使用matrixStats :: rowMedians查找每行的中位数 [英] Find median of every row using matrixStats::rowMedians

问题描述

我正在尝试使用matrixStats包中的rowMedians计算数据帧df的行中位数.

I am trying to calculate the row median for a data frame df with rowMedians from matrixStats package.

Abundance Sample1 Sample2 Sample3 Sample4 Sample5 Sample6 Sample7
Species1   2        4       0       0       0       6       0
Species2   3        5       6       4       0       0       0
Species3   3        7       2       5       8       0       0
Species4   0        0       3       8       0       0       8
Species5   7        5       6       0       0       4       4
Species6   4        2       3       0       0       2       1

我想计算每行的中位数，然后将它们附加在新列中.我遇到了错误

I want to calculate the median of every row and append them in a new column. I got an error

参数"x"必须是向量或矩阵

Argument 'x' must be a vector or matrix

所以我尝试将df转换为矩阵. str函数显示该物种的每个值都是数字，所以我尝试了:

so I tried to convert my df to a matrix. The str function shows that every value for the species is numeric, so I tried:

library(matrixStats)
matrix(df, rownames.force = NA)
rowMedians(df)

但是我仍然遇到相同的错误.感谢您的帮助.

but I am still getting the same error. Any help is appreciated.

推荐答案

您不想包括该Abundance列用于中值计算.假设df是您当前的数据帧.

You don't want to include that Abundance column for median calculation. Let df be your current data frame.

library(matrixStats)
rowMedians(as.matrix(df[-1]))

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除了上面的正确代码外，还有一些注释.

---

A few comments besides the correct code above.

1. 您是否检查过matrix(df)是什么?
2. 即使正确返回了一个数字矩阵，它也不会覆盖df.所以rowMedians(df)会给您同样的错误，好像什么也没发生；
3. 作为练习，比较as.matrix(df[-1])和as.matrix(df).

1. Have you checked what matrix(df) is?
2. Even if it correctly returns you a numeric matrix, it does not overwrite df. So rowMedians(df) gives you the same error as if nothing has happened;
3. As an exercise, compare as.matrix(df[-1]) and as.matrix(df).

了解这些问题可防止您将来出错.

Understanding these issues prevents you from getting errors in future.

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