How to print a string in alternating case?

Question

I want to print a string in Python with alternate cases. For example my string is "Python". I want to print it like "PyThOn". How can I do this?

string = "Python" 
for i in string: 
  if (i%2 == 0): 
    (string[i].upper()) 
  else: 
    (string[i].lower()) 
print (string)

Answer 1

It's simply not Pythonic if you don't manage to work a zip() in there somehow:

string = 'Pythonic'

print(''.join(x + y for x, y in zip(string[0::2].upper(), string[1::2].lower())))

OUTPUT

PyThOnIc

Answer 2

mystring="Python"
newstring=""
odd=True
for c in mystring:
  if odd:
    newstring = newstring + c.upper()
  else:
    newstring = newstring + c.lower()
  odd = not odd
print newstring

Answer 3

For random caps and small characters

>>> def test(x):
...    return [(str(s).lower(),str(s).upper())[randint(0,1)] for s in x]
... 
>>> print test("Python")
['P', 'Y', 't', 'h', 'o', 'n']
>>> print test("Python")
['P', 'y', 'T', 'h', 'O', 'n']
>>> 
>>> 
>>> print ''.join(test("Python"))
pYthOn
>>> print ''.join(test("Python"))
PytHon
>>> print ''.join(test("Python"))
PYTHOn
>>> print ''.join(test("Python"))
PytHOn
>>> 

For Your problem code is :

st = "Python"

out = ""
for i,x in enumerate(st):
    if (i%2 == 0):
        out += st[i].upper()
    else:
        out += st[i].lower()
print out

Answer 4

Try it:

def upper(word, n):
    word = list(word)
    for i in range(0, len(word), n):
        word[i] = word[i].upper()
    return ''.join(word)

Answer 5

You can iterate using list comprehension, and force case depending on each character's being even or odd.

example:

s = "This is a test string"
ss = ''.join([x.lower() if i%2 else x.upper() for i,x in enumerate(s)])
print ss

s = "ThisIsATestStringWithoutSpaces"
ss = ''.join([x.lower() if i%2 else x.upper() for i,x in enumerate(s)])
print ss

output:

 ~/so_test $ python so_test.py 
ThIs iS A TeSt sTrInG
ThIsIsAtEsTsTrInGwItHoUtSpAcEs
 ~/so_test $