我有两个数据框：
df1<-structure(list(Name = c("sub7", "sub7", "sub7", "sub7", "sub7",
"sub7", "sub7", "sub7", "sub7", "sub7"), StimulusName = c("Alpha1",
"Alpha1", "Alpha1", "Alpha1", "Alpha1", "Alpha1", "Alpha1", "Alpha1",
"Alpha1", "Alpha1"), PupilLeft = c(10.046, 10.05, 10.062, 10.072,
10.072, 10.056, 10.056, 10.056, 10.066, 10.066)), row.names = c(NA,
-10L), class = c("tbl_df", "tbl", "data.frame"))
df2<-structure(list(Name = c("sub7", "sub7"), StimulusName = c("Alpha1",
"Alpha1"), Row_Num = c(1, 3), Label = c("Onset", "Offset")), row.names = c(NA,
-2L), vars = "Name", drop = TRUE, indices = list(0:1), group_sizes = 2L, biggest_group_size = 2L, labels = structure(list(
Name = "Guilty Subject 07"), row.names = c(NA, -1L), class = "data.frame", vars = "Name", drop = TRUE), class = c("grouped_df",
"tbl_df", "tbl", "data.frame"))
我想从Row_Num中的Label和df2列中获取值，并将它们映射到df1中与{{1 }}。
我想在不使用循环的情况下实现这一目标。也许只是将Row_Num值用作索引？
在这种情况下，最终的数据帧如下所示：
Row_Num答案 0 :(得分：1)
另一种选择是在合并之前在Row_Num中创建df1。
df_out <- merge(
transform(df1, Row_Num = seq_len(nrow(df1))),
df2,
by = c("Name", "StimulusName", "Row_Num"),
all.x = TRUE)
df_out$Row_Num <- ifelse(df_out$Row_Num %in% df2$Row_Num, df_out$Row_Num, NA)
df_out
# Name StimulusName Row_Num PupilLeft Label
#1 sub7 Alpha1 1 10.046 Onset
#2 sub7 Alpha1 NA 10.050 <NA>
#3 sub7 Alpha1 3 10.062 Offset
#4 sub7 Alpha1 NA 10.072 <NA>
#5 sub7 Alpha1 NA 10.072 <NA>
#6 sub7 Alpha1 NA 10.056 <NA>
#7 sub7 Alpha1 NA 10.056 <NA>
#8 sub7 Alpha1 NA 10.056 <NA>
#9 sub7 Alpha1 NA 10.066 <NA>
#10 sub7 Alpha1 NA 10.066 <NA>
答案 1 :(得分：1)
使用tidyverse
library(tidyverse)
df1 %>%
mutate(Row_Num = row_number()) %>%
left_join(df2) %>%
mutate(Row_Num = replace(Row_Num, !Row_Num %in% c(1, 3), NA))
# A tibble: 10 x 5
# Name StimulusName PupilLeft Row_Num Label
# <chr> <chr> <dbl> <dbl> <chr>
# 1 sub7 Alpha1 10.0 1 Onset
# 2 sub7 Alpha1 10.0 NA <NA>
# 3 sub7 Alpha1 10.1 3 Offset
# 4 sub7 Alpha1 10.1 NA <NA>
# 5 sub7 Alpha1 10.1 NA <NA>
# 6 sub7 Alpha1 10.1 NA <NA>
# 7 sub7 Alpha1 10.1 NA <NA>
# 8 sub7 Alpha1 10.1 NA <NA>
# 9 sub7 Alpha1 10.1 NA <NA>
#10 sub7 Alpha1 10.1 NA <NA>
如果要通过行名联接
rownames_to_column(df1, "Row_Num") %>%
mutate(Row_Num = as.numeric(Row_Num)) %>%
left_join(., df2 %>%
ungroup %>%
select(Row_Num, Label), by = "Row_Num") %>%
mutate(Row_Num = replace(Row_Num, !Row_Num %in% c(1, 3), NA))
或使用match中的base R
i1 <- match(row.names(df1), df2$Row_Num)
df1[names(df2)[3:4]] <- lapply(df2[3:4], `[`, i1)
