Home/ Questions/Simple solutions for the error in colmeans(x, na.rm = true) : 'x' must be numeric error
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Jules Forestier
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Asked: 2022-05-12T11:33:49+00:00 In:

Simple solutions for the error in colmeans(x, na.rm = true) : ‘x’ must be numeric error

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I am new to r and searching the “error in colmeans(x, na.rm = true) : ‘x’ must be numeric” to understand it better. It seems it doesn’t work as expected when I used some suggestions before. This is the command line I use:

data(birth.death.rates.1966)
 data2 <- birth.death.rates.1966
 princ <- prcomp(data2)

The error I’m getting is below:

Error in colMeans(x, na.rm = TRUE) : 'x' must be numeric

Please give me the solution for this issue.

error in colmeans
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2 Answers

  1. Best Answer
    Expert

    factor allows you to convert a vector of characters into numeric values. Each unique value then gets an unique integer code. This example shows four values. The numbers 1 through 4 are alphabetical.

    > d = data.frame(country=c("foo","bar","baz","qux"),x=runif(4),y=runif(4))
> d
  country          x         y
1     foo 0.84435112 0.7022875
2     bar 0.01343424 0.5019794
3     baz 0.09815888 0.5832612
4     qux 0.18397525 0.8049514
> d$country = as.numeric(as.factor(d$country))
> d
  country          x         y
1       3 0.84435112 0.7022875
2       1 0.01343424 0.5019794
3       2 0.09815888 0.5832612
4       4 0.18397525 0.8049514

    Then, you can run prcomp

    > prcomp(d)
Standard deviations:
[1] 1.308665216 0.339983614 0.009141194

Rotation:
               PC1          PC2          PC3
country -0.9858920  0.132948161 -0.101694168
x       -0.1331795 -0.991081523 -0.004541179
y       -0.1013910  0.009066471  0.994805345

    It is up to the application whether this makes sense. You might just want to remove prcomp(d[,-1]) from the first column and instead work with numeric data.

    • 15

  2. The character is the first column in the data frame. You can also recode the data frame to create row names like: 

    library(tidyverse)
data2 %>% remove_rownames %>% column_to_rownames(var="country")
princ <- prcomp(data2)

    Alternately, you can also use: 

    data2 <- data2[,-1]
rownames(data2) <- data2[,1]
princ <- prcomp(data2)
    • 18

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