字典样式替换多个项目 [英] Dictionary style replace multiple items

问题描述

目前我正在进行中，我想要转换的字符数据的大量数据，基于通常称为其他语言的字典。它像这样：

  foo<  -  data.frame（snp1 = c（AA，AG，AA ，AA），snp2 = c（AA，AT，AG，AA），snp3 = c（NA，GG，GG，GC），stringsAsFactors = FALSE ）
 foo<  - 替换（foo，foo ==AA，0101）
 foo<  -  replace（foo，foo ==AC，0102）
 foo<  -  replace（foo，foo ==AG，0103）
  

工作正常，但显然不漂亮，似乎很愚蠢地重复替换语句，每次我想替换data.frame中的一个项目。

有没有更好的方法，因为我有一个约25个键/值对的字典？

解决方案

  map = setNames（c（0101，0102，0103），c（AA，AC，AG））
 foo []<  -  map [unlist（foo）] 
   map 涵盖 foo 中的所有案例，p> 
 
 
。如果 foo 是一个矩阵（字符（）），那么
 $ b，这将不太像一个黑客，而且在空间和时间上都更有效率
 $ b 
 矩阵（map [foo]，nrow = nrow（foo），dimnames = dimnames（foo））
  
当有数百万个SNP和数千个样本时，矩阵和数据帧变体都会与R的2 ^ 31-1限制相关。 p> 
I have a large data.frame of character data that I want to convert based on what is commonly called a dictionary in other languages.

Currently I am going about it like so:
foo <- data.frame(snp1 = c("AA", "AG", "AA", "AA"), snp2 = c("AA", "AT", "AG", "AA"), snp3 = c(NA, "GG", "GG", "GC"), stringsAsFactors=FALSE)
foo <- replace(foo, foo == "AA", "0101")
foo <- replace(foo, foo == "AC", "0102")
foo <- replace(foo, foo == "AG", "0103")
This works fine, but it is obviously not pretty and seems silly to repeat the replace statement each time I want to replace one item in the data.frame.

Is there a better way to do this since I have a dictionary of approximately 25 key/value pairs?
解决方案
map = setNames(c("0101", "0102", "0103"), c("AA", "AC", "AG"))
foo[] <- map[unlist(foo)]
assuming that map covers all the cases in foo. This would feel less like a 'hack' and be more efficient in both space and time if foo were a matrix (of character()), then
matrix(map[foo], nrow=nrow(foo), dimnames=dimnames(foo))
Both matrix and data frame variants run afoul of R's 2^31-1 limit on vector size when there are millions of SNPs and thousands of samples.

                        
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