Python 中的字符串替换列表

Question

Is there a far shorter way to write the following code?

my_string = my_string.replace('A', '1')
my_string = my_string.replace('B', '2')
my_string = my_string.replace('C', '3')
my_string = my_string.replace('D', '4')
my_string = my_string.replace('E', '5')

Note that I don't need those exact values replaced; I'm simply looking for a way to turn 5+ lines into fewer than 5

Answer 1

Looks like a good opportunity to use a loop:

mapping = { 'A':'1', 'B':'2', 'C':'3', 'D':'4', 'E':'5'}
for k, v in mapping.iteritems():
    my_string = my_string.replace(k, v)

A faster approach if you don't mind the parentheses would be:

mapping = [ ('A', '1'), ('B', '2'), ('C', '3'), ('D', '4'), ('E', '5') ]
for k, v in mapping:
    my_string = my_string.replace(k, v)

Answer 2

You can easily use string.maketrans() to create the mapping string to pass to str.translate():

import string
trans = string.maketrans("ABCDE","12345")
my_string = my_string.translate(trans)

Answer 3

If you want to get the wrong answer, slowly, then use string.replace in a loop. (Though it does work in this case of no overlap among the patterns and replacements.)

For the general case with possible overlaps or a long subject string, use re.sub:

import re

def multisub(subs, subject):
    "Simultaneously perform all substitutions on the subject string."
    pattern = '|'.join('(%s)' % re.escape(p) for p, s in subs)
    substs = [s for p, s in subs]
    replace = lambda m: substs[m.lastindex - 1]
    return re.sub(pattern, replace, subject)

>>> multisub([('hi', 'bye'), ('bye', 'hi')], 'hi and bye')
'bye and hi'

For the special case of 1-character patterns and 1- or 0-character replacements, use string.maketrans.

Answer 4

Also look into str.translate() . It replaces characters according to a mapping you provide for Unicode strings, or otherwise must be told what to replace each character from chr(0) to chr(255) with.

Answer 5

replaceDict = {'A':'1','B':'2','C':'3','D':'4','E':'5'}       
for key, replacement in replaceDict.items():  
  my_string = my_string.replace( key, replacement )

Answer 6

I think it could be a little more efficient:

mapping = { 'A':'1', 'B':'2', 'C':'3', 'D':'4', 'E':'5'}
my_string = "".join([mapping[c] if c in mapping else c for c in my_string])

I suggest some benchmark with "timeit", with real cases in base of the lenght of "my_string".

Answer 7

You can do it in one line using Pandas.

import pandas as pd

my_string="A B C test"

my_string =pd.DataFrame([my_string])[0].replace(["A","B","C","D","E"],['1','2','3','4','5'],regex=True)[0]

print(my_string)
'1 2 3 test'

Answer 8

One way I do it is with an associated array (a dictionary). Here is an example of the replacements I use when getting a file ready for deployment in LaTeX using regular expressions.

  import re
  def escapeTexString(string): # Returns TeX-friendly string
    rep = { # define desired replacements in this dictionary (mapping)
         '&': '\\&',
         '%': '\\%',
         '#': '\\#',
         '_': '\\_',
         '{': '\\{', # REGEX Special
         '}': '\\}', # REGEX Special
         '~': '\\char"007E{}', # LaTeX Special
         '$': '\\$', # REGEX Special
         '\\': '\\char"005C{}', # REGEX/LaTeX Special
         '^': '\\char"005E{}', # REGEX/LaTeX Special
         '"': '\\char"FF02{}'
        }
    # use these two lines to do the replacement (could be shortened to one line)
    pattern = re.compile("|".join(map(re.escape,rep.keys()))) # Create single pattern object (key to simultaneous replacement)
    new_string = pattern.sub(lambda match: rep[match.group(0)], string)
    return new_string