[英]Create new data frame based on the match of values between a list and a column of data frame
我正在嘗試使用與舊數據框中列的值匹配的值列表來創建新數據框。 同樣對於新數據框,我想保留用於匹配的值列表中的順序。 這是我想要實現的示例:
#A list of values used for matching
time.new <- c(2, 3, 4, 3, 4, 5, 4, 5, 6)
#The old data frame which I would match on the column of **time.old**
old <- data.frame(time.old=1:10, y=rnorm(10))
time.old y
1 0.20320
2 -0.74696
3 -0.73716
4 -0.61959
5 1.12733
6 2.58322
7 -0.08138
8 -0.10436
9 -0.13081
10 -1.20050
#Here is the expected new data frame
time y
2 -0.74696
3 -0.73716
4 -0.61959
3 -0.73716
4 -0.61959
5 1.12733
4 -0.61959
5 1.12733
6 2.58322
嘗試dplyr中的left_join 。 首先,將 time.new 轉換為數據框的一列:
library(tidyverse)
time.new <- c(2, 3, 4, 3, 4, 5, 4, 5, 6)
#The old data frame which I would match on the column of **time.old**
old <- data.frame(time.old=1:10, y=rnorm(10))
time.new <- data.frame(time=time.new)
new_dataframe <- left_join(time.new, old, by=c("time"="time.old"))
在基礎 R 中使用合並:
merge(x = time.new, y = old, by.x = "time", by.y="time.old", all.x = TRUE)
如果要保留 time.new 的順序,則需要在數據中添加輔助行號列,合並,行號排序並刪除 id 列:
time.new <- c(2, 3, 4, 3, 4, 5, 4, 5, 6)
old <- data.frame(time.old=1:10, y=rnorm(10))
time.new <- data.frame(id = 1:length(time.new), time=time.new)
new_dataframe <- merge(x = time.new, y = old, by.x = "time", by.y="time.old", all.x = TRUE)
new_dataframe <- new_dataframe[order(new_dataframe$id), ]
new_dataframe$id <- NULL
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