python - Python，在第 n 次迭代中打印 n 次字母

Question

我似乎被这个难住了。不幸的是，我什至不知道如何解释我想要什么。对于循环中的 n 次迭代，我想打印一个字母 n 次。这是一些入门代码...

n = 1
max = 3

letters = string.lowecase
letters.split

while n <= max:
    for letter in letters:
        print letter #n times
    n = n + 1

我想结束：

a b ... z aa ... zz aaa ... zzz

Answer 1

字符串可以相乘。

>>> 'foo' * 4
'foofoofoofoo'

Answer 2

for i in range(1, 10):
    for j in "abcdefghijklmnopqrstuvwxyz":
        print j * i

Answer 3

>>> import string
>>> letters = string.ascii_lowercase

>>> print("".join( x*n for n in range(1,4) for x in letters  ))

abcdefghijklmnopqrstuvwxyzaabbccddeeffgghhiijjkkllmmnnooppqqrrssttuuvvwwxxyyzzaaabbbcccdddeeefffggghhhiiijjjkkklllmmmnnnooopppqqqrrrssstttuuuvvvwwwxxxyyyzzz

Answer 4

使用另一个循环：

# Prints the letters
for letter in letters:
        print letter

# Prints each letter 3 times:
for letter in letters:
    for i in xrange(3):
        print letter

Answer 5

Other way of doing

(lambda s: ''.join([x*n for n in xrange(4) for x in s]))(letters)

Answer 6

n = 1
max = 3

letters = string.lowecase
letters.split

while n <= max:
    for letter in letters:
        print letter * n
    n = n + 1

乘以字符串