从 pandas.core.series.Series 中删除前导零

Question

我有一个带有数据的 pandas.core.series.Series

0    [00115840, 00110005, 001000033, 00116000...
1    [00267285, 00263627, 00267010, 0026513...
2                             [00335595, 00350750]

我想从系列中删除前导零。我试过了

x.astype('int64')

但收到错误信息

ValueError: setting an array element with a sequence.

你能建议我如何在 python 3.x 中做到这一点吗？

Answer 1

如果想要将strings 列表转换为integerss 列表，请使用list comprehension：

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s = pd.Series([[int(y) for y in x] for x in s], index=s.index)\n

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s = s.apply(lambda x: [int(y) for y in x])\n

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样本：

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a = [['00115840', '00110005', '001000033', '00116000'],\n     ['00267285', '00263627', '00267010', '0026513'],\n     ['00335595', '00350750']]\n\ns = pd.Series(a)\nprint (s)\n0    [00115840, 00110005, 001000033, 00116000]\n1      [00267285, 00263627, 00267010, 0026513]\n2                         [00335595, 00350750]\ndtype: object\n\ns = s.apply(lambda x: [int(y) for y in x])\nprint (s)\n0    [115840, 110005, 1000033, 116000]\n1      [267285, 263627, 267010, 26513]\n2                     [335595, 350750]\ndtype: object\n

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编辑：

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如果integer只想要，您可以展平值并转换为ints：

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s = pd.Series([item for sublist in s for item in sublist]).astype(int)\n

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替代解决方案：

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import itertools\ns = pd.Series(list(itertools.chain(*s))).astype(int)\n\nprint (s)\n0     115840\n1     110005\n2    1000033\n3     116000\n4     267285\n5     263627\n6     267010\n7      26513\n8     335595\n9     350750\ndtype: int32\n

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时间安排：

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a = [['00115840', '00110005', '001000033', '00116000'],\n     ['00267285', '00263627', '00267010', '0026513'],\n     ['00335595', '00350750']]\n\ns = pd.Series(a)\ns = pd.concat([s]*1000).reset_index(drop=True)\n

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In [203]: %timeit pd.Series([[int(y) for y in x] for x in s], index=s.index)\n100 loops, best of 3: 4.66 ms per loop\n\nIn [204]: %timeit s.apply(lambda x: [int(y) for y in x])\n100 loops, best of 3: 5.13 ms per loop\n\n#c\xe1\xb4\x8f\xca\x9f\xe1\xb4\x85s\xe1\xb4\x98\xe1\xb4\x87\xe1\xb4\x87\xe1\xb4\x85 sol\nIn [205]: %%timeit\n     ...: v = pd.Series(np.concatenate(s.values.tolist()))\n     ...: v.astype(int).groupby(s.index.repeat(s.str.len())).agg(pd.Series.tolist)\n     ...: \n1 loop, best of 3: 226 ms per loop\n\n#Wen solution\nIn [211]: %timeit pd.Series(s.apply(pd.Series).stack().astype(int).groupby(level=0).apply(list))\n1 loop, best of 3: 1.12 s per loop\n

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扁平化解决方案（@c\xe1\xb4\x8f\xca\x9f\xe1\xb4\x85s\xe1\xb4\x98\xe1\xb4\x87\xe1\xb4\x87\xe1\xb4\x85的想法）：

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In [208]: %timeit pd.Series([item for sublist in s for item in sublist]).astype(int)\n100 loops, best of 3: 2.55 ms per loop\n\nIn [209]: %timeit pd.Series(list(itertools.chain(*s))).astype(int)\n100 loops, best of 3: 2.2 ms per loop\n\n#c\xe1\xb4\x8f\xca\x9f\xe1\xb4\x85s\xe1\xb4\x98\xe1\xb4\x87\xe1\xb4\x87\xe1\xb4\x85 sol\nIn [210]: %timeit pd.Series(np.concatenate(s.values.tolist()))\n100 loops, best of 3: 7.71 ms per loop\n

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