删除符合条件的数据框行的一半 [英] Deleting half of dataframe rows which meet condition

问题描述

我正在基于条件提取数据帧的子集.假设

I'm looking to extract a subset of a dataframe based on a condition. Let's say

 df = pd.Dataframe({'Col1': [values1], 'Col2' = [values2], 'Col3' = [values3]})

我想按Col2排序.在Col2中否定的条目(如果有)中，我想删除最大的一半.因此，如果values2 = [-5,10,13，-3，-1，-2]，那么我想删除对应于值-5和-3的行.

I'd like to sort by Col2. Of the entries in Col2 that are negative (if any), I'd like to drop the largest half. So if values2 = [-5,10,13,-3,-1,-2], then I'd want to drop the rows corresponding to the values -5 and -3.

如果我想在排序后只丢弃整个数据帧的一半，我(认为)可以做到

If I wanted to simply drop half the entire dataframe after sorting, I (think) could do

df = df.iloc[(df.shape[0]/2):]

不确定如何引入仅减少负值一半的条件.我的大部分经验是在numpy中使用-仍然习惯于根据数据帧进行思考.预先感谢.

Not sure how to introduce the conditionality of dropping half of only the negative values. The vast majority of my experience is in numpy - still getting used to thinking in terms of dataframes. Thanks in advance.

推荐答案

一种简单的方法，首先，您希望对数据帧进行排序:

A straight-forward approach, first, you wanted your data-frame sorted:

In [16]:  df = pd.DataFrame({'Col1': values1, 'Col2':values2, 'Col3': values3})
In [17]: df
Out[17]:
   Col1  Col2 Col3
0     1    -5    a
1     2    10    b
2     3    13    c
3     4    -3    d
4     5    -1    e
5     6    -2    f

In [18]: df.sort_values('Col2', inplace=True)

In [19]: df
Out[19]:
   Col1  Col2 Col3
0     1    -5    a
3     4    -3    d
5     6    -2    f
4     5    -1    e
1     2    10    b
2     3    13    c

然后，为负值创建一个布尔掩码，使用np.where获取索引，将索引切成一半，然后删除这些索引:

Then, create a boolean mask for the negative values, use np.where to get the indices, cut the indices and half, then drop those indices:

In [20]: mask = (df.Col2 < 0)

In [21]: idx, = np.where(mask)

In [22]: df.drop(df.index[idx[:len(idx)//2]])
Out[22]:
   Col1  Col2 Col3
5     6    -2    f
4     5    -1    e
1     2    10    b
2     3    13    c

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