Replace Every N'Th Occurrence in Huge Line in a Loop

replace every nth occurrence of a pattern using awk

Your code is almost right, i modified it.

To replace every nth occurrence, you need a modular expression.

So for better understanding with brackets, you need an expression like ((i % n) == 0)

awk -v p="@" -v n="5" ' $0~p { i++ } ((i%n)==0) { sub(/^@/, "\n#\n@") }{ print }' in.bib > out.bib

How to use sed or awk or something similar to replace every odd occurrence of character?

With any sed that supports EREs via -E, e.g. GNU sed and OSX/BSD sed:

$ echo "1,0,2,0,3,0,4,0,5,0,6,0,13,05,24233,55" | sed -E 's/,([^,]+(,|$))/.\1/g'
1.0,2.0,3.0,4.0,5.0,6.0,13.05,24233.55

The above was inspired by @PesaThe's comment to my original answer.

Replace nth occurrence of substring in string

I use simple function, which lists all occurrences, picks the nth one's position and uses it to split original string into two substrings. Then it replaces first occurrence in the second substring and joins substrings back into the new string:

import re

def replacenth(string, sub, wanted, n):
where = [m.start() for m in re.finditer(sub, string)][n-1]
before = string[:where]
after = string[where:]
after = after.replace(sub, wanted, 1)
newString = before + after
print(newString)

For these variables:

string = 'ababababababababab'
sub = 'ab'
wanted = 'CD'
n = 5

outputs:

ababababCDabababab

Notes:

The where variable actually is a list of matches' positions, where you pick up the nth one. But list item index starts with 0 usually, not with 1. Therefore there is a n-1 index and n variable is the actual nth substring. My example finds 5th string. If you use n index and want to find 5th position, you'll need n to be 4. Which you use usually depends on the function, which generates our n.

This should be the simplest way, but maybe it isn't the most Pythonic way, because the where variable construction needs importing re library. Maybe somebody will find even more Pythonic way.

Sources and some links in addition:

  • where construction: How to find all occurrences of a substring?
  • string splitting: https://www.daniweb.com/programming/software-development/threads/452362/replace-nth-occurrence-of-any-sub-string-in-a-string
  • similar question: Find the nth occurrence of substring in a string

Python - replace every nth occurrence of string

The code you got from the previous question is a nice starting point, and only a minimal adaptation is required to have it change every nth occurence:

def nth_repl_all(s, sub, repl, nth):
find = s.find(sub)
# loop util we find no match
i = 1
while find != -1:
# if i is equal to nth we found nth matches so replace
if i == nth:
s = s[:find]+repl+s[find + len(sub):]
i = 0
# find + len(sub) + 1 means we start after the last match
find = s.find(sub, find + len(sub) + 1)
i += 1
return s

Execute statement every N iterations in Python

How about keeping a counter and resetting it to zero when you reach the wanted number? Adding and checking equality is faster than modulo.

printcounter = 0

# Whatever a while loop is in Python
while (...):
...
if (printcounter == 1000000):
print('Progress report...')
printcounter = 0
...
printcounter += 1

Although it's quite possible that the compiler is doing some sort of optimization like this for you already... but this may give you some peace of mind.

Iterate over every nth element in string in loop - python

If you want to do something every nth step, and something else for other cases, you could use enumerate to get the index, and use modulus:

sample = "This is a string"
n = 3 # I want to iterate over every third element
for i,x in enumerate(sample):
if i % n == 0:
print("do something with x "+x)
else:
print("do something else with x "+x)

Note that it doesn't start at 1 but 0. Add an offset to i if you want something else.

To iterate on every nth element only, the best way is to use itertools.islice to avoid creating a "hard" string just to iterate on it:

import itertools
for s in itertools.islice(sample,None,None,n):
print(s)

result:

T
s
s

r
g


Related Topics



Leave a reply



Submit