列数不等的两个data.frame合并的问题

januslian

我现在有两个数据框，其中一个有11列，一个有8列，具体地如下：

second1[1:3,]

code coding foredate period before1 volume ratiob ratior

1 759 2 2006-04-07 2006-03-31 0 255.2852 1.00000000 0.0000000

2 2014 5 2006-10-17 2006-12-31 0 152.1729 0.13208252 0.3258520

3 897 5 2006-10-19 2006-12-31 0 1338.7747 0.07899333 0.4179136

ratioo tradevalue lnvalue

1 1.5471 19044276 16.76228

2 2.8572 13254260 16.39983

3 3.1130 88359130 18.29692

second2[1:3,]

code coding volume ratiob ratior ratioo tradevalue lnvalue

1 600069 4 930.0066 0.3749956 0.5005418 3.6538 36404717 33.43234

2 600079 4 861.6510 0.6322675 0.8475378 3.9797 41997653 17.55312

3 594 5 2805.5808 0.2221042 1.0000000 8.5160 116805679 18.57602

我现在想把两个数据框合并在一起，合并后数据框的列数和second1的列数相同，也就是

code coding foredate period before1 volume ratiob ratior

1 759 2 2006-04-07 2006-03-31 0 255.2852 1.00000000 0.0000000

2 2014 5 2006-10-17 2006-12-31 0 152.1729 0.13208252 0.3258520

3 897 5 2006-10-19 2006-12-31 0 1338.7747 0.07899333 0.4179136

1 600069 4 NA NA NA 930.0066 0.3749956 0.5005418

2 600079 4 NA NA NA 861.6510 0.6322675 0.8475378

3 594 5 NA NA NA 2805.5808 0.2221042 1.0000000

ratioo tradevalue lnvalue

1 1.5471 19044276 16.76228

2 2.8572 13254260 16.39983

3 3.1130 88359130 18.29692

4 3.6538 36404717 33.43234

5 3.9797 41997653 17.55312

6 8.5160 116805679 18.57602

请问应该怎么合并？

karlqi

我写了一个小函数"join"可以将两个data frame连接起来，希望你可以用得上：

"join" <- function(dfr1, dfr2)

{

if(!(is.data.frame(dfr1)&&is.data.frame(dfr2))){

print("the input objects must be data.frame")

return()

}else{

ind <- unique(c(dimnames(dfr1)[[2]],dimnames(dfr2)[[2]]))

ind1 <- match(dimnames(dfr1)[[2]], ind)

ind2 <- match(dimnames(dfr2)[[2]], ind)

jo <- matrix(NA, ncol =length(ind), nrow =nrow(dfr1) + nrow(dfr2))

jo[1:nrow(dfr1), ind1] <- as.matrix(dfr1)

jo[(1:nrow(bb))+nrow(dfr1), ind2] <- as.matrix(dfr2)

dimnames(jo)[[2]] <- ind

return(as.data.frame(jo))

}

}

## examples :

aa <- data.frame(x = seq(6), y = rep(2,6), z = rep(1,6))

bb <- data.frame(w = rep(3,2), x = rep(2,2), z = seq(2))

join(aa,bb)

x y z w

1 1 2 1 NA

2 2 2 1 NA

3 3 2 1 NA

4 4 2 1 NA

5 5 2 1 NA

6 6 2 1 NA

7 2 NA 1 3

8 2 NA 2 3

陈放

xiexie

januslian

哇，太好了

非常感谢！

houliping

非常感谢，程序中有个小错误。

修正后的程序为

"join" <- function(dfr1, dfr2)

{

if(!(is.data.frame(dfr1)&&is.data.frame(dfr2))){

print("the input objects must be data.frame")

return()

}else{

ind <- unique(c(dimnames(dfr1)[[2]],dimnames(dfr2)[[2]]))

ind1 <- match(dimnames(dfr1)[[2]], ind)

ind2 <- match(dimnames(dfr2)[[2]], ind)

jo <- matrix(NA, ncol =length(ind), nrow =nrow(dfr1) + nrow(dfr2))

jo[1:nrow(dfr1), ind1] <- as.matrix(dfr1)

jo[(1:nrow(dfr2))+nrow(dfr1), ind2] <- as.matrix(dfr2)

dimnames(jo)[[2]] <- ind

return(as.data.frame(jo))

