How to check if a string is pangram?

Question

I want to create a function that takes a string as input and checks if the string is pangram or not (pangram is a piece of text that contains every letter of the alphabet).

I wrote the following code that works, but I am looking for an alternative way to do this, I hope, briefly.

import string def is_pangram (gram): gram = gram.lower() gram_list_old = sorted([c for c in gram if c != ' ']) gram_list = [] for c in gram_list_old: if c not in gram_list: gram_list.append(c) if gram_list == list(string.ascii_lowercase): return True else: return False 

I feel this question may be against the rules of this website, but hopefully this is not the case. I am just curious and would like to see alternative ways to do this.

Answer 1

 is_pangram = lambda s: not set('abcdefghijklmnopqrstuvwxyz') - set(s.lower()) >>> is_pangram('abc') False >>> is_pangram('the quick brown fox jumps over the lazy dog') True >>> is_pangram('Does the quick brown fox jump over the lazy dog?') True >>> is_pangram('Do big jackdaws love my sphinx of quartz?') True 

The test string s is a pangram, if we start with the alphabet, delete every letter found in the test string, and all letters of the alphabet are deleted.

Explanation

Using lambda is a way to create a function, so one line is equivalent to writing def like:

  def is_pangram(s): return not set('abcdefghijklmnopqrstuvwxyz') - set(s.lower()) 

set() creates a data structure in which there can be no duplicates, and here:

• The first set is the (English) letters of the alphabet, lowercase
• The second set is the characters from the test string, also in lower case. And all the duplicates also disappeared.

Subtracting things like set(..) - set(..) returns the contents of the first set, minus the contents of the second set. set('abcde') - set('ace') == set('bd') .

In this pangram test:

• we take the characters in the test string from the alphabet
• If nothing is left, then the test line contains all the letters of the alphabet and should be pangram.

• If something remains, then the test line does not contain all the letters of the alphabet, so it should not be a pangram.

• any spaces, punctuation marks from the set of test strings were never in the alphabetical set, so they do not matter.

set(..) - set(..) will return an empty set or a set with content. If we insert sets into the simplest True / False values ​​in Python, then containers with the contents of "True" and empty containers of "False".

So, we use not to check "is there anything left?" forcing the result to True / False, depending on whether there are any leftovers or not.

not also changes True → False, and False → True. This is useful here because (the alphabet used) is → an empty set that is False , but we want is_pangram return True in this case. Conversely, (the alphabet has some leftovers) -> a set of letters that is True , but we want is_pangram return False for this.

Then return the result True / False.

 is_pangram = lambda s: not set('abcdefghijklmnopqrstuvwxyz') - set(s.lower()) # Test string `s` #is a pangram if # the alphabet letters # minus # the test string letters # has NO leftovers 

Answer 2

You can use something simple like:

 import string is_pangram = lambda s: all(c in s.lower() for c in string.ascii_lowercase) 

Answer 3

Great membership testing kits:

 >>> import string >>> candidate = 'ammdjri * itouwpo ql ? k @ finvmcxzkasjdhgfytuiopqowit' >>> ascii_lower = set(string.ascii_lowercase) 

Separate the space and punctuation from the candidate, then check:

 >>> candidate_lower = ascii_lower.intersection(candidate.lower()) >>> ascii_lower == candidate_lower False 

Find out what is missing:

 >>> ascii_lower.symmetric_difference(candidate_lower) set(['b', 'e']) 

Try again, but add the missing letters:

 >>> candidate = candidate + 'be' >>> candidate_lower = ascii_lower.intersection(candidate.lower()) >>> ascii_lower == candidate_lower True >>> 

Answer 4

  def pangram(word): return all(chr(c+97) in word for c in range(25)) 

Answer 5

How to simply check if each of the lower alphabet is in a sentence:

 text = input() s = set(text.lower()) if sum(1 for c in s if 96 < ord(c) < 123) == 26: print ('pangram') else: print ('not pangram') 

or in function:

 def ispangram(text): return sum(1 for c in set(text.lower()) if 96 < ord(c) < 123) == 26 

