Write an expression whose value is true if and only if x is NOT a letter

Question

Assume that x is a string variable that has been given a value. Write an expression whose value is true if and only if x is NOT a letter.

I tried notx.true() recently and it says Remarks: ⇒ Unexpected identifiers: notx, true More Hints: ⇒ You almost certainly should be using: not ⇒ You almost certainly should be using: x — Kelly Edwards, Oct 08 '15 at 00:32
`x != 'a'` tests if `x` is not a. `x != 'a' and x != 'b'` tests if it is neither 'a' nor 'b' ... This should be your first attempt. Then you learn about the `in` operator and/or comparisons. Then you might deviate from the right path and learn regular expressions. — Kijewski, Oct 08 '15 at 00:50

LetzerWille · Answer 1 · 2015-10-08T00:59:16.427

3

def isLetter(ch):
    import string
    return len(ch) == 1 and ch in string.ascii_letters


print(isLetter('A'))
True

edited Oct 08 '15 at 00:59

answered Oct 08 '15 at 00:55

LetzerWille

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*Write an expression whose value is true if and only if x is NOT a letter*, this is the most roundabout way possible to do that. – Padraic Cunningham Oct 08 '15 at 22:53

score 2 · Answer 2 · edited May 23 '17 at 11:52

If you want to check the type of the variable x, you can use the following:

if type(x) is str:
    print 'is a string'

In python, String and Char will have the same type and same output unlike languages like java.

type(chr(65)) == str
type('A') == str

EDIT:

As @Kay suggested, you should use isinstance(foo, Bar) instead of type(foo) is bar since isinstance is checking for inheritance while type does not.

See this for more details about isinstance vs type

Using isinstance will also support unicode strings.

isinstance(u"A", basestring)
>>> true

# Here is an example of why isinstance is better than type
type(u"A") is str
>>> false
type(u"A") is basestring
>>> false
type(u"A") is unicode
>>> true

EDIT 2:

Using regex to validate ONLY ONE letter

import re

re.match("^[a-zA-Z]$", "a") is not None
>>> True

re.match("^[a-zA-Z]$", "0") is not None
>>> False

Don't do `type(foo) is Bar`. Use `isinstance(foo, Bar)` instead. — Kijewski, Oct 08 '15 at 00:46

score 2 · Answer 3 · edited Oct 08 '15 at 22:52

2

Turns out the answer to it is not((x>='A' and x<='Z') or (x>='a' and x<='z'))

edited Oct 08 '15 at 22:52

Padraic Cunningham

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answered Oct 08 '15 at 01:00

Kelly Edwards

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Actually Python allows you to write comparisons like a mathematical notation: `a < b < c` <-> `a < b and b < c`, but `a`, `b` and `c` are only evaluated once. – Kijewski Oct 08 '15 at 01:02
Thank you for telling me that. I will have to try it out – Kelly Edwards Oct 08 '15 at 01:03
No need to overly complicate it `return not x.isalpha()`, it if it is not alphabetic it cannot be a letter – Padraic Cunningham Oct 08 '15 at 22:52

Ryan Holleran · Answer 4 · 2015-10-08T00:50:46.367

0

Use regular expression. http://www.pythex.org is a good learning spot, official docs here: https://docs.python.org/2/library/re.html

something like this should work though:

if x != '[a-zA-Z]':

edited Oct 08 '15 at 00:50

answered Oct 08 '15 at 00:39

Ryan Holleran

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This is not Perl. Also digit != letter. – Kijewski Oct 08 '15 at 00:45

Write an expression whose value is true if and only if x is NOT a letter

4 Answers4