Check if space is in a string

word = ' '
while True:
    if ' ' in word:
        word = raw_input("Please enter a single word: ")
    else:
        print "Thanks"
        break

This is more idiomatic python – comparison against True or False is not necessary – just use the value returned by the expression ' ' in word.

Also, you don’t need to use pastebin for such a small snippet of code – just copy the code into your post and use the little 1s and 0s button to make your code look like code.

You can say word.strip(" ") to remove any leading/trailing spaces from the string – you should do that before your if statement. That way if someone enters input such as " test " your program will still work.

That said, if " " in word: will determine if a string contains any spaces. If that does not working, can you please provide more information?

Write if " " in word: instead of if " " in word == True:.

Explanation:

• In Python, for example a < b < c is equivalent to (a < b) and (b < c).
• The same holds for any chain of comparison operators, which include in!
• Therefore ' ' in w == True is equivalent to (' ' in w) and (w == True) which is not what you want.

There are a lot of ways to do that :

t = s.split(" ")
if len(t) > 1:
  print "several tokens"

To be sure it matches every kind of space, you can use re module :

import re
if re.search(r"s", your_string):
  print "several words"

Use this:

word = raw_input("Please enter a single word : ")
while True:
    if " " in word:
        word = raw_input("Please enter a single word : ")
    else:
        print "Thanks"
        break

You can try this, and if it will find any space it will return the position where the first space is.

if mystring.find(' ') != -1:
    print True
else:
    print False

# The following would be a very simple solution.

print("")
string = input("Enter your string :")
noofspacesinstring = 0
for counter in string:
    if counter == " ":
       noofspacesinstring += 1
if noofspacesinstring == 0:
   message = "Your string is a single word" 
else:
   message = "Your string is not a single word"
print("")   
print(message)   
print("")

You can use the ‘re’ module in Python 3.
If you indeed do, use this:

re.search('s', word)

This should return either ‘true’ if there’s a match, or ‘false’ if there isn’t any.

def word_in(s):
   return " " not in s 

You can see whether the output of the following code is 0 or not.

'import re
x='  beer   '
len(re.findall('s', x))

You mentioned whitespace in general, rather than just spaces. I stumbled upon a solution with isidentifier. Per W3 schools:

A string is considered a valid identifier if it only contains alphanumeric letters (a-z) and (0-9), or underscores (_). A valid identifier cannot start with a number, or contain any spaces.

So, if this matches your requirements, isidentifier is quick and easy to use.

Somebody mentioned efficiency of regex, and I was curious:

import timeit

setup='import re; rs="s"; rc=re.compile(rs); s="applebananacanteloupe"'
stm1='re.search(rs,s)'
stm2='re.search(rc,s)'
stm3='" " in s'
stm4='s.isidentifier()'

timeit.repeat(stm1,setup)
# result: [0.9235025509842671, 0.8889087940042373, 0.8771460619755089, 0.8753634429886006, 1.173506731982343]

timeit.repeat(stm2,setup)
# results: [1.160843407997163, 1.1500899779784959, 1.1857644470001105, 1.1485740720236208, 1.2856045850203373]
# compiled slower than uncompiled? Hmm, I don't get regex...

timeit.repeat(stm3,setup)
# [0.039073383988579735, 0.03403249100665562, 0.03481135700712912, 0.034628107998287305, 0.03392893000273034]

timeit.repeat(stm4,setup)
# [0.08866660299827345, 0.09206177099258639, 0.08418851799797267, 0.08478381999884732, 0.09471498697530478]

So, isidentifier is almost as fast as in, and 10x faster than regex. Note that there is technically no guarantee that python’s idea of what an identifier is won’t change – but it’s also likely that if it did, your code would need some rework anyway.