How to check if string is a pangram?

Question

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

I wrote the following code, which works, but I am looking for an alternative way to do it, hopefully a shorted way.

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 like this question might be against the rules of this website but hopefully it isn't. I am just curious and would like to see alternative ways to do this.

I didn't think of that. In the code above, I delete all whitespace, so I would do the same to punctuation. — , Jul 16 '14 at 03:29

TessellatingHeckler · Accepted Answer · 2016-11-26T04:02:39.073

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

Test string s is a pangram if we start with the alphabet, remove every letter found in the test string, and all the alphabet letters get removed.

Explanation

The use of 'lambda' is a way of creating a function, so it's a one line equivalent to writing a def like:

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

set() creates a data structure which can't have any duplicates in it, and here:

The first set is the (English) alphabet letters, in lowercase
The second set is the characters from the test string, also in lowercase. And all the duplicates are gone as well.

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 away from the alphabet
If there's nothing left, then the test string contained all the letters of the alphabet and must be a pangram.
If there's something leftover, then the test string did not contain all the alphabet letters, so it must not be a pangram.
any spaces, punctuation characters from the test string set were never in the alphabet set, so they don't matter.

set(..) - set(..) will return an empty set, or a set with content. If we force sets into the simplest True/False values in Python, then containers with content are 'True' and empty containers are 'False'.

So we're using not to check "is there anything leftover?" by forcing the result into a True/False value, depending on whether there's any leftovers or not.

not also changes True -> False, and False -> True. Which is useful here, because (alphabet used up) -> an empty set which is False, but we want is_pangram to return True in that case. And vice-versa, (alphabet has some leftovers) -> a set of letters which is True, but we want is_pangram to return False for that.

Then return that True/False result.

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

Wow, this method is great! I still don't understand it at the moment but will read the docs and hopefully understand it. — , Jul 16 '14 at 03:31

dwitvliet · Answer 2 · 2018-09-02T01:01:11.743

4

You can use something as simple as:

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

edited Sep 02 '18 at 01:01

answered Jul 16 '14 at 02:48

dwitvliet

7,242
7
36
62

1

Nice! I had a feeling it could be done much shorter. – Jul 16 '14 at 02:51
By the way, what does `all` do here? – Jul 16 '14 at 02:52
1

`all()` returns `True` if all elements in the list are `True`. See [documentation](https://docs.python.org/2/library/functions.html#all) for definition. – dwitvliet Jul 16 '14 at 02:53
brackets can be removed in the `all()` – monkut Jul 16 '14 at 04:13
@monkut Why? Because of lambda expression? – NAND Jun 09 '20 at 19:38
`all()` without brackets will create/use a generator expression, providing the `[]` will create a list evaluated by all. There are discussions of which is faster in certain cases, but imho if you data is small the bracketless solution is cleaner. – monkut Jun 10 '20 at 00:21

wwii · Answer 3 · 2014-07-16T12:00:33.370

Sets are excellent for membership testing:

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

Strip the whitespace and punctuation from the candidate then test:

>>> 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 it again but add the missing letters:

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

Wheat · Answer 4 · 2018-10-01T02:01:16.397

2

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

edited Oct 01 '18 at 02:01

answered Oct 26 '17 at 21:40

Wheat

31
6

score 1 · Answer 5 · answered Jan 03 '16 at 16:47

How about simply check whether each one of the lowercased alphabet is in the 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 a function:

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

score 1 · Answer 6 · answered Apr 14 '17 at 19:13

1

Here is another definition:

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

answered Apr 14 '17 at 19:13

morevitamins

11
3

score 1 · Answer 7 · answered Sep 17 '18 at 13:16

I came up with the easiest and without using module programe.

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)

Trent Roberts · Answer 8 · 2015-08-28T01:41:18.760

I see this thread is a little old, but I thought I'd throw in my solution 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

score 0 · Answer 9 · answered Nov 19 '17 at 08:02

0

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

answered Nov 19 '17 at 08:02

Parth Shah

1

Care: You write `pharseletter=""`. then `phraseletter= phraseletter +...` next bug! – F. Hauri - Give Up GitHub Nov 19 '17 at 10:57

score 0 · Answer 10 · answered Apr 08 '18 at 23:21

0

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

or more simpler way

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

answered Apr 08 '18 at 23:21

Tatheer Hussain mir

45
9

This is the most perfect answer ! – Hearner Apr 10 '18 at 16:21

score 0 · Answer 11 · answered Jun 19 '18 at 13:41

0

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

answered Jun 19 '18 at 13:41

Alex M.

1

3

Please provide explanation to your code and justify why it works. – chevybow Jun 19 '18 at 13:58

score 0 · Answer 12 · answered Jul 25 '18 at 06:18

0

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

answered Jul 25 '18 at 06:18

ravi tanwar

598
5
16

score 0 · Answer 13 · edited Aug 06 '18 at 22:06

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

Execute the command:

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

Output: "it's pangram." Hint: string.ascii_lowercase returns output

abcdefghijklmnopqrstuvwxyz

score -1 · Answer 14 · answered Dec 13 '15 at 09:37

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");
        }
    }
}

score -1 · Answer 15 · answered Dec 01 '16 at 12:29

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"

How to check if string is a pangram?

15 Answers15

or more simpler way

Linked