df1
# A tibble: 10 x 5
# Name StimulusName PupilLeft Row_Num Label
# <chr> <chr> <dbl> <dbl> <chr>
# 1 sub7 Alpha1 10.0 1 Onset
# 2 sub7 Alpha1 10.0 NA <NA>
# 3 sub7 Alpha1 10.1 3 Offset
# 4 sub7 Alpha1 10.1 NA <NA>
# 5 sub7 Alpha1 10.1 NA <NA>
# 6 sub7 Alpha1 10.1 NA <NA>
# 7 sub7 Alpha1 10.1 NA <NA>
# 8 sub7 Alpha1 10.1 NA <NA>
# 9 sub7 Alpha1 10.1 NA <NA>
#10 sub7 Alpha1 10.1 NA <NA>
答案 2 :(得分：1)
如果我们仅通过Row_Num加入，那么我们可以这样做：
rownames(df2) <- df2$Row_Num
merge(df1, df2, by=0, all.x=TRUE)
Row.names Name.x StimulusName.x PupilLeft Name.y StimulusName.y Row_Num Label
1 1 sub7 Alpha1 10.046 sub7 Alpha1 1 Onset
2 10 sub7 Alpha1 10.066 <NA> <NA> NA <NA>
3 2 sub7 Alpha1 10.050 <NA> <NA> NA <NA>
4 3 sub7 Alpha1 10.062 sub7 Alpha1 3 Offset
5 4 sub7 Alpha1 10.072 <NA> <NA> NA <NA>
6 5 sub7 Alpha1 10.072 <NA> <NA> NA <NA>
7 6 sub7 Alpha1 10.056 <NA> <NA> NA <NA>
8 7 sub7 Alpha1 10.056 <NA> <NA> NA <NA>
9 8 sub7 Alpha1 10.056 <NA> <NA> NA <NA>
10 9 sub7 Alpha1 10.066 <NA> <NA> NA <NA>
或者，您可能要使用：
merge(df1, df2, by="row.names", all.x=TRUE)
要使by参数的含义更加隐秘。
答案 3 :(得分：0)
merge怎么样？
df2 <- data.frame(df2, stringsAsFactors = F)
df3 <- merge(df1,df2)
> df3
Name StimulusName PupilLeft Row_Num Label
1 sub7 Alpha1 10.046 1 Onset
2 sub7 Alpha1 10.046 3 Offset
3 sub7 Alpha1 10.050 1 Onset
4 sub7 Alpha1 10.050 3 Offset
5 sub7 Alpha1 10.062 1 Onset
6 sub7 Alpha1 10.062 3 Offset
7 sub7 Alpha1 10.072 1 Onset
8 sub7 Alpha1 10.072 3 Offset
9 sub7 Alpha1 10.072 1 Onset
10 sub7 Alpha1 10.072 3 Offset
11 sub7 Alpha1 10.056 1 Onset
12 sub7 Alpha1 10.056 3 Offset
13 sub7 Alpha1 10.056 1 Onset
14 sub7 Alpha1 10.056 3 Offset
15 sub7 Alpha1 10.056 1 Onset
16 sub7 Alpha1 10.056 3 Offset
17 sub7 Alpha1 10.066 1 Onset
18 sub7 Alpha1 10.066 3 Offset
19 sub7 Alpha1 10.066 1 Onset
20 sub7 Alpha1 10.066 3 Offset
编辑：使用建议的方法，将行名作为索引：
df1$id <- 1:nrow(df1)
df2$id <- 1:nrow(df2)
df4 <- merge(df1,df2, by="id", all.x=T)
> df4
id Name.x StimulusName.x PupilLeft Name.y StimulusName.y Row_Num Label
1 1 sub7 Alpha1 10.046 sub7 Alpha1 1 Onset
2 2 sub7 Alpha1 10.050 sub7 Alpha1 3 Offset
3 3 sub7 Alpha1 10.062 <NA> <NA> NA <NA>
4 4 sub7 Alpha1 10.072 <NA> <NA> NA <NA>
5 5 sub7 Alpha1 10.072 <NA> <NA> NA <NA>
6 6 sub7 Alpha1 10.056 <NA> <NA> NA <NA>
7 7 sub7 Alpha1 10.056 <NA> <NA> NA <NA>
8 8 sub7 Alpha1 10.056 <NA> <NA> NA <NA>
9 9 sub7 Alpha1 10.066 <NA> <NA> NA <NA>
10 10 sub7 Alpha1 10.066 <NA> <NA> NA <NA>