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')
但得到了错误信息
ValueError: setting an array element with a sequence.
你能建议我在python 3.x中怎么做吗?
解决方法:
如果想要将字符串列表转换为整数列表,请使用list comprehension:
s = pd.Series([[int(y) for y in x] for x in s], index=s.index)
s = s.apply(lambda x: [int(y) for y in x])
样品:
a = [['00115840', '00110005', '001000033', '00116000'],
['00267285', '00263627', '00267010', '0026513'],
['00335595', '00350750']]
s = pd.Series(a)
print (s)
0 [00115840, 00110005, 001000033, 00116000]
1 [00267285, 00263627, 00267010, 0026513]
2 [00335595, 00350750]
dtype: object
s = s.apply(lambda x: [int(y) for y in x])
print (s)
0 [115840, 110005, 1000033, 116000]
1 [267285, 263627, 267010, 26513]
2 [335595, 350750]
dtype: object
编辑:
如果只想要整数,你可以将值展平并转换为整数:
s = pd.Series([item for sublist in s for item in sublist]).astype(int)
替代方案:
import itertools
s = pd.Series(list(itertools.chain(*s))).astype(int)
print (s)
0 115840
1 110005
2 1000033
3 116000
4 267285
5 263627
6 267010
7 26513
8 335595
9 350750
dtype: int32
时序:
a = [['00115840', '00110005', '001000033', '00116000'],
['00267285', '00263627', '00267010', '0026513'],
['00335595', '00350750']]
s = pd.Series(a)
s = pd.concat([s]*1000).reset_index(drop=True)
In [203]: %timeit pd.Series([[int(y) for y in x] for x in s], index=s.index)
100 loops, best of 3: 4.66 ms per loop
In [204]: %timeit s.apply(lambda x: [int(y) for y in x])
100 loops, best of 3: 5.13 ms per loop
#cᴏʟᴅsᴘᴇᴇᴅ sol
In [205]: %%timeit
...: v = pd.Series(np.concatenate(s.values.tolist()))
...: v.astype(int).groupby(s.index.repeat(s.str.len())).agg(pd.Series.tolist)
...:
1 loop, best of 3: 226 ms per loop
#Wen solution
In [211]: %timeit pd.Series(s.apply(pd.Series).stack().astype(int).groupby(level=0).apply(list))
1 loop, best of 3: 1.12 s per loop
flatenning的解决方案(@cᴏʟᴅsᴘᴇᴇᴅ的想法):
In [208]: %timeit pd.Series([item for sublist in s for item in sublist]).astype(int)
100 loops, best of 3: 2.55 ms per loop
In [209]: %timeit pd.Series(list(itertools.chain(*s))).astype(int)
100 loops, best of 3: 2.2 ms per loop
#cᴏʟᴅsᴘᴇᴇᴅ sol
In [210]: %timeit pd.Series(np.concatenate(s.values.tolist()))
100 loops, best of 3: 7.71 ms per loop
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