R - 从data.frame中舍弃双精度向量 [英] R - Exctracting a vector of doubles from data.frame

问题描述

我有这个问题使用 read.table（）有或没有 header = T ，试图提取从 data.frame 与 as.double（as.character（））（见？factor ）。

但这只是如何我意识到，逻辑。因此，您不会看到 read.table 在下面的代码中，只有必要的部分。

1. 使用 header = T 等价：

    （a<  -  data.frame（array（c（0.5,0.5,0.5,0.5） ），c（1,4）））
    as.character（a）
   ＃[1]0.50.50.50.5
     




2. 没有 header = T 等效：

   
   
   <$ c $ p> b < - data.frame（array（c（a，0.5，b，0.5，c，0.5，d 0.5，c（2,4）））
（a <-b [2，]）
as.character（a）
＃ 11

（a< - data.frame（a，row.names = NULL））＃现在甚至没有视觉差别
as.character（a）
＃[1]1111


解决方案

问题出在 data.frame 的默认设置，选项 stringsAsFactors 设置为 TRUE 。这是您的场景中的一个问题，因为当您使用 header = FALSE 时，该行中的字符值的存在将整个列强制为字符，然后将其转换为因子除非您设置 stringsAsFactors = FALSE ）。

以下是一些要玩的例子：

  ##两个类似的`data.frame`s  - 只有一个参数不同
 
b<  -  data.frame （c（a，0.5，b，0.5，c，0.5，d，0.5），c（2,4）））
 b2 < c（a，0.5，b，0.5，c，0.5，d，0.5），c（2,4）），
 stringsAsFactors = FALSE）
 
 ##首先与b
 
 as.character（b [2，]）
＃[1]1111
 
 sapply（b [2，]，as.character）
＃X1 X2 X3 X4 
＃0.50.50.50.5
 asmatrix b）[2，] 
＃X1 X2 X3 X4 
＃0.50.50.50.5
 as.double（asmatrix（b）[2， ）
＃[1] 0.5 0.5 0.5 0.5 
 
 ##现在用b2
 
 as.character（b2 [2，]）
 ＃[1]0.50.50.50.5
 as.double（as.character（b2 [2，]））
＃[1] 0.5 0.5 0.5 0.5 
  

I got this question using read.table() with or without header=T, trying to extract a vector of doubles from the resulting data.frame with as.double(as.character()) (see ?factor).

But that's just how I realized that I don't understand R's logic. So you won't see e.g. read.table in the code below, only the necessary parts. Could you tell me what's the difference between the following options?

1. With header=T equivalent:

   (a <- data.frame(array(c(0.5,0.5,0.5,0.5), c(1,4))))
   as.character(a)
   # [1] "0.5" "0.5" "0.5" "0.5"
   

2. Without header=T equivalent:

   b <- data.frame(array(c("a",0.5,"b",0.5,"c",0.5,"d",0.5), c(2,4)))
   (a <- b[2,])
   as.character(a)
   # [1] "1" "1" "1" "1"
   
   (a <- data.frame(a, row.names=NULL)) # now there's not even a visual difference
   as.character(a)
   # [1] "1" "1" "1" "1"
   

解决方案

The problem lies in the default setting of data.frame, where one of the options, stringsAsFactors is set to TRUE. This is a problem in your scenario because when you use header = FALSE, the presence of character values in that row coerces the entire column to characters, which is then converted to factors (unless you set stringsAsFactors = FALSE).

Here are some examples to play with:

## Two similar `data.frame`s -- just one argument different

b <- data.frame(array(c("a",0.5,"b",0.5,"c",0.5,"d",0.5), c(2,4)))
b2 <- data.frame(array(c("a",0.5,"b",0.5,"c",0.5,"d",0.5), c(2,4)),
                stringsAsFactors = FALSE)

## First with "b"

as.character(b[2, ])
# [1] "1" "1" "1" "1"

sapply(b[2, ], as.character)
#    X1    X2    X3    X4 
# "0.5" "0.5" "0.5" "0.5"
as.matrix(b)[2, ]
#    X1    X2    X3    X4 
# "0.5" "0.5" "0.5" "0.5"
as.double(as.matrix(b)[2, ])
# [1] 0.5 0.5 0.5 0.5

## Now with "b2"

as.character(b2[2, ])
# [1] "0.5" "0.5" "0.5" "0.5"
as.double(as.character(b2[2, ]))
# [1] 0.5 0.5 0.5 0.5

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