用Python中另一个字符串的第n个字符创建一个字符串

Question

I would like to split a string with the result being a string consisting only of every nth character of the original string. My first approach looks like this (and works):

#Divide the cipher text in the estimated number of columns
for i in range(0,keyLengthEstimator):
    cipherColumns.append(cipherstring[i])

for i in range(keyLengthEstimator,len(cipherstring)):
    if i % keyLengthEstimator == 0:
        cipherColumns[0] += cipherstring[i]
    elif i % keyLengthEstimator == 1:
        cipherColumns[1] += cipherstring[i]
    elif i % keyLengthEstimator == 2:
        cipherColumns[2] += cipherstring[i]
    elif i % keyLengthEstimator == 3:
        cipherColumns[3] += cipherstring[i]
    elif i % keyLengthEstimator == 4:
        cipherColumns[4] += cipherstring[i]

I have the feeling, that there is a more efficient way to do it. In Matlab there would be the reshape function, is there a similar function in Python?

Answer 1

Why not simply:

#Divide the cipher text in the estimated number of columns
for i in range(0,keyLengthEstimator):
    cipherColumns.append(cipherstring[i])

for i in range(keyLengthEstimator,len(cipherstring)):
    cipherColumns[i % keyLengthEstimator] += cipherstring[i]

Answer 2

You could slice the string N times, specifying a step of N for each one:

>>> s = "Hello I am the input string"
>>> columns = 5
>>> seq = [s[i::columns] for i in range(columns)]
>>> print seq
['H  i n', 'eItnsg', 'l hpt', 'laeur', 'om ti']
>>> print "\n".join(seq)
H  i n
eItnsg
l hpt
laeur
om ti