I have a string and 2 arrays like below:
st="a1b2c3d"
arr1 = ['1','2','3']
arr2 = ['X','Y','Z']
I want to replace all the value of '1', '2', '3' to 'X', 'Y', 'Z'. The final string will look like:
'aXbYcZd'
So I wrote this for loop:
for i in range(0, len(arr1)):
st.replace(str(arr1[i]),str(arr2[i]))
The result is:
'aXb2c3d'
'a1bYc3d'
'a1b2cZd'
How to correctly do what I want above?
Thanks!
Use zip() to iterate through two lists simultaneously to replace values:
st = "a1b2c3d"
arr1 = ['1','2','3']
arr2 = ['X','Y','Z']
for x, y in zip(arr1, arr2):
st = st.replace(x, y)
print(st)
# aXbYcZd
str.replace() does not replace a string in-place. You need to assign returned value back to a variable.
If you're replacing characters, instead of the inefficient replace loop use str.translate with str.maketrans:
>>> table = str.maketrans('123', 'XYZ')
>>> result = 'a1b2c3d'.translate(table)
>>> result
'aXbYcZd'
maketrans requires 2 strings as arguments. If you really have a list, you can use ''.join(l) to make it into a suitable string. You need to make the table only once.
The efficiency is but one point. str.translate is the way to do this correctly in cases where you will map a => b and b => something else. If you want to replace strings then you might need to use re.sub instead.
Calling replace over and over means you have to iterate through the entire string for each replacement, which is O(m * n). Instead:
rep = dict(zip(arr1, arr2)) # make mapping, O(m)
result = ''.join(rep.get(ch, ch) for ch in st)
The first line is O(m), where m is the length of arr1 and arr2.
The second line is O(n), where n is the length of st.
In total this is O(m + n) instead of O(m * n), which is a significant win if either m or n is large.
