Python Title Case, but leave a pre-existing uppercase

Question

I am looking for a very pythonic path (Python 3.x) to accomplish the following, but haven't come up with yet. If I have the following line:

string = 'this is a test string'

I can write it with:

string.title()

Result:

'This Is A Test String'

However, I want to convert if I have the following line:

string = 'Born in the USA'

Using the sample header results in:

string = 'Born in the USA'

Should the results:

'Born In The USA'

I am looking for a way to make a headline but not adjust the existing text in uppercase. Is there any way to do this?

Answer 1

It is not clear what result you expect.

If you want to ignore the entire line because it contains uppercase words, check if the line is the first line of the string:

if string.islower():
    string = string.title()

, , , :

string = ' '.join([w.title() if w.islower() else w for w in string.split()])

:

>>> string = 'Born in the USA'
>>> ' '.join([w.title() if w.islower() else w for w in string.split()])
'Born In The USA'

Answer 2

:

string = 'born in the USA'

title = "".join([a if a.isupper() else b for a,b in zip(string,string.title())])
print title

:

Born In The USA

, , .

Answer 3

Looks like this is what you want:

def smart_title(s):
    return ' '.join(w if w.isupper() else w.capitalize() for w in s.split())

You can use it, as in the following example:

>>> smart_title('Born in the USA')
'Born In The USA'

Answer 4

There is no built-in method, but easy enough to simulate:

def cap(word):
    return word[0].upper() + word[1:]

def title_preserving_caps(string):
    return " ".join(map(cap, string.split(' ')))

string = title_preserving_caps('Born in the USA')

Answer 5

Here's a quickie, just ASCII.

In [3]: data = "in UK"

In [4]: titled = data.title()

In [5]: titled
Out[5]: 'In Uk'

In [8]: "".join(map(min, zip(data, titled)))
Out[8]: 'In UK'