Python 列表解析

1. 簡單列表解析

假設咱們須要建立一個列表爲:[0,0,0,0,0,0,  0,0,0,  0](size=10)spa

顯然這樣寫0很費勁。因此有一種叫作列表解析的東西能夠快速生成:code

>>> [0 for i in range(10)]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
# 還能夠按序生成
>>> [i for i in range(10)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

2. 帶條件列表解析

假設咱們須要建立一個列表:[0,2,0, 4, 0, 6, 0, 8, 0,  10] (size=10, 奇數爲0,偶數不變)blog

>>> [i+1 if i%2 == 1 else 0 for i in range(10)]
[0, 2, 0, 4, 0, 6, 0, 8, 0, 10]

note: 當條件子句在for前時必須帶上else,此時else表示不符合if條件時列表元素的取值;當條件子句在for後時不能帶上else。否則會報錯!input

>>> [i+1 if i%2 == 1 for i in range(10)]
  File "<input>", line 1
    [i+1 if i%2 == 1 for i in range(10)]  # if在前面時,必須有else
                       ^
SyntaxError: invalid syntax

>>> [i+1 for i in range(10) if i%2 == 1 else 0]  
  File "<input>", line 1
    [i+1 for i in range(10) if i%2 == 1 else 0]  # if在後面時,不能有else
                                           ^
SyntaxError: invalid syntax

3. 多重循環的列表解析

假設咱們須要根據建立一個列表:[(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)](能夠當作(1, 2) 和 (1, 2, 3)的全排列)class

>>> [(i, j) for i in [1, 2] for j in [1, 2, 3]]
[(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)]  # perfect!

矩陣降維或者表量化:循環

>>> matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> array = [i for row in matrix for i in row]
[1, 2, 3, 4, 5, 6, 7, 8, 9]

notes: 注意兩個 for 的前後順序:高維在前!di

4.字典解析

假設咱們想把列表:[(1, 201), (2, 202), (3, 205)],變成字典 {1: 201, 2: 201, 3: 205}co

>>> {k: v for k, v in [(1, 201), (2, 201), (3, 205)]}
{1: 201, 2: 201, 3: 205}  # perfect!

 

任何大神都是從小白當起! 大神

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