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问题:
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
回答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)
回答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)
回答3:
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.
回答4:
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.
回答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 )
回答6:
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