Python replace 3 random characters in a string with no duplicates

Question

I need change 3 random characters in a string using Python, example string:

Adriano Celentano 
Luca Valentina

I need to replace 3 characters, not replacing with the same character or number, not replacing space. How can I do this using Python ?

Need output like this :

adraano cettntano
lacr vilenntina

I don't know from where i can start to make this.

My code so far:

for i in xrange(4):
    for n in nume :
        print n.replace('$', random.choice(string.letters)).replace('#', random.choice(string.letters))

why are you looping 4 times and how do '$' and '#' fit into it? — Padraic Cunningham
– Padraic Cunningham, Commented Dec 17, 2015 at 16:06

Padraic Cunningham · Accepted Answer · 2015-12-17 19:51:07Z

6

If you just want to change chars that are not whitespace and not the same char in regards to index, you can first pull the indexes where the non-whitespace chars are:

import random
inds = [i for i,_ in enumerate(s) if not s.isspace()]

print(random.sample(inds,3))

Then use those indexes to replace.

s = "Adriano Celentano"
import random
inds = [i for i,_ in enumerate(s) if not s.isspace()]
sam = random.sample(inds, 3)
from string import ascii_letters

lst = list(s)
for ind in sam:
    lst[ind] = random.choice(ascii_letters)

print("".join(lst))

If you want a unique char each time to replace with also:

s = "Adriano Celentano"
import random
from string import ascii_letters
inds = [i for i,_ in enumerate(s) if not s.isspace()]

sam = random.sample(inds, 3)

letts =  iter(random.sample(ascii_letters, 3))
lst = list(s)
for ind in sam:
    lst[ind] = next(letts)

print("".join(lst))

output:

Adoiano lelenhano

edited Dec 17, 2015 at 19:51

answered Dec 17, 2015 at 15:56

Padraic Cunningham

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4 Comments

Andrea Corbellini Over a year ago

Instead of excluding numbers and spaces, I think it's safer to include only letters via set('your string...') & set('abcdef...')

Padraic Cunningham Over a year ago

@AndreaCorbellini, it depends on what the OP wants to replace, there is some ambiguity in the question in regard to what they actually want to replace

dwanderson Over a year ago

set('your string') & set(string.ascii_uppercase + string.ascii_lowercase) even

Mark K Over a year ago

in case the string contains space and punctuation, how about "inds = [i for i,_ in enumerate(s) if _.isalpha()]"?

Alex · Accepted Answer · 2015-12-17 20:24:29Z

0

You can do this in two stages. In the first stage you pick 3 random positions in your string that meet your search criteria (isalnum):

import random
import string
replacement_chars='abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'
# replacement_chars = string.letters + string.digits
input = 'Adriano Celentano'
input_list = list(input) 
input_dic = dict(enumerate(input_list))
valid_positions=[key for key in input_dic if input_dic[key].isalnum()]
random_positions=random.sample(valid_positions,3)

In the second part you generate 3 random characters and replace the characters in the previously selected positions. I have added a while loop to generate a new random character if it matches the existing value

random_chars = random.sample(replacement_chars,len(random_positions)) 
char_counter = 0
for position in random_positions:   
    #check if the replacement character matches the existing one
    #and generate another one if needed
    while input_list[position]==random_chars[char_counter]:
        random_chars[char_counter] = random.choice(replacement_chars)    
    input_list[position]=random_chars[char_counter]    
    char_counter = char_counter + 1
print "".join(input_list).lower()

edited Dec 17, 2015 at 20:24

answered Dec 17, 2015 at 16:08

Alex

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1 Comment

Padraic Cunningham Over a year ago

Why use random.sample when you want a single element?

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