INTERVIEW QUESTION - Count the frequency of words appearing in a string Using Python

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4 жыл бұрын

INTERVIEW QUESTION - Count the frequency of words appearing in a string Using Python
GitHub Link :- github.com/netsetos/python_co...
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Пікірлер: 45

@khetrabasitechvideo362 2 жыл бұрын

Thanks for quality videos

@sharmishthadevi7637 3 жыл бұрын

Awesome video

@pannagabm2100 6 ай бұрын

Thank you❤

@siddhigolatkar8558 8 ай бұрын

Thank you

@shailesh__learning3848 3 жыл бұрын

Thanks ..

@SumitKumar-ls6bq 2 жыл бұрын

Well, if you want you may not add, "d.keys()", instead you can use only d.

@aminemixes9304 3 жыл бұрын

you can just do this : str = input("Enter a string: ") li=str.split() for val in li: con=li.count(val) print(f" {val} : {con}")

@learnvik 2 жыл бұрын

this doesn't work, it prints the occurrence twice of a single word. like if 'hello' is twice: it will show hello : 2 hello : 2

@kiranhinge22 Жыл бұрын

very easy...it works

@safariyakm 5 ай бұрын

d={} s=str(input("enter a string:" )) sp=s.split() for i in sp: d[i]=sp.count(i) print(d)

@Honey12682 5 ай бұрын

a=input().split(" ") c=dict() for i in a: c[i]=a.count(i) print(c)

@MrMukulpandey 2 жыл бұрын

mam thats great...please keep making this videos as it will help us alot.

@techiewithcamera 11 ай бұрын

My Solution: def f_words(s): dict={} for i in s: if i in dict.keys(): dict[i]+=1 else: dict[i]=1 print(dict)

@rohittiwari7019 9 ай бұрын

I am not able to understand this solution

@sumedhasen4534 3 жыл бұрын

It's a good tutorial. But ma'am can we consider fullstop as a word? Or can any other punctuation marks be considered as word?? please reply

@hbTrendsIn 7 ай бұрын

import re sentence = input("Enter a sentence: ") words = re.findall(r'\b\w+\b', sentence) word_counts = {} for word in words: count = sentence.count(word) word_counts[word] = count word_counts = {word: count for word, count in word_counts.items() if count > 1} print(word_counts)

@shantanuwanare5486 2 жыл бұрын

To count frequency of letters : - def frequency_wor(): s = input("Enter the Letter : ") d = {} for i in s: if i in d: d[i] = d[i] + 1 else: d[i] = 1 print(d) frequency_wor()

@necronlord52 6 ай бұрын

Yes, this one seems more elegant. Key initialization is an overkill even for beginners.

@kamleshnizare4633 3 ай бұрын

s = input('Enter your string') output = { key:s.count(key) for key in s.split() }

@shantanuwanare5486 2 жыл бұрын

def frequency_wor(): str = input("Enter the words : ") sp = str.split() d = {} for i in sp: if i in d: d[i] = d.get(i) + 1 else: d[i] = 1 print(d) frequency_wor()

@Kumar-tv8vp 3 ай бұрын

Nice bro

@PerfectBlue1412 3 жыл бұрын

yeah I really like the tutorial but what is the intro for?

@ideatv2389 7 ай бұрын

details explanation before go into short way.

@jekhaarchana8207 3 жыл бұрын

How to print this without colan?? I need to print as sheela1loves2mango2.... and so on..

@harishparamathmananda1385 5 ай бұрын

#Let's split this words based on space and store it in a list Ls = Sentence.split(' ') #remove the dot and other symbols Found_words = [] Repeated_times = [] for word in ls: if word not in Found_words: Found_words.append(word) Repeated_times.append(1) else: Idx= Found_words.index(word) Repeated_times[Idx] = Repeated_times[Idx] + 1 # now accessing the added words and based on the count greater than one, we are printing the result. For index, word in enumerate(Found_words): if Repeated_times[index] > 1: print(f"{Found_words[index]} founded {Repeated_times[index]}") ❤Happy coding 😊

@user-qx8xn7et5l 6 ай бұрын

a = input('enter string:') c={} for i in a: if i in c: c[i] += 1 else: c[i] =1 print(c) this is working in simple way

@shaktipattnayak 2 ай бұрын

You need to split the string before iterating.

@andypandy1ify Жыл бұрын

Two liner: def freq_words(): l=[i.strip(".").strip(".") for i in input("Enter a string: ").split()] for w in set(l): print(f"'{w}' : {l.count(w)}", end=", ")

@necronlord52 6 ай бұрын

Not a Pythonian way - explicit should be preferred over implicit, simple over complicted. It will work, probably. But you shouldn't nest operations without any necessity.

@andypandy1ify 6 ай бұрын

@@necronlord52 I disagree. The fewer the lines of code, the better. The time complexity and the space complexity of my solution would be smaller than a longer solution - my solution solves the problem quicker and uses less resources

@nothymasango8331 8 ай бұрын

def Counting (sentence : str): results = {} string_list = sentence.strip().split() for i in string_list: results[i] = string_list.count(i) return results input = " Nothile loves eating mango apple , banana and aslo apple" print( Counting(input))