How to downcase the first character of a string in Python ?

Question

Hello,

There is a function to capitalize a string, I would like to be able to change the first character of a string to be sure it will be lowercase.

How can I do that in Python ?

Answer 1

def first_lower(s):
   if len(s) == 0:
      return s
   else:
      return s[0].lower() + s[1:]

print first_lower("HELLO")  # Prints "hELLO"
print first_lower("")       # Doesn't crash  :-)

Answer 2

Simplest way:

>>> mystring = 'ABCDE'
>>> mystring[0].lower() + mystring[1:]
'aBCDE'
>>> 

Update

See this answer (by @RichieHindle) for a more foolproof solution, including handling empty strings. That answer doesn't handle None though, so here is my take:

>>> def first_lower(s):
   if not s: # Added to handle case where s == None
   return 
   else:
      return s[0].lower() + s[1:]

>>> first_lower(None)
>>> first_lower("HELLO")
'hELLO'
>>> first_lower("")
>>> 

Answer 3

s = "Bobby tables"
s = s[0].lower() + s[1:]

Answer 4

Interestingly, none of these answers does exactly the opposite of capitalize(). For example, capitalize('abC') returns Abc rather than AbC. If you want the opposite of capitalize(), you need something like:

def uncapitalize(s):
  if len(s) > 0:
    s = s[0].lower() + s[1:].upper()
  return s

Answer 5

I'd write it this way:

def first_lower(s):
    if s == "":
        return s
    return s[0].lower() + s[1:]

This has the (relative) merit that it will throw an error if you inadvertently pass it something that isn't a string, like None or an empty list.

Answer 6

No need to handle special cases (and I think the symmetry is more Pythonic):

def uncapitalize(s):
    return s[:1].lower() + s[1:].upper()

Answer 7

One-liner which handles empty strings and None:

func = lambda s: s[:1].lower() + s[1:] if s else ''

>>> func(None)
>>> ''
>>> func('')
>>> ''
>>> func('MARTINEAU')
>>> 'mARTINEAU'