Writing a Python function to check whether a string is pangram or not

You can use set for it.

A set object is an unordered collection of distinct hashable objects.

import stringdef is_pangram(s, alphabet=string.ascii_lowercase):return set(alphabet) == set(s.replace(" ", "").lower())text = "The quick brown fox jumps over the lazy dog"assert is_pangram(text) is Truetext = "Some text for testing"assert is_pangram(text) is False

 import stringalphabet = set(string.ascii_lowercase) def ispangram(str):return not set(alphabet) - set(str)string = 'the quick brown fox jumps over the lazy dog'if(ispangram(string) == True): print("Yes") else: print("No") 

Your strategy will fail. For each character in the string, "combine", you are checking if you have seen that character before. A pangram can have repeated characters. What you want to do is check that the input string "s" has each character of the alphabet. The following code is simple and fast and demonstrates two very useful tools: string and set.

import stringdef is_pangram(s):s = s.lower().replace(" ","")return set(s.lower()) >= set(string.ascii_lowercase)

def is_panagram(sentence):alp='abcdefghijklmnopqrstuvwxyz' # define a alp with all alphabets(alp)for letter in alp: # checking every letter is present in alp.if letter in sentence.lower(): # string.lower to avoid case differencereturn True # return only if all alp is present in sentencereturn False # return false if it fails

Panagram- Where every word in the alphabets are present in the given sentence.

create a alphabet strings (alp)Get the run-time input with function(sentence)check every letter in alp is present in the given sentence after converting to lower using string.lower()return True,False-- accordingly

I'm new,code in reference to geeksforgeeks.org .. happy to learn

Here's my contribution.

print("Yes" if len(set(filter(lambda x: x.isalpha(), input().lower()))) == 26 else "No")

import stringdef ispangram(str1, alphabet=string.ascii_lowercase): list1 = []characters = str1.split() x = ''.join(characters) #to remove the spaces from the stringfor i in range(len(x)) : if x[i] in alphabet :if not x[i] in list1 :list1.append(x[i])if len(list1) == len(alphabet):return True return False

The check for equality would work as essentially what you are checking is whether all letters from the alphabet are found in the string. No matter whether you have got duplicates.

 def isPangram(text, alphabet = string.ascii_lowercase):modified = text.replace(" ", "").lower()return set(alphabet) == set(modified)

Try this:

import stringdef is_pangram(sentence):return set(string.ascii_lowercase) <= set(sentence.lower())

import stringimport redef pangram(s):s = s.lower() # convert the string to lowercases = re.compile('[^a-z]').sub('',s) # string after removing non English alphabetsreturn len(set(s)) == len(string.ascii_lowercase)

I hope this solves your problem.

Here we convert the string to lower case and then eliminates all the unwanted characters. then finds out the unique values and check if the length is same as the number of characters in English alphabet.

# Using Set print("Pangram" if len(set([ord(s[i])-ord('a') for i in range(len(s)) if s[i]!=' ']))==26 else "Not Pangram")```

import redef pangram(s):s = s.lower()s = re.compile('[^a-z]').sub('',s) # string after removing non English alphabetsreturn len(set(s)) == 26

Try this code.Here we convert the string to lower case and then eliminates all the unwanted characters.then finds out the unique values and check if the length is same as the number of characters in English alphabet.