切片 #

[start:end:step]

遍历 #

enumerate #

a = ["a", "b", "c", "d"]

# index & value
for i in xrange(len(a)):
    print i, a[i]

# iterate with index
for i, el in enumerate(a):
    print i, el

# i 从 1 开始
for i, el in enumerate(a, 1):
    print i, el

step 为负数 #

最后一个字符到下标是 -1，即从右到左，是从 -1 开始，然后 -1、-2、-3、-4、-5

字符串反转 #

letter[::-1]

为什么 s[:-1:-1] 是空 #

只需要记住，s[start:end:step] 就行，包括 start，不包括 end，所以，如果要实现：去掉最后一个元素，然后再倒序，应该是：

倒数第二个作为 start，end 默认，step 是 -1
也就是：s[-2::-1]

s[i:j:k]

If i or j is negative, the index is relative to the end of sequence s:
- len(s) + i or len(s) + j is substituted.
- But note that -0 is still 0.
The slice of s from i to j with step k
- is defined as the sequence of items with index x = i + n*k
- such that 0 <= n < (j-i)/k.
the indices are i, i+k, i+2k, i+3k and so on,
- stopping when j is reached (but never including j).
When k is positive, i and j are reduced to len(s) if they are greater.
When k is negative, i and j are reduced to len(s) - 1 if they are greater.
If i or j are omitted or None, they become “end” values (which end depends on the sign of k).
k cannot be zero.
- If k is None, it is treated like 1.
s[:]
- 等同于 s[::]
- 等同于 s[None:None:None]
- 等同于 s[0:len(s):1]
s[::-1]
- 等同于 s[None:None:-1]
- 等同于 s[len(s)-1:-len(s)-1:-1]
  - 包括 start，所以 start 从 len(s)-1 开始, 就是 s[-1]
  - 不包括 end, 所以 end 到 -len(s)-1, 实际上就是到 -len(s), 也就是 s[0]
s[:-1:-1]
- 等同于 s[len(s)-1:-1:-1]
- 等同于 s[len(s)-1:len(s)-1:-1]

参考：

sequence 乘法 #

s * n 或 n * s
- equivalent to adding s to itself n times
  - items in the sequence s are not copied
  - they are referenced multiple times
- Values of n less than 0 are treated as 0
  - which yields an empty sequence of the same type as s

>>> lists = [[]] * 3
>>> lists
[[], [], []]
>>> lists[0].append(3)
>>> lists
[[3], [3], [3]]