efosky1246:


Bro this might work but it's quite inefficient.why looping thrice...

quite inefficient... see as u talk am self...
No be for FUN? tsk...

If point-and-kill is what u prefer, have it!

def counter(word):
word = list(word.lower())
for i in word:
if (i.isalpha()) or (i.isdigit()):
count = word.count(i)
if (count > 1):
print "{} ={}" %(i,count)
while(word.count(i) > 1):
word.remove(i)

that's all....
code attached...

solution in Python(2)
<code>
def counter(word):
numberList = [] # empty list for numbers
alphaList = [] # empty list for alphabets

#sorting num and alpha from word into resp. list
for ch in word:
if ch.isdigit():
numberList.append(ch)
elif ch.isalpha():
alphaList.append(ch.lower()) #converting to lower case since matching is case-INsensitive

#counting and print only num/alpha occurring more than once
for x in numberList:
count = numberList.count(x)
if count >1:
print "{} --> {} ".format(x,count)
while numberList.count(x) > 1 : #weeding tested num
numberList.remove(x)

for i in alphaList:
count = alphaList.count(i)and
if count > 1:
print" {} --> {}".format(i, count)
while alphaList.count(i) > 1: #weeding tested alpha
alphaList.remove(i)
</code>

call function with your string as argument e.g
counter("pEpper62532622")
That's all... can download attachment or visit link below if code is not well formatted...

https://gist.github.com/nny326/93d34e5d63b023d17aa5fa4534a9fb4a#file-duplicatecounter-py

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