Python: Convert a given string to camelcase
Python String: Exercise-96 with Solution
From Wikipedia,
Camel case (sometimes stylized as camelCase or CamelCase; also known as camel caps or more formally as medial capitals) is the practice of writing phrases without spaces or punctuation, indicating the separation of words with a single capitalized letter, and the first word starting with either case.
Write a Python program to convert a given string to camelcase.
- Use re.sub() to replace any - or _ with a space, using the regexp r"(_|-)+".
- Use str.title() to capitalize the first letter of each word and convert the rest to lowercase.
- Finally, use str.replace() to remove spaces between words.
Sample Solution:
Python Code:
from re import sub
def camel_case(s):
s = sub(r"(_|-)+", " ", s).title().replace(" ", "")
return ''.join([s[0].lower(), s[1:]])
print(camel_case('JavaScript'))
print(camel_case('Foo-Bar'))
print(camel_case('foo_bar'))
print(camel_case('--foo.bar'))
print(camel_case('Foo-BAR'))
print(camel_case('fooBAR'))
print(camel_case('foo bar'))
Sample Output:
javascript fooBar fooBar foo.Bar fooBar foobar fooBar
Flowchart:

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Next: Write a Python program to convert a given string to snake case.
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Python: Tips of the Day
Check if a given key already exists in a dictionary:
In is the intended way to test for the existence of a key in a dict.
d = {"key1": 10, "key2": 23} if "key1" in d: print("this will execute") if "nonexistent key" in d: print("this will not")
If you wanted a default, you can always use dict.get():
d = dict() for i in range(100): key = i % 10 d[key] = d.get(key, 0) + 1
and if you wanted to always ensure a default value for any key you can either use dict.setdefault() repeatedly or defaultdict from the collections module, like so:
from collections import defaultdict d = defaultdict(int) for i in range(100): d[i % 10] += 1
but in general, the in keyword is the best way to do it.
Ref: https://bit.ly/2XPMRyz
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