}

}

注：

jo[(1:nrow([u]bb[/u]))+nrow(dfr1), ind2] <- as.matrix(dfr2)应该改为jo[(1:nrow([u]dfr2[/u]))+nrow(dfr1), ind2] <- as.matrix(dfr2)

bjt

> merge(second1,second2,all = T)

    code coding    volume     ratiob    ratior ratioo tradevalue  lnvalue   foredate     period before1<br />
1    594      5 2805.5808 0.22210420 1.0000000 8.5160  116805679 18.57602       <NA>       <NA>      NA<br />
2    759      2  255.2852 1.00000000 0.0000000 1.5471   19044276 16.76228 2006-04-07 2006-03-31       0<br />
3    897      5 1338.7747 0.07899333 0.4179136 3.1130   88359130 18.29692 2006-10-19 2006-12-31       0<br />
4   2014      5  152.1729 0.13208252 0.3258520 2.8572   13254260 16.39983 2006-10-17 2006-12-31       0<br />
5 600069      4  930.0066 0.37499560 0.5005418 3.6538   36404717 33.43234       <NA>       <NA>      NA<br />
6 600079      4  861.6510 0.63226750 0.8475378 3.9797   41997653 17.55312       <NA>       <NA>      NA<br />

karlqi

hehe 谢谢指正马虎了

另外，楼上说的merge我也尝试过，是不行的

## test ##

> aa <- data.frame(x = seq(6), y = rep(2,6), z = rep(1,6))

> bb <- data.frame(w = rep(3,2), x = rep(2,2), z = seq(2))

> merge(aa,bb,all = T)

x z y w

1 1 1 2 NA

2 2 1 2 3

3 2 2 NA 3

4 3 1 2 NA

5 4 1 2 NA

6 5 1 2 NA

7 6 1 2 NA

不是想要的结果

zwzcosname

merge(aa, bb, all = T,incomparables = NA, sort=F)

karlqi

> aa <- data.frame(x = seq(6), y = rep(2,6), z = rep(1,6))

> bb <- data.frame(w = rep(3,2), x = rep(2,2), z = seq(2))

> merge(aa, bb, all = T,incomparables = NA, sort=F)

x z y w

1 2 1 2 3

2 1 1 2 NA

3 3 1 2 NA

4 4 1 2 NA

5 5 1 2 NA

6 6 1 2 NA

7 2 2 NA 3

也不对的，至少是8行啊，连行数都不对的

houliping

CRAN有一个叫reshape的package，提供一些数据合并及转换方面的函数。其中就包括rbind.fill()函数，可以用来合并多个data.frame，缺失值自动用NA补齐。

###Example###

install.packages("reshape")

library(reshape)

aa <- data.frame(x = seq(6), y = rep(2,6), z = rep(1,6))

bb <- data.frame(w = rep(3,2), x = rep(2,2), z = seq(2))

cc <- data.frame(m = rep(9,2), y = rep(0,2), n = seq(2))

rbind.fill(aa, bb, cc)

x y z w m n

1 1 2 1 NA NA NA

2 2 2 1 NA NA NA

3 3 2 1 NA NA NA

4 4 2 1 NA NA NA

5 5 2 1 NA NA NA

6 6 2 1 NA NA NA

7 2 NA 1 3 NA NA

8 2 NA 2 3 NA NA

9 NA 0 NA NA 9 1

10 NA 0 NA NA 9 2

zwzcosname

> aa <- data.frame(x=seq(6), y=rep(2,6), z=rep(1,6))

> bb <- data.frame(w=rep(3,2), x=rep(2,2), z=seq(2))

> merge(aa, bb, all=T, incomparables=NA, sort=F)

x z y w

1 1 1 2 NA

2 2 1 2 NA

3 3 1 2 NA

4 4 1 2 NA

5 5 1 2 NA

6 6 1 2 NA

7 2 1 NA 3

8 2 2 NA 3

>

结果还是8行呀？

zwzcosname

不过确实有个问题，重复的记录不会重复打印出来，：）

merge(aa, aa, all=F, incomparables=NA, sort=F)

karlqi

我用的是2.6.0版的R，经过验证（系统的merge函数没有被覆盖），输出确实如上文所说的7行，可能是那个版本的merge函数中有bug吧。

zwzcosname

从你的输出结果来看，感觉2.6版本的 merge 不支持 sort 和 incomparables 参数项

bjt

Changes in R Version 2.7.0

New features

The ’incomparables’ argument to duplicated(),

unique() and match() is now implemented,

and passed to match() from merge().