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40個你可能不知道的Python技巧附代碼

2020-02-15 21:28:45
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1、拆箱

>>> a, b, c = 1, 2, 3>>> a, b, c(1, 2, 3)>>> a, b, c = [1, 2, 3]>>> a, b, c(1, 2, 3)>>> a, b, c = (2 * i + 1 for i in range(3))>>> a, b, c(1, 3, 5)>>> a, (b, c), d = [1, (2, 3), 4]>>> a1>>> b2>>> c3>>> d4

2、使用拆箱進行變量交換

>>> a, b = 1, 2>>> a, b = b, a>>> a, b(2, 1)

3、擴展的拆箱(Python 3支持)

>>> a, *b, c = [1, 2, 3, 4, 5]>>> a1>>> b[2, 3, 4]>>> c5

4、負數索引

>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]>>> a[-1]10>>> a[-3]8

5、列表切片(a[start:end])

>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]>>> a[2:8][2, 3, 4, 5, 6, 7]

6、負數索引的列表切片

>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]>>> a[-4:-2][7, 8]

7、帶步數的列表切片(a[start:end:step])

>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]>>> a[::2][0, 2, 4, 6, 8, 10]>>> a[::3][0, 3, 6, 9]>>> a[2:8:2][2, 4, 6]

8、負數步數的列表切片

>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]>>> a[::-1][10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]>>> a[::-2][10, 8, 6, 4, 2, 0]

9、列表切片賦值

>>> a = [1, 2, 3, 4, 5]>>> a[2:3] = [0, 0]>>> a[1, 2, 0, 0, 4, 5]>>> a[1:1] = [8, 9]>>> a[1, 8, 9, 2, 0, 0, 4, 5]>>> a[1:-1] = []>>> a[1, 5]

10、切片命名(slice(start, end, step))

>>> a = [0, 1, 2, 3, 4, 5]>>> LASTTHREE = slice(-3, None)>>> LASTTHREEslice(-3, None, None)>>> a[LASTTHREE][3, 4, 5]

11、遍歷列表索引和值(enumerate)

>>> a = ["Hello", "world", "!"]>>> for i, x in enumerate(a):... print "{}: {}".format(i, x)...0: Hello1: world2: !

12、遍歷字典的KEY和VALUE(dict.iteritems)

>>> m = {"a": 1, "b": 2, "c": 3, "d": 4}>>> for k, v in m.iteritems():... print "{}: {}".format(k, v)...a: 1c: 3b: 2d: 4 # 注意:Python 3中要使用dict.items

13、壓縮 & 解壓列表和可遍歷對象

>>> a = [1, 2, 3]>>> b = ["a", "b", "c"]>>> z = zip(a, b)>>> z[(1, "a"), (2, "b"), (3, "c")]>>> zip(*z)[(1, 2, 3), ("a", "b", "c")]

14、使用zip分組相鄰列表項

>>> a = [1, 2, 3, 4, 5, 6] >>> # Using iterators>>> group_adjacent = lambda a, k: zip(*([iter(a)] * k))>>> group_adjacent(a, 3)[(1, 2, 3), (4, 5, 6)]>>> group_adjacent(a, 2)[(1, 2), (3, 4), (5, 6)]>>> group_adjacent(a, 1)[(1,), (2,), (3,), (4,), (5,), (6,)]  >>> # Using slices>>> from itertools import islice>>> group_adjacent = lambda a, k: zip(*(islice(a, i, None, k) for i in range(k)))>>> group_adjacent(a, 3)[(1, 2, 3), (4, 5, 6)]>>> group_adjacent(a, 2)[(1, 2), (3, 4), (5, 6)]>>> group_adjacent(a, 1)[(1,), (2,), (3,), (4,), (5,), (6,)]            
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