如何在R中的数据框中插入行数较少的列
How to insert a column with less rows in a dataframe in R
例如，我有以下数据框：
Date Price
2020-01-01 500
2020-01-02 550
2020-01-03 480
2020-01-04 420
2020-01-05 450
2020-01-06 390
2020-01-07 430
2020-01-08 480
2020-01-09 490
2020-01-10 485
并且我想在前一个数据框中插入 "Prince2" 列，以便我重复该行直到下一个类似日期：
Date Price2
2020-01-01 50
2020-01-04 20
2020-01-06 90
2020-01-07 30
2020-01-10 80
会是这样的：
Date Price Price2
2020-01-01 500 50
2020-01-02 550 50
2020-01-03 480 50
2020-01-04 420 20
2020-01-05 450 20
2020-01-06 390 90
2020-01-07 430 30
2020-01-08 480 30
2020-01-09 490 30
2020-01-10 485 80
有人可以帮助我吗？
我们可以使用 merge 和 zoo
中的 na.locf0
library(zoo)
transform(merge(df1, df2, by = 'Date',all.x = TRUE), Price2 = na.locf0(Price2))
或 data.table
library(data.table)
setDT(df1)[df2, Price2 := Price2, on = .(Date)][, Price2 := na.locf0(Price2)][]
或 tidyverse
library(dplyr)
library(tidyr)
left_join(df1, df2, by = 'Date') %>%
fill(Price2)
# Date Price Price2
#1 2020-01-01 500 50
#2 2020-01-02 550 50
#3 2020-01-03 480 50
#4 2020-01-04 420 20
#5 2020-01-05 450 20
#6 2020-01-06 390 90
#7 2020-01-07 430 30
#8 2020-01-08 480 30
#9 2020-01-09 490 30
#10 2020-01-10 485 80
或转换为 Date class 并执行 complete
df1$Date <- as.Date(df1$Date)
df2$Date <- as.Date(df2$Date)
df2 %>%
complete(Date = seq(min(df1$Date), max(df1$Date), by = '1 day')) %>%
fill(Price2) %>%
left_join(df1)
数据
df1 <- structure(list(Date = c("2020-01-01", "2020-01-02", "2020-01-03",
"2020-01-04", "2020-01-05", "2020-01-06", "2020-01-07", "2020-01-08",
"2020-01-09", "2020-01-10"), Price = c(500L, 550L, 480L, 420L,
450L, 390L, 430L, 480L, 490L, 485L)), class = "data.frame", row.names = c(NA,
-10L))
df2 <- structure(list(Date = c("2020-01-01", "2020-01-04", "2020-01-06",
"2020-01-07", "2020-01-10"), Price2 = c(50L, 20L, 90L, 30L, 80L
)), class = "data.frame", row.names = c(NA, -5L))
例如，我有以下数据框：
Date Price
2020-01-01 500
2020-01-02 550
2020-01-03 480
2020-01-04 420
2020-01-05 450
2020-01-06 390
2020-01-07 430
2020-01-08 480
2020-01-09 490
2020-01-10 485
并且我想在前一个数据框中插入 "Prince2" 列，以便我重复该行直到下一个类似日期：
Date Price2
2020-01-01 50
2020-01-04 20
2020-01-06 90
2020-01-07 30
2020-01-10 80
会是这样的：
Date Price Price2
2020-01-01 500 50
2020-01-02 550 50
2020-01-03 480 50
2020-01-04 420 20
2020-01-05 450 20
2020-01-06 390 90
2020-01-07 430 30
2020-01-08 480 30
2020-01-09 490 30
2020-01-10 485 80
有人可以帮助我吗？
我们可以使用 merge 和 zoo
na.locf0
library(zoo)
transform(merge(df1, df2, by = 'Date',all.x = TRUE), Price2 = na.locf0(Price2))
或 data.table
library(data.table)
setDT(df1)[df2, Price2 := Price2, on = .(Date)][, Price2 := na.locf0(Price2)][]
或 tidyverse
library(dplyr)
library(tidyr)
left_join(df1, df2, by = 'Date') %>%
fill(Price2)
# Date Price Price2
#1 2020-01-01 500 50
#2 2020-01-02 550 50
#3 2020-01-03 480 50
#4 2020-01-04 420 20
#5 2020-01-05 450 20
#6 2020-01-06 390 90
#7 2020-01-07 430 30
#8 2020-01-08 480 30
#9 2020-01-09 490 30
#10 2020-01-10 485 80
或转换为 Date class 并执行 complete
df1$Date <- as.Date(df1$Date)
df2$Date <- as.Date(df2$Date)
df2 %>%
complete(Date = seq(min(df1$Date), max(df1$Date), by = '1 day')) %>%
fill(Price2) %>%
left_join(df1)
数据
df1 <- structure(list(Date = c("2020-01-01", "2020-01-02", "2020-01-03",
"2020-01-04", "2020-01-05", "2020-01-06", "2020-01-07", "2020-01-08",
"2020-01-09", "2020-01-10"), Price = c(500L, 550L, 480L, 420L,
450L, 390L, 430L, 480L, 490L, 485L)), class = "data.frame", row.names = c(NA,
-10L))
df2 <- structure(list(Date = c("2020-01-01", "2020-01-04", "2020-01-06",
"2020-01-07", "2020-01-10"), Price2 = c(50L, 20L, 90L, 30L, 80L
)), class = "data.frame", row.names = c(NA, -5L))