 3:30 -7:00
print(add_zeros(line4)) # F 4:00-5:00

Explanation:

I give re.sub the first argument within the group. re.split gives a list with odd-indexed elements being separators . With the pattern I used in re.split the seperators are "ready" hours (which do not need zero-padding). I then use re.sub on every even-indexed element of list (the non "ready" hours), treating every 1 or 2 digit number as an hour and replacing it with the number followed by :00

Answer 2

您可以使用负向后看和负向前看(?<!(:)\\d)\\d(?!(:|\\d)) https://regex101.com/r/nAQh3e/4这将选择数字之前或之后没有数字并且还没有的数字:

Answer 3

I think it will be much easier to find the incorrect time after replacing the correct time with a placeholder. Then we can correct the incorrect time format and then again substitute the placeholder with actual times

Here is simple implementation, you can tweak this to match your need

import re

texts = ["Mon-Wed 6-7:30pm",
"Tuesday/Thurs 5:00 - 6:30",
"M/T/W 3:30 -7",
"F 4-5",]

def get_placeholder_replacer(replaced_strings):
    def replace_with_placeholder(x):
        replaced_strings.append(x[0])
        return "{}"
    return replace_with_placeholder


ptrn_correct_time = re.compile(r"\d+:\d+")
ptrn_incorrect_time = re.compile(r"\d{1,2}")

for text in texts:
    replaced_strings = []
    placeholder_replacer = get_placeholder_replacer(replaced_strings)
    new_text = ptrn_correct_time.sub(placeholder_replacer,text)
    new_text = ptrn_incorrect_time.sub(lambda x: "{}:00".format(x[0]), new_text)

    print(new_text.format(*replaced_strings))

## Output
# Mon-Wed 6:00-7:30pm
# Tuesday/Thurs 5:00 - 6:30
# M/T/W 3:30 -7:00
# F 4:00-5:00