Checking if first letter of string is in uppercase

Question

I want to create a function that would check if first letter of string is in uppercase. This is what I've came up with so far:

def is_lowercase(word):
    if word[0] in range string.ascii_lowercase:
        return True
    else:
        return False

When I try to run it I get this error:

    if word[0] in range string.ascii_lowercase
                             ^
SyntaxError: invalid syntax

Can someone have a look and advise what I'm doing wrong?

`range` is a builtin function that returns a range of numbers, and has no place whatsoever in that code. — Wooble, Sep 08 '11 at 20:22
avoid the pattern `if [bool]: return True: else return False`. you can simply use `return [bool]` — wim, Sep 09 '11 at 02:10
It's just simply `if word[0] in string.ascii_lowercase`, there is no need for range. — David Callanan, Apr 06 '18 at 09:21

score 81 · Accepted Answer · edited Jun 11 '18 at 16:05

81

Why not use str.isupper();

In [2]: word = 'asdf'   
In [3]: word[0].isupper()
Out[3]: False

In [4]: word = 'Asdf'   
In [5]: word[0].isupper()
Out[5]: True

edited Jun 11 '18 at 16:05

Xantium

answered Sep 08 '11 at 20:22

AlG

This is a good answer but I was looking to learn what I did wrong in the code I wrote rather than just look for a different solution to the same problem. Thank you anyway. – Blücher Sep 08 '11 at 20:48
@gameFace Good point. Sometimes the best solution is a different solution. ;) – AlG Sep 09 '11 at 11:41
@Blücher In that case, I suggest `Code Review`, instead of `Stack Overflow`. – Ryan Jun 10 '18 at 05:28
one way could be is to use split. if you are sure there would be multiple words with space as separator then this might be a solution: `w="Hello world w.split()[0].istitle()` – Swapnil B. Apr 30 '21 at 01:24

score 35 · Answer 2 · answered Sep 08 '11 at 20:21

35

This is built-in for strings:

word = "Hello"
word.istitle() # True

but note that str.istitle looks whether every word in the string is title-cased, so this might give you a surprise:

"Hello world".istitle() # returns False!

If you just want to check the very first character of a string use this:

word = "Hello world"
word[0].isupper() # True

answered Sep 08 '11 at 20:21

orlp

1

Visceral reaction: in the original post, before you edited, you didn't make it clear that phrases like "Hello world" and "HELlo" fail. istitle checks that every word satisfies the format of , and it wasn't clear if that was a restriction given by the OP. Rolled back downvote – Foo Bah Sep 08 '11 at 20:26
Thank you, I've seen this but wanted to create something myself rather than use istitle(). – Blücher Sep 08 '11 at 20:26
1

@gameFace: Why? If this is for an excersice then I doubt it's usefulness, but if it's for real code than it's downright bad. – orlp Sep 08 '11 at 20:27
2

@nightcracker It's for an exercise, I feel I learn more by creating something new (even if it's usefulness is doubtful) rather than by blindly using ready made solutions - at least at the stage I'm right now. – Blücher Sep 08 '11 at 20:35
1

Never use `istitle()` function. It can lead to strange hard to find bug. For example try it `"Father's".istitle()` and then imagine that you have to debug the wrong behavior without knowing what this is the problem. – Salvador Dali Sep 11 '15 at 09:59

score -1 · Answer 3 · answered Sep 08 '11 at 20:21

-1

The syntax error stems from the fact that you need parentheses:

range(string.ascii_lowercase)

But in fact you shouldn't use range. It's as simple as:

if word[0] in string.ascii_lowercase

answered Sep 08 '11 at 20:21

Foo Bah

2

Testing if it is in `ascii_lowercase` could lead to localization problems down the road -- better to use isupper(). For example, what if the word is in Cyrillic? The first character would be upper case, but it would not be in `ascii_lowercase`. – Alex Martini Sep 08 '11 at 20:25
1

`string.ascii_lowercase` is not an integer, and isn't a valid argument to `range()`. – Wooble Sep 08 '11 at 20:26
@Wooble explaining why the syntax error arose in the first place. Hence I wrote "you shouldn't use range" afterwards, to make sure I wasn't advocating using that solution – Foo Bah Sep 08 '11 at 20:27

3 Answers3