用冒号替换每个奇怪的逗号 - 正则表达式

Question

I need to write a regex that converts every odd comma to a colon in python.

For example

"[2, 0.2520474110789976, 8, 0.25215388264234934, 3, 0.3560689678084889, 1, 0.3573715347893714, 4, 0.5626369616327825, 5, 0.793617535995843]"

gets converted to

"[2: 0.2520474110789976, 8: 0.25215388264234934, 3: 0.3560689678084889, 1: 0.3573715347893714, 4: 0.5626369616327825, 5: 0.793617535995843]"

I did go through other questions on StackOverflow and found the below question. However, the JS version doesn't seem to work in Python.

Regex - replace all odd numbered occurrences of a comma

I did the following based on the link above

pattern = "(?=(?:[^\"]*\"[^\"]*\")*[^\"]*$)(,)(.*?,|)(?=.*?(?:,|$))"
stringa = re.sub(pattern,": ",flat_list_string)

and got an output

"2:  8:  3:  1:  4:  5:  0.793617535995843"

instead of the one mentioned earlier.

I'm pretty new to Regex, so haven't tried much myself. Would appreciate any help. Thanks.

Update1: Pasted my incorrect output

Answer 1

You seem to just be using the regex incorrectly. First, you should use a raw string literal r"..." so that you don't have to escape the backslashes:

pattern = r"(?=(?:[^\"]*\"[^\"]*\")*[^\"]*$)(,)(.*?,|)(?=.*?(?:,|$))"

Next, you should change the replacement string to r":\2" , which means : followed by group 2. The regex matches every odd comma, and also all the characters after it until the next even comma. It puts all this into group 2. Replacing with just : will replace all those matched characters too.

stringa = re.sub(pattern, r":\2",flat_list_string)

The JS regex also handles commas in quotes that the OP of the other post doesn't want to consider, such as:

"hello, world", 1, "bye, world", 2
      ^.               ^
these should not be counted as commas

If you do want to count these commas, then you can use this regex:

,([^,]+(?:,|$))

And replace with :\1 .

Answer 2

You can achieve the same result with this simple regex

re.sub(r'(\b\d{1,2}),',r'\g<1>:',search_string)