R:pmatch来完成更困难的任务

本文介绍了R:pmatch来完成更困难的任务的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

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