在R中使用data.table时，中位数返回错误 [英] Median returning an error when using data.table in R

问题描述

我有以下数据集

> head(DT)
    V1 V2 V3   V4   V5     V6 V7
1:  2  1  2 0.91 0.02 880.00  1
2:  3  2  1 0.02 0.00   2.24  2
3:  1  1  1 0.15 0.01   3.41  3
4:  1  2  1 3.92 0.05 268.67  2
5:  1  1  2 0.10 0.01   1.59  3
6:  0  1  1 1.20 0.04   1.43  3

> sapply(DT, class)
       V1        V2        V3        V4        V5        V6        V7 
"integer" "integer" "integer" "numeric" "numeric" "numeric"  "factor" 

可扩展数千行。我正在尝试计算由变量V7定义的8个组中V1-V6的中位数。

which expands for thousands of rows. I am trying to calculate the median values of V1-V6 within the 8 groups defined by the factor Variable V7

> levels(DT$V7)
[1] "1" "2" "3" "4" "5" "6" "7" "8"

此刻，我正在使用以下命令，该命令会返回错误：

At the moment I am using the following command, which returns an error:

> DT[, lapply(.SD, median), by = V7]
 Error in `[.data.table`(DF, , lapply(.SD, median), by = V7) : 
 Column 1 of result for group 4 is type 'integer' but expecting type 'double'. Column types must be consistent for each group.

我在某处读到一种解决方法是使用 as.double（median （X））。但这适用于单个列： DT [，as.double（median（X）），by = V7] ，但不适用于考虑所有列的情况： DT [，lapply（.SD，as.double（median）），by = V7] （正如预期的那样，因为您必须将输入传递给中值）

I read somewhere that a way around this was using as.double(median(X)). But this works for individual columns: DT[, as.double(median(X)), by = V7], but not for when considering all columns: DT[, lapply(.SD, as.double(median)), by = V7] (as expected, because you have to pass an input to median)

我可以使用汇总

> aggregate(DT[,c(1:6), with = FALSE], by = list(DF$V7), FUN = median)
  Group.1 V1 V2 V3     V4   V5      V6
   1       1  0  1  1  1.285 0.04 401.500
   2       2  1  2  1  3.565 0.06   6.400
   3       3  0  1  1  0.360 0.03  11.200
   4       4  1  1  1 74.290 0.26 325.960
   5       5  2  1  0  1.145 0.04   1.415
   6       6  0  1  1 10.100 0.18  93.000
   7       7  1  1  0  0.740 0.04   1.080
   8       8  1  1  0  7.970 0.40   0.050

但是我想知道是否有一种方法可以解决上述错误，并使用data.table进行计算

But I'd like to know if there is a way to solve the error described above and do this calculation using data.table

推荐答案

中位数很不寻常，因为它可以为相同的输入类型返回不同类型的返回值：

median is unusual because it can return different types of return values for the same input type:

默认方法返回与x，
类型相同的长度为1的对象，除非x是偶数长度的整数，那么结果将是
的两倍。

The default method returns a length-one object of the same type as x, except when x is integer of even length, when the result will be double.

但是，data.table需要一致的返回值类型。您有两种可能：

However, data.table needs a consistent return value type. You have two possibilities:

将所有列转换为数字：

DT[, paste0("V", 1:6) := lapply(.SD, as.numeric), by = V7]

或转换返回值中位数：

DT[, lapply(.SD, function(x) as.numeric(median(x))), by = V7]

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