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up vote 0 down vote favorite

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!

share|improve this question
    
It is failing [subbing the empty string] because you made all the components optional with *. What is the minimum string that will be on a line that you want to replace? If it is .x then change the second \d* to \d+ –  beroe Nov 21 '14 at 18:58
    
Problem is they can have the 3 forms I mentioned: x.y, .y and x.. I want (and need) to cover all these cases. –  MBR Nov 23 '14 at 10:47

3 Answers 3

up vote 0 down vote accepted

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.

share|improve this answer
up vote 0 down vote

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 ;

share|improve this answer
    
Please read my edit, the strings I want to replace have a more general form, this was just an example. –  MBR Nov 21 '14 at 18:38
up vote 0 down vote 
(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)
share|improve this answer
    
The string I want to replace can have a much general form than x_[a-Z] = .... The test in my question is what it is: a simple test :) –  MBR Nov 23 '14 at 10:48
    
@MBR regex101.com/r/yP3iB0/6 ??????/ –  vks Nov 23 '14 at 12:08
    
nope, because I don't want to modify only the "0.1" but actualy any real number, and I don't have all my expressions ending with a semi-column. and I want to replace the original string by its value, not a dummy 0.1. –  MBR Nov 23 '14 at 17:06

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