如何在python中打印一个没有三分之一字符的字符串？

Question

我正在尝试这个挑战，它给我一个字符串，比如'Python'，我需要删除每三个字符，或者索引可以被3整除的每个字符。在本例中，输出将是yton。到目前为止，每三个字符都有word[::3]部分来查找，但是如何从字符串中删除它们呢？在

代码

word = str(input())
newWord = word[::3]
print(newWord) #for testing

Answer 1

输入字符串是不可变的，但将其转换为列表，您可以对其进行编辑：

>>> word = list(input())  # Read in a word
abcdefghijklmnop
>>> del word[::3]         # delete every third character
>>> ''.join(word)         # join the characters together for the result
'bcefhiklno'

从不同的字符开始：

^{pr2}$

Answer 2

看看这个：

>>> word = 'Python For All'
>>> new_word = ''.join(character for index, character in enumerate(word) if index%3 != 0)
>>> new_word
'ytonFo Al'

Answer 3

我的答案是：

s = '0123456'
print s[::3]
# 036

# first way(understandable) create a new str
s_new = ''
for i in xrange(len(s)):
    if i % 3 != 0:
        s_new += s[i]

print s_new
# 1245

# second way
s_lst = [c if i % 3 else '' for i, c in enumerate(s)]
print s_lst
# ['', '1', '2', '', '4', '5', '']
s_new = ''.join(s_lst)
print s_new
# 1245

# you can put it in single line
s_new = ''.join([c if i % 3 else '' for i, c in enumerate(s)])
print s_new
# 1245

# third way
s_idx = filter(lambda x: x[0] % 3, enumerate(s))
print s_idx
# [(1, '1'), (2, '2'), (4, '4'), (5, '5')]
print ''.join([x[1] for x in s_idx])
# 1245