“可渗透"在数据帧切片中返回索引的方法 [英] "Pandorable" way to return index in dataframe slicing
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问题描述
是否有一种可取的方法来仅获取数据帧切片中的索引? 换句话说,是否有更好的方法来编写以下代码:
Is there a pandorable way to get only the index in dataframe slicing? In other words, is there a better way to write the following code:
df.loc [df ['A']> 5] .index
df.loc[df['A'] >5].index
谢谢!
推荐答案
是的,最好只过滤索引值,而不是所有DataFrame,然后选择索引:
Yes, better is filter only index values, not all DataFrame and then select index:
#filter index
df.index[df['A'] >5]
#filter DataFrame
df[df['A'] >5].index
性能上也存在差异:
np.random.seed(1245)
df = pd.DataFrame({'A':np.random.randint(10, size=1000)})
print (df)
In [40]: %timeit df.index[df['A'] >5]
208 µs ± 11.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [41]: %timeit df[df['A'] >5].index
428 µs ± 6.42 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [42]: %timeit df.loc[df['A'] >5].index
466 µs ± 40.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
如果性能很重要,请使用numpy
-通过 values
到numpy数组:
If performance is important use numpy
- convert values of index and column by values
to numpy array:
In [43]: %timeit df.index.values[df['A'] >5]
157 µs ± 8.71 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [44]: %timeit df.index.values[df['A'].values >5]
8.91 µs ± 196 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
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