合并连续行熊猫

Question

我想将 Pandas 数据框中的连续行合并为一行。 这是我目前拥有的表：

<头>

id_number	document_number	价值	日志日期	co	延迟（天）
4100000739	106782327	0	1/13/2017 14:23	A
4100000739	106788192	1	1/13/2017 16:39	A	0
4100000740	106787500	0	1/13/2017 16:14	A
4100000740	106788227	F	1/13/2017 16:40	A	0
4100000743	109334630	N	2/13/2017 14:22	B
4100000743	109358034	0	2/14/2017 9:24	B	0
4100000743	109358735	1	2/14/2017 9:37	B	0
4100000743	109334630	N	2/13/2017 14:22	C
4100000743	109358034	0	2/14/2017 9:24	C	0
4100000743	109358735	1	2/14/2017 9:37	C	0
4100000743	109334630	N	2/13/2017 14:22	C
4100000743	109358034	0	2/14/2017 9:24	C	0
4100000743	109358735	1	2/14/2017 9:37	C	0
4100000743	109334630	N	2/13/2017 14:22	D
4100000743	109358034	0	2/14/2017 9:24	D	0
4100000743	109358735	1	2/14/2017 9:37	D	0

期望输出：

<头>

id_number	document_numb	value1	value2	log_date1	log_date2	co	延迟（天）
4100000739	106782327	0	1	1/13/2017 14:23	1/13/2017 16:39	A	0
4100000740	106787500	0	F	1/13/2017 16:14	1/13/2017 16:14	A	0
4100000743	109334630	N	0	2/13/2017 14:22	2/14/2017 9:24	B	0
4100000743	109358034	0	1	2/14/2017 9:24	2/14/2017 9:37	B	0

等等。 本质上，第一个表中的延迟列包含 row(X+1) 和 row(X) 上的 log_date 之间的日期差异。所以第二列的延迟应该包含行 row(x+1) 的值。 行的合并是基于 id_number 和 co 完成的。我真的希望这很清楚

Answer 1

• groupby() id_number 用于逻辑
• concat(axis=1) 第 n 行和第 (n+1) 行
• rename(columns=...) 输出你想要的列名
• 使用 head() 删除组中没有 (n+1) 行的最后一行

import pandas as pd
import io

df = pd.read_csv(io.StringIO("""id_number~document_number~value~log_date~co~delay(days)
4100000739~106782327~0~1/13/2017 14:23~A~
4100000739~106788192~1~1/13/2017 16:39~A~0
4100000740~106787500~0~1/13/2017 16:14~A~
4100000740~106788227~F~1/13/2017 16:40~A~0
4100000743~109334630~N~2/13/2017 14:22~B~
4100000743~109358034~0~2/14/2017 9:24~B~0
4100000743~109358735~1~2/14/2017 9:37~B~0
4100000743~109334630~N~2/13/2017 14:22~C~
4100000743~109358034~0~2/14/2017 9:24~C~0
4100000743~109358735~1~2/14/2017 9:37~C~0
4100000743~109334630~N~2/13/2017 14:22~C~
4100000743~109358034~0~2/14/2017 9:24~C~0
4100000743~109358735~1~2/14/2017 9:37~C~0
4100000743~109334630~N~2/13/2017 14:22~D~
4100000743~109358034~0~2/14/2017 9:24~D~0
4100000743~109358735~1~2/14/2017 9:37~D~0"""), sep="~")


def mergerows(df):
    return pd.concat([df.rename(columns={"value":"value1","log_date":"log_date1"}).drop(columns="delay(days)"), 
           df.shift(-1).loc[:,["value","log_date","delay(days)"]].rename(columns={"value":"value2","log_date":"log_date2"})
                     ], axis=1).head(len(df)-1)


dfc = df.groupby("id_number", as_index=False).apply(mergerows).reset_index(drop=True)

<头>

	id_number	document_number	value1	log_date1	co	value2	log_date2	延迟（天）
0	4100000739	106782327	0	1/13/2017 14:23	A	1	1/13/2017 16:39	0
1	4100000740	106787500	0	1/13/2017 16:14	A	F	1/13/2017 16:40	0
2	4100000743	109334630	N	2/13/2017 14:22	B	0	2/14/2017 9:24	0
3	4100000743	109358034	0	2/14/2017 9:24	B	1	2/14/2017 9:37	0
4	4100000743	109358735	1	2/14/2017 9:37	B	N	2/13/2017 14:22	nan
5	4100000743	109334630	N	2/13/2017 14:22	C	0	2/14/2017 9:24	0
6	4100000743	109358034	0	2/14/2017 9:24	C	1	2/14/2017 9:37	0
7	4100000743	109358735	1	2/14/2017 9:37	C	N	2/13/2017 14:22	nan
8	4100000743	109334630	N	2/13/2017 14:22	C	0	2/14/2017 9:24	0
9	4100000743	109358034	0	2/14/2017 9:24	C	1	2/14/2017 9:37	0
10	4100000743	109358735	1	2/14/2017 9:37	C	N	2/13/2017 14:22	nan
11	4100000743	109334630	N	2/13/2017 14:22	D	0	2/14/2017 9:24	0
12	4100000743	109358034	0	2/14/2017 9:24	D	1	2/14/2017 9:37	0