如何打印不包含特定字母字符串的单词？

Question

I need to write a program to prompt the user to enter a string of forbidden letters and then print the number of words that don't contain any of them

This is what I have:

fin=open("demo.txt",'r')
letters=str(raw_input("Enter the key you want to exclude"))
def avoid ():   
    for line in fin:
        word = line.strip()
        for l in letters:
            if l not in word: 
                print word          
avoid()

If I use the in keyword it prints all the words that have the particular strings. But it's not working the same when I am using not in .

Answer 1

Something like this should work:

for l in letters:
    if l in word:
        break  # we found the letter in the word, so stop looping
else:
    # This will execute only when the loop above didn't find
    # any of the letters. 
    print(word)

Answer 2

Simplest way is to make the letters a set, split each line into words and sum how many times there were words with no forbidden letters:

with open("demo.txt", 'r') as f:
    letters = set(raw_input("Enter the key you want to exclude"))
    print(sum(not letters.intersection(w) for line in f for w in line.split()))

You could also use str.translate ,checking if the word length has changed after translating:

with open("demo.txt", 'r') as f:
    letters = raw_input("Enter the key you want to exclude")
    print(sum(len(w.translate(None, letters)) == len(w) for line in f for w in line.split()))

If the word is the same length after trying to remove any letters then the word does not contain any letter from letters.

Or using any :

with open("in.txt", 'r') as f:
    letters = raw_input("Enter the key you want to exclude")
    print(sum(not any(let in w for let in letters) for line in f for w in line.split()))

any(let in w for let in letters) will check every letter in letters and see if any letter is in each word, if it finds a forbidden letter it will short circuit and return True or else return False if no letter appeared in the word then move on to the next word .

You cannot use if l in word unless you only have one word per line, you need to split into individual words.

Using your own code you just need to break when you find a letter in the word, else print the word if we looped over all the letters and found no match:

for line in fin:
    word = line.strip()
    for l in letters:
        if l  in word:
            break 
    else:
         print word

The pythonic way to do what you want in your loop would be to use any:

for line in fin:
    word = line.strip()
    if not any(l not in word for l in letters): 
         print word

Which is equivalent to the break/else just a lot nicer.

If you want the sum then you need to keep track as you go:

  total = 0
  for line in fin:
      word = line.strip()
      if not any(l not in word  for l in letters): 
           total += 1
 print(total)

Which is a less efficient way of doing:

print(sum(not any(let in w.rstrip() for let in letters) for word in f))

Answer 3

Just another way.

exclude = set(raw_input("Enter the key you want to exclude"))
with open("demo.txt") as f:
    print len(filter(exclude.isdisjoint, f.read().split()))

Answer 4

This is better:

    forbidden='ab'
    listofstring=['a','ab','acc','aaabb']
    for i in listofstring:
        if not any((c in i) for c in forbidden):
            print i