Answer 6

Here is another definition:

 def is_pangram(s): return len(set(s.lower().replace(" ", ""))) == 26 

Answer 7

I see that this thread is a little old, but I thought that I would give up my decision anyway.

 import string def panagram(phrase): new_phrase=sorted(phrase.lower()) phrase_letters = "" for index in new_phrase: for letter in string.ascii_lowercase: if index == letter and index not in phrase_letters: phrase_letters+=letter print len(phrase_letters) == 26 

or for the last line:

  print phrase_letters == string.ascii_lowercase 

Answer 8

Using ASCII aproach:

 def checkPan(s): alphaT= sum([i for i in range(97,123)]) + 32 s = [i.lower() for i in s] s = list(set(s)) sn = sum([ord(i) for i in s]) if sn==alphaT: return("pangram") else: return("not pangram") 

Answer 9

 def panagram(phrase): alphabet="abcdefghiklmnopqrstuvwxyz" pharseletter="" for char in phrase: if char in aphabet: phraseletter= phraseletter + char for char in aplhabet: if char not in phrase: return false 

Answer 10

 import string def ispangram(str, alphabet=string.ascii_lowercase): alphabet = set(alphabet) return alphabet <= set(str.lower()) 

or easier way

 def ispangram(str): return len(set(str.lower().replace(" ", ""))) == 26 

Answer 11

 import string def is_pangram(phrase, alpha=string.ascii_lowercase): num = len(alpha) count=0 for i in alpha: if i in phrase: count += 1 return count == num 

Answer 12

 def panagram(str1): str1=str1.replace(' ','').lower() s=set(str1) l=list(s) if len(l)==26: return True return False str1='The quick brown fox jumps over the dog' q=panagram(str1) print(q) 

true

Answer 13

 import string def ispangram(str1,alphabet=string.ascii.lowercase): for myalphabet in alphabet: if myalphabet not in str1: print(it not pangram) break else: print(it pangram) 

Run the command:

 ispangram("The quick brown fox jumps over the lazy dog") 

Conclusion: "this is a pangram." Hint: string.ascii_lowercase returns output

 abcdefghijklmnopqrstuvwxyz 

Answer 14

I came up with the simplest and not using a program module.

 def checking(str_word): b=[] for i in str_word: if i not in b: b.append(i) b.sort() #print(b) #print(len(set(b))) if(len(set(b))>=26): print(b) print(len(b)) print(" String is pangram .") else: print(" String isn't pangram .") #b.sort() #print(b) str_word=input(" Enter the String :") checking(str_word) 

Answer 15

I am new to Python and coding. and I made my code to check the pangram, and the person who received the error can check what is wrong in my code. thank you in advance

 import string def ispangram(s): al=string.ascii_lowercase q=0 k=[] s.lower() k=list(s) e=[] e=list(al) x=0 for x in k: if k[x] in e and q<27: q=q+1 elif q==26: return True ispangram("The quick brown fox jumps over the lazy dog") 

Answer 16

 import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { String s; char f; Scanner in = new Scanner(System.in); s = in.nextLine(); char[] charArray = s.toLowerCase().toCharArray(); final Set set = new HashSet(); for (char a : charArray) { if ((int) a >= 97 && (int) a <= 122) { f = a; set.add(f); } } if (set.size() == 26){ System.out.println("pangram"); } else { System.out.println("not pangram"); } } } 

Answer 17

 import string import re list_lower= list(string.lowercase); list_upper=list(string.uppercase); list_total=list_lower + list_upper ; def is_panagram(temp): for each in temp: if each not in list_total : return 'true' sample=raw_input("entre the string\n"); string2=re.sub('[^A-Za-z0-9]+', '', sample) ram=is_panagram(string2); if ram =='true': print "sentence is not a panagram" else:`enter code here` print"sentece is a panagram"