pandas 基于整数索引拆分数据框
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split data frame based on integer index
提问by user2426361
In pandas how do I split Series/dataframe into two Series/DataFrames where odd rows in one Series, even rows in different? Right now I am using
在Pandas中，如何将系列/数据帧拆分为两个系列/数据帧，其中一个系列中的奇数行，不同的偶数行？现在我正在使用
rng = range(0, n, 2)
odd_rows = df.iloc[rng]
This is pretty slow.
这很慢。
回答by Andy Hayden
Use slice:
使用切片：
In [11]: s = pd.Series([1,2,3,4])
In [12]: s.iloc[::2] # even
Out[12]:
0 1
2 3
dtype: int64
In [13]: s.iloc[1::2] # odd
Out[13]:
1 2
3 4
dtype: int64
回答by Jeff
Here's some comparisions
这是一些比较
In [100]: df = DataFrame(randn(100000,10))
simple method (but I think range makes this slow), but will work regardless of the index (e.g. does not have to be a numeric index)
简单的方法（但我认为范围会使这个变慢），但不管索引如何都可以工作（例如不必是数字索引）
In [96]: %timeit df.iloc[range(0,len(df),2)]
10 loops, best of 3: 21.2 ms per loop
The following require an Int64Indexthat is range based (which is easy to get, just reset_index()).
以下需要一个Int64Index基于范围的（这很容易获得，只是reset_index()）。
In [107]: %timeit df.iloc[(df.index % 2).astype(bool)]
100 loops, best of 3: 5.67 ms per loop
In [108]: %timeit df.loc[(df.index % 2).astype(bool)]
100 loops, best of 3: 5.48 ms per loop
make sure to give it index positions
确保给它索引位置
In [98]: %timeit df.take(df.index % 2)
100 loops, best of 3: 3.06 ms per loop
same as above but no conversions on negative indicies
同上，但没有负指数的转换
In [99]: %timeit df.take(df.index % 2,convert=False)
100 loops, best of 3: 2.44 ms per loop
This winner is @AndyHayden soln; this only works on a single dtype
这位获胜者是@AndyHayden soln；这仅适用于单个 dtype
In [118]: %timeit DataFrame(df.values[::2],index=df.index[::2])
10000 loops, best of 3: 63.5 us per loop
