Split String Every Nth Character

Split string every nth character from the right?

You can adapt the answer you linked, and use the beauty of mod to create a nice little one-liner:

>>> s = '1234567890'
>>> '/'.join([s[0:len(s)%3]] + [s[i:i+3] for i in range(len(s)%3, len(s), 3)])
'1/234/567/890'

and if you want this to auto-add the dot for the cases like your first example of:

s = '100243'

then you can just add a mini ternary use or as suggested by @MosesKoledoye:

>>> '/'.join(([s[0:len(s)%3] or '.']) + [s[i:i+3] for i in range(len(s)%3, len(s), 3)])
'./100/243'

This method will also be faster than reversing the string before hand or reversing a list.

split-string-every-nth-character with nth+1 separator '0'

You can use a regular expression with capture groups.

import re

instr = '01110100001101001001110100'
outlist = list(sum(re.findall(r'(\d{8})(\d)', instr), ()))

print(outlist)

re.findall() returns a list of tuples, list(sum(..., ()) flattens it into a single list.

Python split string every n character

Those are called n-grams.

This should work :)

text = "BANANA"
n = 2
chars = [c for c in text]
ngrams = []
for i in range(len(chars)-n + 1):
  ngram = "".join(chars[i:i+n])
  ngrams.append(ngram)
print(ngrams)

output: ['BA', 'AN', 'NA, 'AN', 'NA']

How to insert a character every nth character in a pandas dataframe column

Let us try findall with map (.. means N = 2)

df.mac_address.str.findall('..').map(':'.join)
Out[368]: 
0    00;03;E6;A5;84;C2
1    00;03;E6;A5;84;CC
2    00;03;E6;A5;84;DA
3    00;03;E6;A5;84;DC
4    00;03;E6;A5;84;E4
Name: mac_address, dtype: object

Select every nth character from a string

To follow-up on OP's idea ("use the row numbers"). Split the string, fill a matrix with 10 rows, select the first row.

matrix(strsplit(x, "")[[1]], nrow = 10)[1, ]
# [1] "h" "d" "r" "." "j" "x"

You will get a recycling warning, but that will not affect us because we select the first row.

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Good'ol charToRaw:

rawToChar(charToRaw(x)[c(TRUE, rep(FALSE, 9))])
# [1] "hdr.jx"

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