Python正则表达式：替换，忽略空字符串

Question

I am trying to replace a given pattern with regular expressions in Python, using re . Here is the piece of Python code I wrote:

import re

fname = './prec.f90'
f = open(fname)
lines = f.readlines()
f.close()
for i, line in enumerate(lines):
    search = re.findall('([\d*]?\.[\d*]?)+?[^dq\_]', line)
    if search != []: 
        print('Real found in line #%d: ' %i)
        print search
        print('The following line:\n %s' %line)
        print('will be replace by:')
        newline = re.sub('([\d*]?\.[\d*]?)+?[^dq\_]', r'\g<1>d0\g<2>', line)
        print('%s' %newline)

And the prec.f90 contains something like that (it is just an example, it does not means that all the strings I want to replace have the form [x]_[yz] = ...; ):

  x_pr = 0.1; y_pr = 0.2; z_pr = 0.1q0
  x_sp = 0.1; y_sp = 0.1d0; z_sp = 0.1q0
  x_dp = 0.1; y_dp = 0.1d0; z_dp = 0.1q0
  x_qp = .1; y_qp = 0.1d0; z_qp = 0.1q0
  x_db = 0.; y_db = 0.1d0; y_db = 0.1q0

My goal is to modify all the pattern like 0.1 , .1 and 0. , to get something like 0.1d0 ; I don't want to modify the other patterns. The problem is that re.findall('[\\d*]?\\.[\\d*]?)+?([^dq\\_]') matches the pattern I am looking for, but also returns an empty string for the other ones. Therefore, when I run this piece of code, it fails, being unable to replace match the first and second groups in the re.sub() for the empty strings.

I guess one solution would be to ignore empty string in the re.sub , or to have something like a conditional argument in it, but I could not figure out how.

Any help would be appreciated!

Answer 1

You can simplify the sub as

>>> str="x_db = 0.; y_db = 0.1d0; y_db = 0.1q"
>>> re.sub(r'(0\.1|\.1|0\.)(?=;)', r'\g<1>0d0', str)
'x_db = 0.0d0; y_db = 0.1d0; y_db = 0.1q'

The regex (0\\.1|\\.1|0\\.)(?=;) would match 0.1 , .1 and 0. followed by as ;

Answer 2

(x_[a-zA-Z]{2}\s*=)\s+[^;]+

Try this.Replace by \\1 0.1d0 .See demo.

http://regex101.com/r/qZ6sE3/2

import re
p = re.compile(ur'(x_[a-zA-Z]{2}\s*=)\s+[^;]+')
test_str = u"x_pr = 0.1; y_pr = 0.2; z_pr = 0.1q0\nx_sp = 0.1; y_sp = 0.1d0; z_sp = 0.1q0\nx_dp = 0.1; y_dp = 0.1d0; z_dp = 0.1q0\nx_qp = .1; y_qp = 0.1d0; z_qp = 0.1q0\nx_db = 0.; y_db = 0.1d0; y_db = 0.1q0"
subst = u"\1 0.1d0"

result = re.sub(p, subst, test_str)

Answer 3

I finally came up with this piece of code that is working as intended:

import re

fname = './prec.f90'
f = open(fname)
lines = f.readlines()
f.close()
# If there was no end of the line character (\n) we would need to check if 
# this is the end of the line (something like ([^dq\_0-9]|$)
regex = re.compile(r'(\d*\.\d*)([^dq\_0-9])')
for i, line in enumerate(lines):
    search = regex.findall(line)
    if search != []: 
        print('Real found in line #%d: ' %i)
        print search
        print('The following line:\n %s' %line)
        print('will be replace by:')
        newline = regex.sub(r'\g<1>d0\g<2>', line)
        print('%s' %newline)

I first came up with the more complicated regex ([\\d*]?\\.[\\d*]?)+?[^dq\\_] because else I always match the first part of any string ending with d , q or _ . It seemed to be due to the fact that \\d* was not greedy enough; to add 0-9 in the "ignore" set solves the problem.