我有以下数据帧DF1和DF2。我正在尝试用DF2中Close列中的值填充DF1。但是事实证明，运行循环时输出为零。不知道出什么问题了，但是似乎它没有读取Asset列值。
DF1：
Data Asset Close
1 1986-11-27 ABC 6 5.95
2 1986-12-01 ABC 6 5.90
3 1986-12-03 ABC 6 5.90
4 1986-12-04 ABC 6 5.90
5 1986-12-05 ABC 6 5.00
6 1986-12-08 ABC 6 5.00
7 1986-12-09 ABC 6 4.78
8 1986-10-31 ABC 8 3.90
9 1986-11-03 ABC 8 3.70
10 1986-11-04 ABC 8 3.70
. . . .
. . . .
DF2：
ABC 6 ABC 8
1986-11-27 NA NA
1986-12-01 NA NA
1986-12-03 NA NA
1986-12-04 NA NA
1986-12-05 NA NA
1986-12-08 NA NA
1986-12-09 NA NA
1986-12-10 NA NA
1986-12-11 NA NA
1986-12-12 NA NA
. . .
. . .
for (i in 1:length(DF2))
{
for (m in 1:nrow(DF2))
{
for (n in 1:nrow(DF1))
{
if ((names(DF2[i]) == DF1[n,2]) & (row.names(DF2[m,0])==as.character(DF1[n,1])))
{
DF2[m,i] <- DF1[n,3]
} else{DF2[m,i] <- 0}
}
}
}
输出：
ABC 6 ABC 8
1986-11-27 0 0
1986-12-01 0 0
1986-12-03 0 0
1986-12-04 0 0
1986-12-05 0 0
1986-12-08 0 0
1986-12-09 0 0
1986-12-10 0 0
1986-12-11 0 0
1986-12-12 0 0
. . .
. . .
答案 0 :(得分：0)
我不确定这是否是您想要实现的目标。使用base R的解决方案如下：
DF2 <- cbind(DF2[1],sapply(names(DF2)[-1],
function(v) DF1$Close[match(DF2$Date,subset(DF1,Asset==v)$Date)]))
如此
> DF2
Date ABC 6 ABC 8
1 1986-11-27 5.95 NA
2 1986-12-01 5.90 NA
3 1986-12-03 5.90 NA
4 1986-12-04 5.90 NA
5 1986-12-05 5.00 NA
6 1986-12-08 5.00 NA
7 1986-12-09 4.78 NA
8 1986-12-10 NA NA
9 1986-12-11 NA NA
10 1986-12-12 NA NA
数据
DF1 <- structure(list(Date = structure(c(4L, 5L, 6L, 7L, 8L, 9L, 10L,
1L, 2L, 3L), .Label = c("1986-10-31", "1986-11-03", "1986-11-04",
"1986-11-27", "1986-12-01", "1986-12-03", "1986-12-04", "1986-12-05",
"1986-12-08", "1986-12-09"), class = "factor"), Asset = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L), .Label = c("ABC 6", "ABC 8"
), class = "factor"), Close = c(5.95, 5.9, 5.9, 5.9, 5, 5, 4.78,
3.9, 3.7, 3.7)), row.names = c(NA, -10L), class = "data.frame")
DF2 <- structure(list(Date = structure(1:10, .Label = c("1986-11-27",
"1986-12-01", "1986-12-03", "1986-12-04", "1986-12-05", "1986-12-08",
"1986-12-09", "1986-12-10", "1986-12-11", "1986-12-12"), class = "factor"),
"ABC 6" = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), "ABC 8" = c(NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA)), class = "data.frame", row.names = c(NA,
-10L))
答案 1 :(得分：0)
如果我理解正确，OP将将数据从长格式更改为宽格式。
这是一种非常常见的操作，之前，例如here已经问过很多次问题了。
但是，仅出于演示目的，这是提供预期结果的解决方案的集合：
reshape(DF1, idvar = "Data", timevar = "Asset", v.names = "Close", direction = "wide")
Data Close.ABC 6 Close.ABC 8 1 1986-11-27 5.95 NA 2 1986-12-01 5.90 NA 3 1986-12-03 5.90 NA 4 1986-12-04 5.90 NA 5 1986-12-05 5.00 NA 6 1986-12-08 5.00 NA 7 1986-12-09 4.78 NA 8 1986-10-31 NA 3.9 9 1986-11-03 NA 3.7 10 1986-11-04 NA 3.7
reshape2包reshape2::dcast(DF1, Data ~ Asset)
Using Close as value column: use value.var to override. Data ABC 6 ABC 8 1 1986-10-31 NA 3.9 2 1986-11-03 NA 3.7 3 1986-11-04 NA 3.7 4 1986-11-27 5.95 NA 5 1986-12-01 5.90 NA 6 1986-12-03 5.90 NA 7 1986-12-04 5.90 NA 8 1986-12-05 5.00 NA 9 1986-12-08 5.00 NA 10 1986-12-09 4.78 NA
data.table包library(data.table)
dcast(setDT(DF1), Data ~ Asset)
Using 'Close' as value column. Use 'value.var' to override Data ABC 6 ABC 8 1: 1986-10-31 NA 3.9 2: 1986-11-03 NA 3.7 3: 1986-11-04 NA 3.7 4: 1986-11-27 5.95 NA 5: 1986-12-01 5.90 NA 6: 1986-12-03 5.90 NA 7: 1986-12-04 5.90 NA 8: 1986-12-05 5.00 NA 9: 1986-12-08 5.00 NA 10: 1986-12-09 4.78 NA
tidyr包library(tidyr)
DF1 %>%
pivot_wider(names_from = Asset, values_from = Close)
# A tibble: 10 x 3 Data `ABC 6` `ABC 8` <chr> <dbl> <dbl> 1 1986-11-27 5.95 NA 2 1986-12-01 5.9 NA 3 1986-12-03 5.9 NA 4 1986-12-04 5.9 NA 5 1986-12-05 5 NA 6 1986-12-08 5 NA 7 1986-12-09 4.78 NA 8 1986-10-31 NA 3.9 9 1986-11-03 NA 3.7 10 1986-11-04 NA 3.7
DF1 <- as.data.frame(readr::read_table("rn Data Asset Close
1 1986-11-27 ABC 6 5.95
2 1986-12-01 ABC 6 5.90
3 1986-12-03 ABC 6 5.90
4 1986-12-04 ABC 6 5.90
5 1986-12-05 ABC 6 5.00
6 1986-12-08 ABC 6 5.00
7 1986-12-09 ABC 6 4.78
8 1986-10-31 ABC 8 3.90
9 1986-11-03 ABC 8 3.70
10 1986-11-04 ABC 8 3.70", col_types = "_ccd"))