正则表达式 python：仅查找第二个和第三个字符串模式

Question

for the following text I want to get second and third pattern of "Error:....." (see bold text).

Basically I need to print out from text below:

Error: Calculated last average: 150.34365623971, read last average: 132, Tolerance: 12

Following text:

Error: Register 3.5.0 -------------- [Device type: Am500T_Elektra; Device Id: 00000000]

Error: Calculated last average: 150.34365623971, read last average: 132 -------------- [Device type: Am500T_Elektra; Device Id: 00000000]

Error: Tolerance: 12 -------------- [Device type: Am500T_Elektra; Device Id: 00000000]

I got stuck here:

re.compile(r'Error: (.+ -)').findall

Any hint or advice how to continue?

Answer 1

Assuming that only the values change:

regex = re.compile(r'Error: Calculated last average: \d+\.?\d+?, read last average: \d+, Tolerance: \d+')
re.match(regex, text) 

The \d+\.?\d+? means match any digit at one or more times, followed by an optional dot (the decimal point), followed by another optional digit match. This way, even if the average is not a floating point number, it will still match.

For the other values, I have kept the regex as \d+ , assuming that these will always be integers. You may need to change these as well, as the case may be.

Answer 2

Could try using a re.sub

>>> import re
>>> text = '''
... Error: Register 3.5.0 -------------- [Device type: Am500T_Elektra; Device Id: 00000000]
...
... Error: Calculated last average: 150.34365623971, read last average: 132 -------------- [Device type: Am500T_Elektra; Device Id: 00000000]
...
... Error: Tolerance: 12 -------------- [Device type: Am500T_Elektra; Device Id: 00000000]
... '''
>>>
>>> regex = re.compile(r"(?m)(?:\A\s*.*\s*|^Error:\s*|(\s*--+.*\s*(?=^Error:))|\s*--+.*\s*|(?![\S\s]*?(?:^Error:|--+))[\S\s]+)")
>>>
>>> print ( "Error: " + regex.sub(lambda x: ", " if (x.group(1)) else "", text ) )
Error: Calculated last average: 150.34365623971, read last average: 132, Tolerance: 12