How to use string.replace() in python 3.x

The string.replace() is deprecated on python 3.x. What is the new way of doing this?

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

As in 2.x, use str.replace().

Example:

>>> 'Hello world'.replace('world', 'Guido')
'Hello Guido'

Method 2

replace() is a method of <class 'str'> in python3:

>>> 'hello, world'.replace(',', ':')
'hello: world'

Method 3

The replace() method in python 3 is used simply by:

a = "This is the island of istanbul"
print (a.replace("is" , "was" , 3))

#3 is the maximum replacement that can be done in the string#

>>> Thwas was the wasland of istanbul

# Last substring 'is' in istanbul is not replaced by was because maximum of 3 has already been reached

Method 4

You can use str.replace() as a chain of str.replace(). Think you have a string like 'Testing PRI/Sec (#434242332;PP:432:133423846,335)' and you want to replace all the '#',':',';','/' sign with '-'. You can replace it either this way(normal way),

>>> string = 'Testing PRI/Sec (#434242332;PP:432:133423846,335)'
>>> string = string.replace('#', '-')
>>> string = string.replace(':', '-')
>>> string = string.replace(';', '-')
>>> string = string.replace('/', '-')
>>> string
'Testing PRI-Sec (-434242332-PP-432-133423846,335)'

or this way(chain of str.replace())

>>> string = 'Testing PRI/Sec (#434242332;PP:432:133423846,335)'.replace('#', '-').replace(':', '-').replace(';', '-').replace('/', '-')
>>> string
'Testing PRI-Sec (-434242332-PP-432-133423846,335)'

Method 5

Try this:

mystring = "This Is A String"
print(mystring.replace("String","Text"))

Method 6

Official doc for str.replace of Python 3

official doc: Python 3’s str.replace

str.replace(old, new[, count])

Return a copy of the string with all occurrences of substring old replaced by new. If the optional argument count is given, only the first count occurrences are replaced.

corresponding VSCode‘s syntax notice is:

str.replace(self: str, old, new, count) -> str

Two method to use str.replace

• Method 1: use builtin str’s replace -> str.replace(strVariable, old, new[, count])

replacedStr1 = str.replace(originStr, "from", "to")

• Method 2: use str variable’s replace -> strVariable.replace(old, new[, count])

replacedStr2 = originStr.replace("from", "to")

Full demo

code:

originStr = "Hello world"

# Use case 1: use builtin str's replace -> str.replace(strVariable, old, new[, count])
replacedStr1 = str.replace(originStr, "world", "Crifan Li")
print("case 1: %s -> %s" % (originStr, replacedStr1))

# Use case 2: use str variable's replace -> strVariable.replace(old, new[, count])
replacedStr2 = originStr.replace("world", "Crifan Li")
print("case 2: %s -> %s" % (originStr, replacedStr2))

output:

case 1: Hello world -> Hello Crifan Li
case 2: Hello world -> Hello Crifan Li

screenshot:

My related (Chinese) post: 【详解】Python 3中字符串的替换str.replace

Method 7

FYI, when appending some characters to an arbitrary, position-fixed word inside the string (e.g. changing an adjective to an adverb by adding the suffix -ly), you can put the suffix at the end of the line for readability. To do this, use split() inside replace():

s="The dog is large small"
ss=s.replace(s.split()[3],s.split()[3]+'ly')
ss
'The dog is largely small'

Method 8

Simple Replace:         .replace(old, new, count) .

text = "Apples taste Good."
print(text.replace('Apples', 'Bananas'))          # use .replace() on a variable
Bananas taste Good.          <---- Output

print("Have a Bad Day!".replace("Bad","Good"))    # Use .replace() on a string
Have a Good Day!             <----- Output

print("Mom is happy!".replace("Mom","Dad").replace("happy","angry"))  #Use many times
Dad is angry!                <----- Output

Method 9

ss = s.replace(s.split()[1], +s.split()[1] + 'gy')
# should have no plus after the comma --i.e.,
ss = s.replace(s.split()[1], s.split()[1] + 'gy')

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0