Question

我想找到数据集中每个数据块的开始索引和结束索引。数据就像：

index     A  wanted_column1 wanted_column2
2000/1/1  0                    0
2000/1/2  1   2000/1/2         1
2000/1/3  1                    1
2000/1/4  1                    1
2000/1/5  0                    0
2000/1/6  1   2000/1/6         2
2000/1/7  1                    2
2000/1/8  1                    2
2000/1/9  0                    0

如数据所示，index和A是给定的列，wanted_column1和wanted_column2是我想要的。这个想法是有不同的连续数据块。我想检索每个数据块的起始索引，并想要增加数据中有多少块的计数。

我尝试使用shift(-1)，但是无法区分开始索引和结束索引之间的差异。

Answer 1

是您需要的吗？

    index       A   wanted_column1  wanted_column2
0   2000/1/1    0   None            0
1   2000/1/2    1   2000/1/2        1
2   2000/1/3    1   None            1
3   2000/1/4    1   None            1
4   2000/1/5    0   None            0
5   2000/1/6    1   2000/1/6        2
6   2000/1/7    1   None            2
7   2000/1/8    1   None            2
8   2000/1/9    0   None            2

得出：

gehbiszumeis

编辑：性能比较

19.9 ms的解决方案：my
4.07 ms解决方案：--watchContentBase

Answer 2

假设数据框为df，则可以在df['A'] != 0处找到索引。之前的索引是块的最后一个索引，在块的第一个索引之后。稍后，您对找到的索引数进行计数以计算数据块的数量

import pandas as pd

# Read your data
df = pd.read_csv('my_txt.txt', sep=',')
df['wanted_column1'] = None     # creating already dummy columns
df['wanted_column2'] = None

# Find indices after each index, where 'A' is not 1, except of it is the last value 
#  of the dataframe
first = [x + 1 for x in df[df['A'] != 1].index.values if x != len(df)-1]
# Find indices before each index, where 'A' is not 1, except of it is the first value 
#  of the dataframe
last = [x - 1 for x in df[df['A'] != 1].index.values if x != 0]

# Set the first indices of each chunk at its corresponding position in your dataframe
df.loc[first, 'wanted_column1'] = df.loc[first, 'index']
# You can set also the last indices of each chunk (you only mentioned this in the text, 
#  not in your expected-result-listed). Uncomment for last indices.
# df.loc[last, 'wanted_column1'] = df.loc[last, 'index']   

# Count the number of chunks and fill it to wanted_column2
for i in df.index: df.loc[i, 'wanted_column2'] = sum(df.loc[:i, 'wanted_column1'].notna())
# Some polishing of the df after to match your expected result 
df.loc[df['A'] != 1, 'wanted_column2'] = 0

这给

      index  A wanted_column1  wanted_column2
0  2000/1/1  0           None               0
1  2000/1/2  1       2000/1/2               1
2  2000/1/3  1           None               1
3  2000/1/4  1           None               1
4  2000/1/5  0           None               0
5  2000/1/6  1       2000/1/6               2
6  2000/1/7  1           None               2
7  2000/1/8  1           None               2
8  2000/1/9  0           None               0

并适用于df的所有长度和数据中的块数

熊猫数据块索引

2 个答案: