将值分配给数字因子水平[重复]

Question

我已经浏览过不同的链接，例如：How to convert a factor to an integer\numeric without a loss of information?

但无法解决问题

我有一个数据框

 SYMBOL             PVALUE1             PVALUE2
1   10-Mar   0.813027629406118    0.78820189558684
2   10-Sep 0.00167287722066533 0.00167287722066533
3   11-Mar    0.21179810441316   0.464576340307205
4   11-Sep 0.00221961024320294 0.00221961024320294
5   12-Sep   0.934667427815304   0.986884425214009
6   15-Sep 0.00167287722066533 0.00167287722066533
7    1-Dec   0.464576340307205  0.0911572830792113
8    1-Mar 0.00818426308604705  0.0252302356363697
9    1-Sep    0.60516237199519   0.570568468332992
10   2-Mar  0.0103975819620539 0.00382292568622066
11   2-Sep 0.00167287722066533 0.00167287722066533

当我尝试str()

str(df)
'data.frame':   20305 obs. of  3 variables:
 $ SYMBOL : Factor w/ 21050 levels "","10-Mar","10-Sep",..: 2 3 4 5 6 7 8 9 10 11 ...
 $ PVALUE1: Factor w/ 209 levels "0","0.000109570493049298",..: 169 22 110 24 181 22 139 39 149 44 ...
 $ PVALUE2: Factor w/ 216 levels "0","0.000109570493049298",..: 172 20 141 23 201 20 90 61 150 29 ...

我试试mode()

sapply(df,mode)
SYMBOL   PVALUE1   PVALUE2 
"numeric" "numeric" "numeric" 

当我尝试根据以下条件为两个数字列（2,3）分配值时，

df$Score <- rowSums(ifelse(df[,-1]==0, 0, 
                                       ifelse(df[, -1]<= 0.05, 2, ifelse(df[,-1]>= 0.065,-2,1))))

I get Warning messages:
1: In Ops.factor(left, right) : ‘<=’ not meaningful for factors
2: In Ops.factor(left, right) : ‘<=’ not meaningful for factors
3: In Ops.factor(left, right) : ‘>=’ not meaningful for factors
4: In Ops.factor(left, right) : ‘>=’ not meaningful for factors

输出如下：

SYMBOL             PVALUE1             PVALUE2       Score
1 10-Mar   0.813027629406118    0.78820189558684         NA
2 10-Sep 0.00167287722066533 0.00167287722066533         NA
3 11-Mar    0.21179810441316   0.464576340307205         NA
4 11-Sep 0.00221961024320294 0.00221961024320294         NA
5 12-Sep   0.934667427815304   0.986884425214009         NA
6 15-Sep 0.00167287722066533 0.00167287722066533         NA

如果因子已经是数字，为什么上面的代码不起作用并给出NA。我应该如何进行。

编辑 dput()

structure(list(SYMBOL = structure(1:6, .Label = c("10-Mar", "10-Sep", 
"11-Mar", "11-Sep", "12-Sep", "15-Sep"), class = "factor"), PVALUE1 = structure(c(4L, 
1L, 3L, 2L, 5L, 1L), .Label = c("0.00167287722066533", "0.00221961024320294", 
"0.21179810441316", "0.813027629406118", "0.934667427815304"), class = "factor"), 
    PVALUE2 = structure(c(4L, 1L, 3L, 2L, 5L, 1L), .Label = c("0.00167287722066533", 
    "0.00221961024320294", "0.464576340307205", "0.78820189558684", 
    "0.986884425214009"), class = "factor")), .Names = c("SYMBOL", 
"PVALUE1", "PVALUE2"), row.names = c(NA, 6L), class = "data.frame")

我也试过这个：

  indx <- sapply(df, is.factor)
    df[indx] <- lapply(df[indx], function(x) as.numeric(levels(x))[x])

    indx returns 

    SYMBOL PVALUE1 PVALUE2 
       TRUE    TRUE    TRUE 
Warning message:
In FUN(X[[3L]], ...) : NAs introduced by coercion

Answer 1

使用您的 dput 数据，这工作得很好：

df = structure(list(SYMBOL = structure(1:6, .Label = c("10-Mar", "10-Sep", 
"11-Mar", "11-Sep", "12-Sep", "15-Sep"), class = "factor"), PVALUE1 = structure(c(4L, 
1L, 3L, 2L, 5L, 1L), .Label = c("0.00167287722066533", "0.00221961024320294", 
"0.21179810441316", "0.813027629406118", "0.934667427815304"), class = "factor"), 
    PVALUE2 = structure(c(4L, 1L, 3L, 2L, 5L, 1L), .Label = c("0.00167287722066533", 
    "0.00221961024320294", "0.464576340307205", "0.78820189558684", 
    "0.986884425214009"), class = "factor")), .Names = c("SYMBOL", 
"PVALUE1", "PVALUE2"), row.names = c(NA, 6L), class = "data.frame")

df$PVALUE1 = as.numeric(as.character(df$PVALUE1))
df$PVALUE2 = as.numeric(as.character(df$PVALUE2))

df
#   SYMBOL     PVALUE1     PVALUE2
# 1 10-Mar 0.813027629 0.788201896
# 2 10-Sep 0.001672877 0.001672877
# 3 11-Mar 0.211798104 0.464576340
# 4 11-Sep 0.002219610 0.002219610
# 5 12-Sep 0.934667428 0.986884425
# 6 15-Sep 0.001672877 0.001672877

sapply(df, class)
#    SYMBOL   PVALUE1   PVALUE2 
#  "factor" "numeric" "numeric" 

如果您对整个数据框执行此操作时遇到问题，则可能存在一些不规则行。不过，我也查看了您在 cmets 中提供的 CSV，看起来还不错。

另请注意，这是您链接的重复问题中的几个等效解决方案之一。

要转换除第一列之外的所有列，您可以这样做

df[, 2:ncol(df)] = lapply(df[, -1], function(x) as.numeric(as.character(x)))

请注意，您不希望以这种方式转换日期列或符号列，因为它们不是数字。

同样，要将命名为 PVALUE1 的列转换为 PVALUE47，您可以构造列名，然后对其进行转换：

col_to_convert = paste0("PVALUE", 1:47)
df[, col_to_convert] = lapply(df[, col_to_convert], function(x) as.numeric(as.character(x)))

一般来说，最佳做法是一开始就不将这些列作为因素。但是，您将这些数据输入 R 可能有一种方法来指定列类，例如，colClasses in read.table、read.csv 等。

Answer 2

使用data.table的选项

 library(data.table)
 setDT(df)[, 2:3 := lapply(.SD, function(x)
                    as.numeric(levels(x))[x]), .SDcols=2:3]

或者更快一点的版本是使用set

 indx <- which(sapply(df, is.factor) & grepl('PVALUE', names(df)))
 setDT(df)

 for(j in indx){
   set(df, i=NULL, j=j, value= as.numeric(levels(df[[j]]))[df[[j]]])
 }

我猜您收到警告的原因是因为您创建的“索引”还包括第一列（因为它也是一个因素），但它不是数字的。通过将非数字元素从 factor 转换为 numeric，这些元素将被强制为 NA。

根据?factor

将因子“f”转换为近似于它的 原始数值，推荐使用“as.numeric(levels(f))[f]” 并且比‘as.numeric(as.character(f))’更高效。