根据pandas DataFrame列中的值序列查找行索引 [英] Finding the index of rows based on a sequence of values in a column of pandas DataFrame

问题描述

我有一个DataFrame，其中的一列具有三个唯一的字符串。我需要做的是生成一个包含行索引的列表，该行的索引在良好之后为非常不好，但在不良之后为非常不好。

I have a DataFrame with a column that has three unique character strings. What I need to do is to generate a list containing indexes of rows that has 'very bad' after good, but not 'very bad' after 'bad'.

import random
df = pd.DataFrame({
    'measure': [random.randint(0,10) for _ in range(0,20)],
})

df['status'] = df.apply(
    lambda x: 'good' if x['measure'] > 4 else 'very bad' if x['measure'] < 2  else 'bad',
    axis=1)

    measure    status
0         8      good
1         8      good
2         0  very bad
3         5      good
4         2       bad
5         3       bad
6         9      good
7         9      good
8        10      good
9         5      good
10        1  very bad
11        7      good
12        7      good
13        6      good
14        5      good
15       10      good
16        3       bad
17        0  very bad
18        3       bad
19        5      good

我希望得到这个列表：

[2，10]

对此是否有一线解决方案？

Is there a one line solution to this?

我不想使用数字值，因为它们仅在此处用于生成DataFrame或遍历所有行，这对于我的用例而言在计算上是昂贵的。

I don't want to use numeric values as they are used purely here to generate the DataFrame or loop over all rows which is computationally expensive for my use case.

推荐答案

如果数据框索引是默认范围索引，则可以使用以下方法：

If your dataframe index is default range index, then you can use this:

np.where((df['status'] == 'very bad') & (df['status'].shift() == 'good'))[0]

输出：

array([ 2, 10], dtype=int64)

其他，则可以使用以下命令：

Else, you can use the following:

irow = np.where((df['status'] == 'very bad') & (df['status'].shift() == 'good'))[0]
df.index[irow]

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