Question

我有具有相同列名的不同列。

在第一个数据中；每列有16行。

longdata<-structure(list(X2012 = c(0, 0.4, 0.8, 0.2, 0, 0, 0.2, 1.8, 0.8, 
3.6, 0.2, 0, 0.8, 4, 5.6, 4.2), X2013 = c(0L, 0L, 0L, 0L, 0L, 
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), X2014 = c(0.4, 0.6, 
0.6, 0, 0, 0, 0.4, 0.2, 0.2, 0, 0.2, 0, 0, 0.2, 0, 0)), .Names = c("X2012", 
"X2013", "X2014"), class = "data.frame", row.names = c(NA, -16L
))

在第二个数据中，每列有6行。

shortdata<-structure(list(X2012 = c(0.2, 0.4, 0.6, 0.2, 0.2, 0.2), X2013 = c(0, 
0.2, 0, 0, 0.2, 0), X2014 = c(0L, 0L, 0L, 0L, 0L, 0L)), .Names = c("X2012", 
"X2013", "X2014"), class = "data.frame", row.names = c(NA, -6L
))

这就是我想做的；

finaldata<-structure(list(X2012 = c(0, 0.4, 0.8, 0.2, 0, 0, 0.2, 1.8, 0.8, 
3.6, 0.2, 0, 0.8, 4, 5.6, 4.2), X2012.1 = c(0.2, 0.4, 0.6, 0.2, 
0.2, 0.2, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), X2013 = c(0L, 
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), 
    X2013.1 = c(0, 0.2, 0, 0, 0.2, 0, NA, NA, NA, NA, NA, NA, 
    NA, NA, NA, NA), X2014 = c(0.4, 0.6, 0.6, 0, 0, 0, 0.4, 0.2, 
    0.2, 0, 0.2, 0, 0, 0.2, 0, 0), X2014.1 = c(0L, 0L, 0L, 0L, 
    0L, 0L, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), .Names = c("X2012", 
"X2012.1", "X2013", "X2013.1", "X2014", "X2014.1"), class = "data.frame", row.names = c(NA, 
-16L))

Answer 1

一种dplyr可能是：

longdata %>%
 rowid_to_column() %>%
 left_join(shortdata %>%
            rowid_to_column(), by = c("rowid" = "rowid"), suffix = c("", ".1")) %>%
 select(-rowid)

此处根据行ID执行左连接。

   X2012 X2013 X2014 X2012.1 X2013.1 X2014.1
1    0.0     0   0.4     0.2     0.0       0
2    0.4     0   0.6     0.4     0.2       0
3    0.8     0   0.6     0.6     0.0       0
4    0.2     0   0.0     0.2     0.0       0
5    0.0     0   0.0     0.2     0.2       0
6    0.0     0   0.0     0.2     0.0       0
7    0.2     0   0.4      NA      NA      NA
8    1.8     0   0.2      NA      NA      NA
9    0.8     0   0.2      NA      NA      NA
10   3.6     0   0.0      NA      NA      NA
11   0.2     0   0.2      NA      NA      NA
12   0.0     0   0.0      NA      NA      NA
13   0.8     0   0.0      NA      NA      NA
14   4.0     0   0.2      NA      NA      NA
15   5.6     0   0.0      NA      NA      NA
16   4.2     0   0.0      NA      NA      NA

或与base R相同的想法：

longdata$rowid <- seq_along(longdata$X2012)
shortdata$rowid <- seq_along(shortdata$X2012)

merge(longdata, shortdata, by = "rowid", all = TRUE, suffixes = c("", ".1"))[, -1]

或更简化（由@Ronak Shah提出）：

merge(longdata, shortdata, by = "row.names", all = TRUE, suffixes = c("", ".1"))[, -1]

Answer 2

我们可以使用rowr::cbind.fill，使用make.unique和order来使名称唯一。

df1 <- rowr::cbind.fill(longdata, shortdata, fill = NA)
names(df1) <- make.unique(names(df1))
df1[order(names(df1))]


#   X2012 X2012.1 X2013 X2013.1 X2014 X2014.1
#1    0.0     0.2     0     0.0   0.4       0
#2    0.4     0.4     0     0.2   0.6       0
#3    0.8     0.6     0     0.0   0.6       0
#4    0.2     0.2     0     0.0   0.0       0
#5    0.0     0.2     0     0.2   0.0       0
#6    0.0     0.2     0     0.0   0.0       0
#7    0.2      NA     0      NA   0.4      NA
#8    1.8      NA     0      NA   0.2      NA
#9    0.8      NA     0      NA   0.2      NA
#10   3.6      NA     0      NA   0.0      NA
#11   0.2      NA     0      NA   0.2      NA
#12   0.0      NA     0      NA   0.0      NA
#13   0.8      NA     0      NA   0.0      NA
#14   4.0      NA     0      NA   0.2      NA
#15   5.6      NA     0      NA   0.0      NA
#16   4.2      NA     0      NA   0.0      NA

在基数R中，您可以使用match行号和cbind

cbind(longdata,shortdata[match(seq_len(nrow(longdata)), seq_len(nrow(shortdata))),])

在这种情况下，您可能还需要使用make.unique来使名称唯一。

如何在r中对齐具有相同名称的列？

2 个答案: