其实本来是要reverse一下list的,就去查了一下list[::-1]是什么意思,发现还有很多要注意的地方,所以就记一下。
主要是参照https://docs.python.org/3/library/stdtypes.html?highlight=list#list
首先Sequence type有三种
- list
- tuple
- range
slice
[i:j:k]
表示的是slice of s from i to j with step k, 对三种类型都有用
>>> a = [1, 2, 3]
>>> a[::-1]
[3, 2, 1]
>>> a = (1, 2, 3)
>>> a[::-1]
(3, 2, 1)
>>> a = range(3)
>>> a[::-1]
range(2, -1, -1)
range中参数是range(start, stop[, step])
initialize a list
s * n
表示的是n shallow copies of s concatenated
注意是浅拷贝哦,所以会有如下情况
>>> lists = [[]] * 3
>>> lists
[[], [], []]
>>> lists[0].append(3)
>>> lists
[[3], [3], [3]]
如果元素不是对象的话就没关系
>>> lists = [0] * 3
>>> lists
[0, 0, 0]
>>> lists[0] = 1
>>> lists
[1, 0, 0]
正确的初始化嵌套list的方法应该是
>>> lists = [[] for i in range(3)]
>>> lists[0].append(3)
>>> lists[1].append(5)
>>> lists[2].append(7)
>>> lists
[[3], [5], [7]]
concatenation pitfall
(感觉还是英文说的清楚些,这一点跟Java是一样的)
Concatenating immutable sequences always results in a new object. This means that building up a sequence by repeated concatenation will have a quadratic runtime cost in the total sequence length. To get a linear runtime cost, you must switch to one of the alternatives below:
-
str.join()
或者写入io.StringIO
-
bytes.join()
或者写入io.BytesIO
或者使用可变的bytearray
- 使用
list
代替tuple
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