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我正在使用data.table fread()函数来读取一些具有缺失值的数据,并且它们是在 Excel 中生成的,因此缺失值字符串是“#N/A”。但是,当我使用na.strings命令时str,读取数据的最终结果仍然是字符。为了复制这一点,这里是代码和数据。

数据:

Date,a,b,c,d,e,f,g
1/1/03,#N/A,0.384650146,0.992190069,0.203057232,0.636296656,0.271766148,0.347567706
1/2/03,#N/A,0.461486974,0.500702057,0.234400718,0.072789936,0.060900352,0.876749487
1/3/03,#N/A,0.573541006,0.478062582,0.840918789,0.061495666,0.64301024,0.939575302
1/4/03,#N/A,#N/A,#N/A,#N/A,#N/A,#N/A,#N/A
1/5/03,#N/A,#N/A,#N/A,#N/A,#N/A,#N/A,#N/A
1/6/03,#N/A,0.66678429,0.897482818,0.569609033,0.524295691,0.132941158,0.194114347
1/7/03,#N/A,0.576835985,0.982816576,0.605408973,0.093177815,0.902145012,0.291035649
1/8/03,#N/A,0.100952961,0.205491093,0.376410642,0.775917986,0.882827749,0.560508499
1/9/03,#N/A,0.350174456,0.290225065,0.428637309,0.022947911,0.7422805,0.354776101
1/10/03,#N/A,0.834345466,0.935128099,0.163158666,0.301310627,0.273928596,0.537167776
1/11/03,#N/A,#N/A,#N/A,#N/A,#N/A,#N/A,#N/A
1/12/03,#N/A,#N/A,#N/A,#N/A,#N/A,#N/A,#N/A
1/13/03,#N/A,0.325914633,0.68192633,0.320222677,0.249631582,0.605508964,0.739263677
1/14/03,#N/A,0.715104989,0.639040211,0.004186366,0.351412982,0.243570606,0.098312443
1/15/03,#N/A,0.750380716,0.264929325,0.782035411,0.963814327,0.93646428,0.453694758
1/16/03,#N/A,0.282389354,0.762102103,0.515151803,0.194083842,0.102386764,0.569730516
1/17/03,#N/A,0.367802161,0.906878948,0.848538256,0.538705673,0.707436236,0.186222899
1/18/03,#N/A,#N/A,#N/A,#N/A,#N/A,#N/A,#N/A
1/19/03,#N/A,#N/A,#N/A,#N/A,#N/A,#N/A,#N/A
1/20/03,#N/A,0.79933188,0.214688799,0.37011313,0.189503843,0.294051763,0.503147404
1/21/03,#N/A,0.620066341,0.329949446,0.123685075,0.69027192,0.060178071,0.599825005

(数据保存在 temp.csv 中)代码:

library(data.table)
a <- fread("temp.csv", na.strings="#N/A")

给出(我有更大的数据集,所以忽略观察的数量):

Classes ‘data.table’ and 'data.frame':  144 obs. of  8 variables:
 $ Date: chr  "1/1/03" "1/2/03" "1/3/03" "1/4/03" ...
 $ a   : chr  NA NA NA NA ...
 $ b   : chr  "0.384650146" "0.461486974" "0.573541006" NA ...
 $ c   : chr  "0.992190069" "0.500702057" "0.478062582" NA ...
 $ d   : chr  "0.203057232" "0.234400718" "0.840918789" NA ...
 $ e   : chr  "0.636296656" "0.072789936" "0.061495666" NA ...
 $ f   : chr  "0.271766148" "0.060900352" "0.64301024" NA ...
 $ g   : chr  "0.347567706" "0.876749487" "0.939575302" NA ...
 - attr(*, ".internal.selfref")=<externalptr> 

此代码工作正常

 a <- read.csv("temp.csv", header=TRUE, na.strings="#N/A")

它是一个错误吗?有一些聪明的解决方法吗?

4

1 回答 1

5

来自?freadfor的文档na.strings如下:

na.strings要转换为 NA_character_ 的字符串字符向量。默认情况下,读取为类型字符“,”的列读取为空白字符串 (“”),而“,NA”读取为 NA_character_。典型的替代方案可能是 na.strings=NULL 或 na.strings = c("NA","N/A","")。

我想你应该自己把它们转换成数字。至少这是我从文档中了解到的。

像这样的东西?

cbind(a[, 1], a[, lapply(.SD[, -1], as.numeric)])
于 2013-04-03T10:14:53.613 回答