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day 3 list列表生成式

發布時間:2023/12/13 编程问答 23 豆豆
生活随笔 收集整理的這篇文章主要介紹了 day 3 list列表生成式 小編覺得挺不錯的,現在分享給大家,幫大家做個參考.

1.定義一個list列表,里面元素是0-33

a = []i = 0 while i<33:a.append(i)i+=1print(a) [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]

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2.range (切片)

  1)python2 版本

  range風險:python2 版本中,有時候一次性申請很大的內存,不會給你

#### range 返回list列表 和切片功能相同 In [1]: range(10) Out[1]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]In [2]: range(10,17) Out[2]: [10, 11, 12, 13, 14, 15, 16]In [3]: range(10,17,2) Out[3]: [10, 12, 14, 16] In [5]: range(0,100000)### 運行結果 994,995,996,997,998,999,...]

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In [5]: range(0,1000000000) --------------------------------------------------------------------------- MemoryError Traceback (most recent call last) <ipython-input-5-30124a0b9388> in <module>() ----> 1 range(0,1000000000)MemoryError: ##range風險:python2 版本中,有時候一次性申請很大的內存,不會給你

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  2)python3版本:要一個數字,給你一個,不會全部一次性給

In [1]: range(0,10) Out[1]: range(0, 10)In [2]: range(10) Out[2]: range(0, 10)In [3]: range(0,100000000) Out[3]: range(0, 100000000)

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3.列表生成式

  1)?a = [ i for i in range(0,18) ]

In [6]: a = [i for i in range(0,18)]In [7]: a Out[7]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17]In [8]: a = [22 for i in range(0,18)] #for只負責循環的次數17次 In [9]: a Out[9]: [22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22]

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?  2)?a = ?[ i for i in range(10) if i%2==0 ]

In [10]: a = [i for i in range(10) if i%2==0]In [11]: a Out[11]: [0, 2, 4, 6, 8]

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  3)?d = [ (i,j) for i in range(3) for j in range(2)]

In [15]: d = [ i for i in range(3) for j in range(2)]In [16]: d Out[16]: [0, 0, 1, 1, 2, 2]In [17]: d = [ i,j for i in range(3) for j in range(2)]File "<ipython-input-17-0277977bdeb0>", line 1d = [ i,j for i in range(3) for j in range(2)]^ SyntaxError: invalid syntax
In [
19]: d = [ (i,j) for i in range(3) for j in range(2)]In [20]: d Out[20]: [(0, 0), (0, 1), (1, 0), (1, 1), (2, 0), (2, 1)] #坐標軸

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  4)e = [(i,j,k) for i in range(3) for j in range(2) for k in range(2)]

In [21]: e = [(i,j,k) for i in range(3) for j in range(2) for k in range(2)]In [22]: e Out[22]: [(0, 0, 0),(0, 0, 1),(0, 1, 0),(0, 1, 1),(1, 0, 0),(1, 0, 1),(1, 1, 0),(1, 1, 1),(2, 0, 0),(2, 0, 1),(2, 1, 0),(2, 1, 1)]

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轉載于:https://www.cnblogs.com/venicid/p/7898595.html

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