时间:2021-07-01 10:21:17 帮助过:89人阅读
首先Sequence type有三种
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: