按ID和日期合并行 [英] Merge Rows by ID and Date

问题描述

我是R的新手，我一直在研究如何解决以下问题.

I am newbie at R and I have been searching on how to solve the following problem.

我有一个看起来像df的

I have a df that looks like:

id ------------日期------------ OB1 ------ OB2 ----- OB3
1 ------- 2017-01-01 --------- 1 --------- 0 --------- 0
2 ------- 2006-01-05 --------- 1 --------- 0 --------- 0
2 ------- 2007-04-19 --------- 0 --------- 1 --------- 0
3 ------- 2015-02-23 --------- 0 --------- 0 --------- 1
3 ------- 2015-02-23 --------- 1 --------- 0 --------- 0

id------------Date ------------OB1------ OB2----- OB3
1 ------- 2017-01-01 --------- 1 --------- 0--------- 0
2 ------- 2006-01-05 --------- 1 --------- 0--------- 0
2 ------- 2007-04-19 --------- 0 --------- 1--------- 0
3 ------- 2015-02-23 --------- 0 --------- 0--------- 1
3 ------- 2015-02-23 --------- 1 --------- 0--------- 0

我要实现的目标显示在这里:

What I have to achieve is shown here:

id ------------日期------------ OB1 ------ OB2 ----- OB3
1 ------- 2017-01-01 --------- 1 --------- 0 --------- 0
2 ------- 2006-01-05 --------- 1 --------- 0 --------- 0
2 ------- 2007-04-19 --------- 0 --------- 1 --------- 0
3 ------- 2015-02-23 --------- 1 --------- 0-- ------ 1

id------------Date ------------OB1------ OB2----- OB3
1 ------- 2017-01-01 --------- 1 --------- 0--------- 0
2 ------- 2006-01-05 --------- 1 --------- 0--------- 0
2 ------- 2007-04-19 --------- 0 --------- 1--------- 0
3 ------- 2015-02-23 --------- 1 --------- 0--------- 1

这是按ID和日期组合行.

This is, to combine rows, by id and date.

如果日期中OB3的值为'1'，而同一日期(对于相同的ID)的OB1的值为'1'，则OB1的结果必须为'1'，'OB3的值必须为'1'和一个日期

If there is value '1' for OB3 in a date and value '1' for OB1 in the same date (for the same ID) the result must be value '1' for OB1, value '1' for 'OB3' and a single date

我一直在尝试应用此处说明的一些解决方案: 将具有相同值的行合并到多列中

I have been trying to apply some solutions explained here: Merge rows having same values in multiple columns

但这没用

OB1，OB2，OBS3是布尔值 感谢您的帮助！

OB1, OB2, OBS3 are boolean values Thanks for your help!

aggregate(.〜ID + Date，df，any)有效！

EDIT 2: aggregate(. ~ ID + Date, df, any) works!

输入数据

structure(list(ID = c(-1L, 1L, 1L), Date = c("2008-01-15", "2011-01-21", "2011-01-21"), `OBS1` = c(0, 0, 0), `OBS2` = c(0, 0, 0), `OBS3` = c(0, 0, 0), `OBS4` = c(0, 0, 0), `OBS5` = c(0, 0, 0), `OBS6` = c(0, 1, 0)), .Names = c("ID", "Date", "OBS1", "OBS2", "OBS3", "OBS4", "OBS5", "OBS6"), row.names = c(NA, 3L), class = "data.frame")

输出数据

structure(list(ID = c(-1L, 1L), Date = c("2008-01-15", "2011-01-21"), `OBS1` = c(FALSE, FALSE), `OBS2` = c(FALSE, FALSE), `OBS3` = c(FALSE, FALSE), `OBS4` = c(FALSE, FALSE), `OBS5` = c(FALSE, FALSE), `OBS6` = c(FALSE, TRUE)), .Names = c("ID", "Date", "OBS1", "OBS2", "OBS3", "OBS4", "OBS5", "OBS6"), row.names = c(NA, -2L), class = "data.frame")

推荐答案

已经使用base R的aggregate()函数回答了该问题.

The question already has been answered using base R's aggregate() function.

但是，我感到很难将问题中打印的样本数据集转换为可重现的示例( 之前，OP编辑了问题以包含dput()的结果).

However, I felt challenged to turn the sample dataset as printed in the question into a reproducible example (before the OP edited the question to include the results of dput()).

此外，OP还提到他有一个非常大的df" ，这可能值得尝试使用data.table方法.

In addition, the OP has mentioned he has a "very large df" which might be worthwhile to try a data.table approach.

library(magrittr)
library(data.table)
df <- readr::read_file(
"id------------Date ------------OB1------ OB2----- OB3
1 ------- 2017-01-01 --------- 1 --------- 0--------- 0
2 ------- 2006-01-05 --------- 1 --------- 0--------- 0
2 ------- 2007-04-19 --------- 0 --------- 1--------- 0
3 ------- 2015-02-23 --------- 0 --------- 0--------- 1
3 ------- 2015-02-23 --------- 1 --------- 0--------- 0"
) %>% stringr::str_replace_all("[-]{2,}", " ") %>% 
  fread()
df

   id       Date   OB1   OB2   OB3
1:  1 2017-01-01  TRUE FALSE FALSE
2:  2 2006-01-05  TRUE FALSE FALSE
3:  2 2007-04-19 FALSE  TRUE FALSE
4:  3 2015-02-23 FALSE FALSE  TRUE
5:  3 2015-02-23  TRUE FALSE FALSE

请注意，fread()已自动识别出布尔列.

Note that fread() has recognised automatically the boolean columns.

library(data.table)
setDT(df)[, lapply(.SD, any), by = .(id, Date)]

   id       Date   OB1   OB2   OB3
1:  1 2017-01-01  TRUE FALSE FALSE
2:  2 2006-01-05  TRUE FALSE FALSE
3:  2 2007-04-19 FALSE  TRUE FALSE
4:  3 2015-02-23  TRUE FALSE  TRUE

如果OP期望整数值0和1而不是逻辑值，则可以一次性创建:

In case, the OP expects integer values 0 and 1 instead of logical values, these can be created in one go:

setDT(df)[, lapply(.SD, function(x) as.integer(any(x))), by = .(id, Date)]

   id       Date OB1 OB2 OB3
1:  1 2017-01-01   1   0   0
2:  2 2006-01-05   1   0   0
3:  2 2007-04-19   0   1   0
4:  3 2015-02-23   1   0   1

这篇关于按ID和日期合并行的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！