python正则表达式查找匹配的字符串

Question

I am trying to find the matched string in a string using regex in Python. The string looks like this:

band   1 # energy  -53.15719532 # occ.  2.00000000

ion      s      p      d    tot
  1  0.000  0.995  0.000  0.995
  2  0.000  0.000  0.000  0.000
tot  0.000  0.996  0.000  0.996

band   2 # energy  -53.15719532 # occ.  2.00000000

ion      s      p      d    tot
  1  0.000  0.995  0.000  0.995
  2  0.000  0.000  0.000  0.000
tot  0.000  0.996  0.000  0.996

band   3 # energy  -53.15719532 # occ.  2.00000000

My goal is to find the string after tot . So the matched string will be something like:

['0.000  0.996  0.000  0.996', 
'0.000  0.996  0.000  0.996']

Here is my current code:

pattern = re.compile(r'tot\s+(.*?)\n', re.DOTALL)
pattern.findall(string)

However, the output gives me:

['1  0.000  0.995  0.000  0.995',
 '0.000  0.996  0.000  0.996',
 '1  0.000  0.995  0.000  0.995',
 '0.000  0.996  0.000  0.996']

Any idea of what I am doing wrong?

Answer 1

You don't want the DOTALL flag. Remove it and use MULTILINE instead.

pattern = re.compile(r'^\s*tot(.*)', re.MULTILINE)

This matches all lines that start with tot . The rest of the line will be in group 1.

Citing the documentation , emphasis mine:

re.DOTALL

Make the '.' special character match any character at all, including a newline ; without this flag, '.' will match anything except a newline.

Note that you can easily do this without regex.

with open("input.txt", "r") as data_file:
    for line in data_file:
        items = filter(None, line.split(" "))
        if items[0] == "tot":
            # etc

Answer 2

You are using re.DOTALL, which means that the dot "." will match anything, even newlines, in essence finding both "tot"-s and everything that follows until the next newline:

                            tot
  1  0.000  0.995  0.000  0.995

and

tot  0.000  0.996  0.000  0.996

Removing re.DOTALL should fix your problem.

Edit: Actually, the DOTALL flag is not really the issue (though unnecessary). The problem in the pattern is that the \\s+ matches the newline. Replacing that with a single space solves that issue:

pattern = re.compile(r'tot (.*?)\n')

Answer 3

The alternative solution using re.findall function with specific regex pattern:

# str is your inital string
result = re.findall('tot [0-9 .]+(?=\n|$)', str)
print(result)

The output:

['tot  0.000  0.996  0.000  0.996', 'tot  0.000  0.996  0.000  0.996']