英文:
How do I find the indices given value in base R by object?
问题
这可能对您来说似乎有些奇怪，因为在dplyr中可以轻松解决类似问题。但我仍然想知道如何做到这一点。
举个例子，假设我正在查看员工数据，目标是找出给定员工日期对应有多少记录。
现在的任务是找出所有那些计算出的行数超过1的员工日期对。我还没有在网上找到答案，因为“by”不是一个好的搜索词。我可以做的是类似下面的事情：
但我不确定如何得到(1, "2020-01-01")。
英文:
This might seem whimsical to you, because a similar problem is easily solved in dplyr. But I still want to know how to do it.
To illustrate, imagine I am looking at employee data and the goal is to find how many records are there for a given employee-date pair.
# Mockup employee datadf <- data.frame(person_id = c(1, 2, 1),record_date = as.Date(c("2020-01-01", "2020-01-01", "2020-01-01")),salary = c(100, 110, 109))# By object counts rows for each unique employee-date pairout <- by(data = df,INDICES = df[, c("win", "record_date")],FUN = nrow)
Now the task is to find all those employee-date pairs where the calculated number of rows is more than 1. I couldn't find answers on the web yet, "by" makes a bad search word. What I can do is something like:
out>1# record_date# person_id 2020-01-01# 1 TRUE# 2 FALSE
But I am not sure how to get (1, "2020-01-01").
答案1
得分: 2
如果您只想要具有多于一条记录的人员/ID组合，您可以执行以下操作：
subset(as.data.frame(with(df, table(person_id, record_date))), Freq > 1)#> person_id record_date Freq#> 1 1 2020-01-01 2
或者，如果您想要所有计数，只需删除subset：
as.data.frame(with(df, table(person_id, record_date)))#> person_id record_date Freq#> 1 1 2020-01-01 2#> 2 2 2020-01-01 1
英文:
If you only want the person / id combinations with more than one record, you can do
subset(as.data.frame(with(df, table(person_id, record_date))), Freq > 1)#> person_id record_date Freq#> 1 1 2020-01-01 2
Or if you want all the counts, just remove the subset:
as.data.frame(with(df, table(person_id, record_date)))#> person_id record_date Freq#> 1 1 2020-01-01 2#> 2 2 2020-01-01 1
答案2
得分: 1
您可以使用 ave。
transform(df, flag=ave(person_id, person_id, record_date, FUN=\(x) length(x) > 1))# person_id record_date salary flag# 1 1 2020-01-01 100 1# 2 2 2020-01-01 110 0# 3 1 2020-01-01 109 1
您也可以在 subset 中使用它。
subset(df, ave(person_id, person_id, record_date, FUN=\(x) length(x) > 1) == 1)# person_id record_date salary# 1 1 2020-01-01 100# 3 1 2020-01-01 109
请注意，ave 在内部使用了 by。
英文:
You can use ave.
transform(df, flag=ave(person_id, person_id, record_date, FUN=\(x) length(x) > 1))# person_id record_date salary flag# 1 1 2020-01-01 100 1# 2 2 2020-01-01 110 0# 3 1 2020-01-01 109 1
You can also use it in subset.
subset(df, ave(person_id, person_id, record_date, FUN=\(x) length(x) > 1) == 1)# person_id record_date salary# 1 1 2020-01-01 100# 3 1 2020-01-01 109
Note, that ave internally uses by.
答案3
得分: 0
在 base R 中使用 duplicated：
subset(df, duplicated(person_id) | duplicated(person_id, fromLast = TRUE))
输出：
person_id record_date salary1 1 2020-01-01 1003 1 2020-01-01 109
英文:
Use duplicated in base R
subset(df, duplicated(person_id)|duplicated(person_id, fromLast = TRUE))
-output
person_id record_date salary1 1 2020-01-01 1003 1 2020-01-01 109
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