Python 每 n 行熊猫
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/25055712/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Pandas every nth row
提问by Mixel
Dataframe.resample() works only with timeseries data. I cannot find a way of getting every nth row from non-timeseries data. What is the best method?
Dataframe.resample() 仅适用于时间序列数据。我找不到从非时间序列数据中获取每第 n 行的方法。最好的方法是什么？
采纳答案by chrisb
I'd use iloc, which takes a row/column slice, both based on integer position and following normal python syntax.
我会使用iloc，它需要一个行/列切片，都基于整数位置和遵循正常的 python 语法。
df.iloc[::5, :]
回答by metastableB
Though @chrisb's accepted answer does answer the question, I would like to add to it the following.
虽然@chrisb 接受的答案确实回答了这个问题，但我想添加以下内容。
A simple method I use to get the nthdata or drop the nthrow is the following:
我用来获取nth数据或删除nth行的简单方法如下：
df1 = df[df.index % 3 != 0] # Excludes every 3rd row starting from 0
df2 = df[df.index % 3 == 0] # Selects every 3rd raw starting from 0
This arithmetic based sampling has the ability to enable even more complex row-selections.
这种基于算法的采样能够实现更复杂的行选择。
This assumes, of course, that you have an indexcolumn of ordered, consecutive, integersstarting at 0.
当然，这假设您有一index列从 0 开始的有序、连续的整数。
回答by Steztric
I had a similar requirement, but I wanted the n'th item in a particular group. This is how I solved it.
我有类似的要求，但我想要特定组中的第 n 个项目。我就是这样解决的。
groups = data.groupby(['group_key'])
selection = groups['index_col'].apply(lambda x: x % 3 == 0)
subset = data[selection]
回答by cs95
There is an even simpler solution to the accepted answer that involves directly invoking df.__getitem__.
对于包含直接调用df.__getitem__.
df = pd.DataFrame('x', index=range(5), columns=list('abc'))
df
a b c
0 x x x
1 x x x
2 x x x
3 x x x
4 x x x
For example, to get every 2 rows, you can do
例如，要获取每 2 行，您可以执行
df[::2]
a b c
0 x x x
2 x x x
4 x x x
There's also GroupBy.first/GroupBy.head, you group on the index:
还有GroupBy.first/ GroupBy.head，你在索引上分组：
df.index // 2
# Int64Index([0, 0, 1, 1, 2], dtype='int64')
df.groupby(df.index // 2).first()
# Alternatively,
# df.groupby(df.index // 2).head(1)
a b c
0 x x x
1 x x x
2 x x x
The index is floor-divved by the stride (2, in this case). If the index is non-numeric, instead do
索引按步幅（在本例中为 2）进行地板划分。如果索引是非数字的，请改为
# df.groupby(np.arange(len(df)) // 2).first()
df.groupby(pd.RangeIndex(len(df)) // 2).first()
a b c
0 x x x
1 x x x
2 x x x
