Python Regex

[quest]

I have the following text file:

Output:
S = 0, X are: (1, 0, 1, 1, 0, 0, )S = 0, Z are: (0, 1, 1, 1, 0, 1, )Data bits measurement:0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, Measured l0: 1, Measured l1: 0 S = 0, X are: (1, 1, 0, 0, 0, 1, )S = 0, Z are: (1, 1, 0, 0, 1, 1, )Data bits measurement:1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, Measured l0: 0, Measured l1: 1 S = 0, X are: (1, 1, 0, 0, 1, 1, )S = 0, Z are: (1, 1, 0, 0, 1, 1, )Data bits measurement:1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, Measured l0: 1, Measured l1: 0 S = 0, X are: (1, 1, 1, 0, 1, 1, )S = 0, Z are: (0, 0, 0, 1, 0, 0, )Data bits measurement:0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, Measured l0: 1, Measured l1: 0 S= 0, X are: (1, 1, 0, 1, 1, 0, )S = 0, Z are: (0, 0, 0, 1, 0, 0, )Data bits measurement:0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, Measured l0: 1, Measured l1: 0 S = 0, X are: (1, 1, 0, 0, 1, 1, )S = 0, Z are: (1, 1, 1, 0, 0, 1, )Data bits measurement:1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, Measured l0: 1, Measured l1: 0

Now, I want to count how many same Z value I have and I have many l0 I have for this Z value:
For example:

Output:
011101 - 1 and l0 1 (in total) 110011 - 2 and l0 1 (in total) 000100 - 2 and l0 2 (in total) 111001 - 1 and l0 1 (in total)

I thought regex can be nice but I could not manage to use it:
Here is my code:

with open("f.dat") as file: lines = file.readlines() #Z=re.search("(^Z*)",line) for i in re.findall(r'Z are: (=(\d+)&',lines): print(i)

And I got this error:

Error:
raise source.error("missing ), unterminated subpattern", re.error: missing ), unterminated subpattern at position 17

[deanhystad]

Parentheses have special meaning in regex and must appear in pairs. If you want to include the () characters in the pattern you need to precede with \: \( or \).

I don't understand some of your pattern.

Why does it contain "&"? This character does not appear in your string and has no special meaning in a regex pattern that I know of. Did you mean "*" (fat finger mistake)?

\d (decimal) appears multiple times in the pattern, but there are also commas and spaces, so the repeat is "\d, ", not "\d". Should "(\d+)&" be "(?:\d, )*"

If you have a repeated group like "(?:(\d), )+" you can only get the last match for the group. You cannot use grouping to get "011101" out of "0, 1, 1, 1, 0, 1, ", you could only retrieve the last match for the group, "1". I think you should match everything between the parentheses and remove all ", " from the string.

I think you want to do this:

import re data = "S = 0, X are: (1, 0, 1, 1, 0, 0, )S = 0, Z are: (0, 1, 1, 1, 0, 1, )Data bits measurement:0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, Measured l0: 1, Measured l1: 0" pattern = r"Z are: \((.*)\)" # \( and \) are to have ( and ) as characters in the pattern. (.*) is the group of characters we want returned match = re.search(pattern, data) group = match.group(1) # Get the first group. This will be something like "0, 1, 1, 1, 0, 1, " print(group, group.replace(", ", "")) # Strip out the ", " to get the desired string. #or match = re.findall(pattern, data) group = match[0] print(group, group.replace(", ", ""))

Output:
0, 1, 1, 1, 0, 1, 011101 0, 1, 1, 1, 0, 1, 011101

What does the "1 and 10" mean in your example output "011101 - 1 and 10 1 (in total)"

[quest]

Parentheses have special meaning in regex and must appear in pairs. If you want to include the () characters in the pattern you need to precede with \: \( or \).

I don't understand some of your pattern.

Why does it contain "&"? This character does not appear in your string and has no special meaning in a regex pattern that I know of. Did you mean "*" (fat finger mistake)?

\d (decimal) appears multiple times in the pattern, but there are also commas and spaces, so the repeat is "\d, ", not "\d". Should "(\d+)&" be "(?:\d, )*"

If you have a repeated group like "(?:(\d), )+" you can only get the last match for the group. You cannot use grouping to get "011101" out of "0, 1, 1, 1, 0, 1, ", you could only retrieve the last match for the group, "1". I think you should match everything between the parentheses and remove all ", " from the string.

I think you want to do this:

import re data = "S = 0, X are: (1, 0, 1, 1, 0, 0, )S = 0, Z are: (0, 1, 1, 1, 0, 1, )Data bits measurement:0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, Measured l0: 1, Measured l1: 0" pattern = r"Z are: \((.*)\)" # \( and \) are to have ( and ) as characters in the pattern. (.*) is the group of characters we want returned match = re.search(pattern, data) group = match.group(1) # Get the first group. This will be something like "0, 1, 1, 1, 0, 1, " print(group, group.replace(", ", "")) # Strip out the ", " to get the desired string. #or match = re.findall(pattern, data) group = match[0] print(group, group.replace(", ", ""))

Output:
0, 1, 1, 1, 0, 1, 011101 0, 1, 1, 1, 0, 1, 011101

What does the "1 and 10" mean in your example output "011101 - 1 and 10 1 (in total)"

Thanks a lot for the answer :) It is not 10 but l0: 1. It is the measurement result. I want to keep track of this result too.
Your answer is very nice. I will first try it and try to figure out how much I can go from there. I will back to you if I have any troubles :)
Thanks