CODEWITHSUNDEEP
python
meryam
meryam

Reputation: 43

how to know if a word has repeated letters in python?

I'm working on processing words in python and I want to see if a string has 3 or more occurances of the same letter back to back .Is there a better way to do that other than a nested for loop ? example: "helllooooo"

Upvotes: 2

Views: 2079

Answers (8)

Ynjxsjmh
Ynjxsjmh

Reputation: 30032

Idea is to shift the word and compare,

First, create a Pandas Series from the given word

import pandas as pd

s = pd.Series(list(w))

print(s)

0    h
1    e
2    l
3    l
4    l
5    o
6    o
7    o
8    o
9    o
dtype: object

Then you can group by and shift the Series to find continuous char

m = s.groupby(s).filter(lambda g: g.eq(g.shift().bfill()).sum() >= 3)

print(m)

2    l
3    l
4    l
5    o
6    o
7    o
8    o
9    o
dtype: object

If you want to determine the repeated char and repeated times

print(m.value_counts())

o    5
l    3
dtype: int64

If you want to find the location of repeated chars

i = m.groupby(m).apply(lambda g: g.iloc[[0, -1]].index.tolist()).to_dict()

print(i)

{'l': [2, 4], 'o': [5, 9]}

Upvotes: 0

tdelaney
tdelaney

Reputation: 77347

There's a tool for that. itertools.groupby will group by each change in a primary key (in this case, the next letter in the word). Its second term is an iterator that will emit like valued items (consecutive letters).

>>> import itertools
>>> for letter, seq in itertools.groupby("hellloooooooo"):
...     sz = len(list(seq))
...     if sz >= 3:
...             print(letter, sz)
... 
l 3
o 8

You could break out of the loop early if you only care about finding the first sequence, as in

for _, seq in itertools.groupby("hellloooooooo"):
    if len(list(seq)) >= 3:
        has_3 = True
        break
else:
    has_3 = False

Upvotes: 0

Brendan Abel
Brendan Abel

Reputation: 37539

Can be done pretty succinctly with regular expressions

>>> import re
>>> re.findall(r"((?P<s>[a-zA-Z0-9])(?P=s){2,})", "hellllooooooo")
[('llll', 'l'), ('ooooooo', 'o')]

Upvotes: 0

Adon Bilivit
Adon Bilivit

Reputation: 27008

There's probably a better way (there usually is) but if you're looking for any character that repeats at least n times consecutively in a string then:

def has_repeats(s, n=3):
    for i in range(len(s)-n+1):
        if (a := s[i:i+n]).count(a[0]) == n:
            return True
    return False

print(has_repeats('helllooooo'))

Output:

True

The default value for n is 3. You could override this if needed to any number.

If you're looking specifically for a sequence of 3 characters then this might be faster:

def has_3_repeats(s):
    for t in zip(s, s[1:], s[2:]):
        if t.count(t[0]) == 3:
            return True
    return False

Upvotes: 0

Jonah
Jonah

Reputation: 34

here's a function I've made, it should return True if the word has repeated letters, False if it doesn't.

def repeated_letters(word:str):
    count = 0
    for i in range(len(word) -1):
        if word[i] == word[i+1]:
            count += 1
            if count == 2:
                return True
        else:
            count = 0
    return False

Upvotes: -1

JNevill
JNevill

Reputation: 50034

As another option, looping once but checking the next two letters ahead will work, and the loop is smaller since you don't have to check the last two characters AND you can break as soon as you find a hit:

for i in range(0, len(instring)-2):
    if (instring[i] == instring[i+1] == instring[i+2]):
        print(f"{instring[i]} occurs three or more times.")
        break

Optionally, if you need to know the value or position in the string where these sequences appear, you could generate a list that holds the index of the letter that start a three-in-a-row sequence:

print([[i,c] for i,c in enumerate(instring) if i < len(instring)-2 and (c == instring[i+1] == instring[i+2])])

This will generate a list of lists with the letter and the position in the word where sequence of 3 is found.

Upvotes: 5

molbdnilo
molbdnilo

Reputation: 66371

You don't need a nested loop - just keep track of the last character you saw and use a counter

def atleast3(s):
    count = 0
    last = None
    for c in s:
        if c == last:
            count += 1
            if count >= 3:
                return True
        else:
            count = 1
        last = c
    return False

Upvotes: 3

WSUN000
WSUN000

Reputation: 177

One loop can work

string = 'hellooo'
res = ''
prev_item = ''
for item in string:
    if item == prev_item:
        res += '1'
    else:
        res += '0'
    prev_item = item

print(res) #'0001011'
   
if '11' in res:
    print('Yes')
else:
    print('No')

Upvotes: 1

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