pandas 从分组的数据框中计算连续相等值的长度 [英] Pandas calculate length of consecutive equal values from a grouped dataframe

问题描述

我想做他们在这里的答案中所做的事情:

I want to do what they've done in the answer here: Calculating the number of specific consecutive equal values in a vectorized way in pandas , but using a grouped dataframe instead of a series.

给定一个具有几列的数据框

So given a dataframe with several columns

A    B    C   
------------ 
x    x    0
x    x    5
x    x    2
x    x    0
x    x    0
x    x    3
x    x    0
y    x    1
y    x    10
y    x    0
y    x    5
y    x    0
y    x    0

我想对A和B列进行分组，然后计算C中连续零的数目.之后，我想返回每个零长度出现的次数的计数.所以我想要这样的输出:

I want to groupby columns A and B, then count the number of consecutive zeros in C. After that I'd like to return counts of the number of times each length of zeros occurred. So I want output like this:

A    B    num_consecutive_zeros  count
---------------------------------------
x    x            1                2
x    x            2                1
y    x            1                1
y    x            2                1

我不知道如何调整链接问题的答案以处理分组数据框.

I don't know how to adapt the answer from the linked question to deal with grouped dataframes.

推荐答案

下面是代码， count_consecutive_zeros()使用numpy函数和 pandas.value_counts()来获取结果，然后使用 groupby().apply(count_consecutive_zeros)为每个组调用 count_consecutive_zeros().调用 reset_index()将 MultiIndex 更改为列:

Here is the code, count_consecutive_zeros() use numpy functions and pandas.value_counts() to get the results, and use groupby().apply(count_consecutive_zeros) to call count_consecutive_zeros() for every group. call reset_index() to change MultiIndex to columns:

import pandas as pd
import numpy as np
from io import BytesIO
text = """A    B    C   
x    x    0
x    x    5
x    x    2
x    x    0
x    x    0
x    x    3
x    x    0
y    x    1
y    x    10
y    x    0
y    x    5
y    x    0
y    x    0"""

df = pd.read_csv(BytesIO(text.encode()), delim_whitespace=True)

def count_consecutive_zeros(s):
    v = np.diff(np.r_[0, s.values==0, 0])
    s = pd.value_counts(np.where(v == -1)[0] - np.where(v == 1)[0])
    s.index.name = "num_consecutive_zeros"
    s.name = "count"
    return s

df.groupby(["A", "B"]).C.apply(count_consecutive_zeros).reset_index()

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