感谢@nullglob,
Thanks @nullglob,
我试图再次运行它,但是输出却不同.如果我滥用了您的代码,您介意教我吗?抱歉,我可能误会了它的工作方式.希望您不介意给我更多建议.
I tried to run it again, but my output is different. Could you mind to teach me if I have misuse your code? Sorry that I may have misunderstand the way how it works. I hope you don't mind to give me some more advice.
df1 <- data.frame( A=c("x01","x02","y03","z02","x04", "x33", "z03"), B=c("A01BB01","A02BB02","C02AA05","B04CC10","C01GX02", "yyy", "zzz")) df2 <- data.frame( X=c("a","b","c","d","e", "f"), Y=c("A01BB","A02","C02A","B04","C01GX", "xxx")) with(c(df1,df2),{ i <- pmatch(Y,B) iunmatched <- which(is.na(i)) nunmatched <- length(iunmatched) nexcess <- length(B) - length(X) data.frame(A = c(A,rep(NA,nunmatched)), B = c(B,rep(NA,nunmatched)), X = c(X[i],rep(NA,nexcess),X[iunmatched]), Y = c(Y[i],rep(NA,nexcess),Y[iunmatched])) }) A B X Y 1 1 1 1 1 2 2 2 2 2 3 5 5 3 5 4 6 3 4 3 5 3 4 5 4 6 4 6 NA NA 7 7 7 NA NA 8 NA NA 6 6======================原始问题=====
======================ORIGINAL Question=====
感谢您回答我的上一个问题. (stackoverflow/q/6592214/602276)
Thanks for answers to my previous question. (stackoverflow/q/6592214/602276)
要以此答案为基础,我想做一个更困难的任务的pmatch.
To build upon this answer, I want to do the pmatch for a more difficult task.
df1 <- data.frame( A=c("x01","x02","y03","z02","x04", "x33", "z03") B=c("A01BB01","A02BB02","C02AA05","B04CC10","C01GX02", "yyy", "zzz") ) A B 1 x01 A01BB01 2 x02 A02BB02 3 y03 C02AA05 4 z02 B04CC10 5 x04 C01GX02 6 x33 yyy 7 z03 zzz我的df2修改如下:
df2 <- data.frame( X=c("a","b","c","d","e", "f"), Y=c("A01BB","A02","C02A","B04","C01GX", "xxx") ) X Y 1 a A01BB 2 b A02 3 c C02A 4 d B04 5 e C01GX 6 f xxx困难是由于df1和df2的行数不同,我无法在正确的开头进行cbind
The difficulty is due to df1 and df2 has different no of rows, i cannot do cbind at the right beginning
更糟糕的是,df1和df2之间存在一些不匹配,它们对应的行应相应地得出NA.
Morover, there is some mismatch between df1 and df2, their corresponding line should results NA accordingly.
预期输出如下:
A B X Y 1 x01 A01BB01 a A01BB 2 x02 A02BB02 b A02 3 y03 C02AA05 c C02A 4 z02 B04CC10 d B04 5 x04 C01GX02 e C01GX 6 x33 yyy NA NA 7 z03 zzz NA NA 7 NA NA f xxx您介意教我如何使用R进行操作吗?非常感谢.
Could you mind to teach me how to do it with R? Thanks a lot.
推荐答案这并非完美的解决方案,但似乎可以解决问题:
This is not exactly an elegant solution, but it seems to do the trick:
with(c(df1,df2),{ i <- pmatch(Y,B) iunmatched <- which(is.na(i)) nunmatched <- length(iunmatched) nexcess <- length(B) - length(X) data.frame(A = c(A,rep(NA,nunmatched)), B = c(B,rep(NA,nunmatched)), X = c(X[i],rep(NA,nexcess),X[iunmatched]), Y = c(Y[i],rep(NA,nexcess),Y[iunmatched])) })输出应为:
A B X Y 1 x01 A01BB01 a A01BB 2 x02 A02BB02 b A02 3 y03 C02AA05 c C02A 4 z02 B04CC10 d B04 5 x04 C01GX02 e C01GX 6 x33 yyy <NA> <NA> 7 z03 zzz <NA> <NA> 8 <NA> <NA> f xxx更多推荐
R:pmatch来完成更困难的任务
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