Correlation analysis between specific columns (in groups of three) possible?

Question

It is possible to calculate the correlation coefficients and p-values in groups? For example with this dataset:

df<-read.csv("http://renatabrandt.github.io/EBC2015/data/varechem.csv", row.names=1)

df

It is possible to calculate between the first 3 columns N, P, K, the correlation between the next three columns of Ca, Mg and S and so on?

I couldn't find any arguments in the cor function that would allow me to specify that. I'm no expert, but I suspect I could split my dataset into sub-datasets first. But even then I don't know how it can be implied that in the cor function.

Anyone have an idea how I could go about this, maybe another function? Or do I have to go the long way and just do each correlation separately? My real data set is relatively large, so an automated solution would help me a lot!

The second set is Ca, Mg, S. So i want to calculate between N-P, N-K, P-K and Ca-Mg, Ca-S, Mg-S and so on. — Bellis
– Bellis, Commented Mar 15, 2023 at 16:43
When you say sets of 3, are you doing correlation between 1 to 3 and 4 to 6, then 7 to 9 and 10 to 12 and 13 to 15 and 16 to 19?? — akrun
– akrun, Commented Mar 15, 2023 at 16:47

akrun · Accepted Answer · 2023-03-15 17:08:58Z

1

Perhaps this helps

i1 <- rep(c(TRUE, FALSE), each = 3)
cor(df[i1], df[!i1])
                Ca          Mg           S          Zn          Mo    Baresoil
N        -0.27089189 -0.16377388 -0.26225073 -0.13216191 -0.05773585  0.10592277
P         0.73720757  0.59793864  0.75272609  0.70231212  0.17246511  0.01389203
K         0.66482924  0.62775899  0.84377126  0.60004281  0.06819748  0.16855851
Al       -0.20586385 -0.11839842  0.35975081 -0.05510622  0.51031000 -0.39955068
Fe       -0.33216455 -0.20221718  0.05651145 -0.31205825  0.22059378 -0.45746292
Mn        0.44331615  0.25758427  0.27457837  0.36446210 -0.20497945  0.24617616
Humdepth  0.24364456  0.37140078  0.15821813  0.13964926  0.05860733  0.59244384
pH        0.09142358 -0.09252275 -0.18689505 -0.08692419 -0.17496367 -0.53181541
>

Or may be

lst1 <- split.default(df, as.integer(gl(ncol(df), 6, ncol(df))))
> lapply(Filter(\(x) ncol(x)  == 6, lst1), \(x) cor(x[1:3], x[4:6]))
$`1`
          Ca         Mg          S
N -0.2708919 -0.1637739 -0.2622507
P  0.7372076  0.5979386  0.7527261
K  0.6648292  0.6277590  0.8437713

$`2`
            Zn         Mo   Baresoil
Al -0.05510622  0.5103100 -0.3995507
Fe -0.31205825  0.2205938 -0.4574629
Mn  0.36446210 -0.2049795  0.2461762

Or could be

lapply(Filter(\(x) ncol(x)  == 6, lst1), \(x) list(cor(x[1:3]), cor(x[4:6])))

-output

$`1`
$`1`[[1]]
           N          P          K
N  1.0000000 -0.2511603 -0.1466368
P -0.2511603  1.0000000  0.7540753
K -0.1466368  0.7540753  1.0000000

$`1`[[2]]
          Ca        Mg         S
Ca 1.0000000 0.7982771 0.5395393
Mg 0.7982771 1.0000000 0.6504151
S  0.5395393 0.6504151 1.0000000


$`2`
$`2`[[1]]
           Al         Fe         Mn
Al  1.0000000  0.8242146 -0.4704864
Fe  0.8242146  1.0000000 -0.4360448
Mn -0.4704864 -0.4360448  1.0000000

$`2`[[2]]
                 Zn         Mo   Baresoil
Zn       1.00000000 0.28200265 0.04450428
Mo       0.28200265 1.00000000 0.03124277
Baresoil 0.04450428 0.03124277 1.00000000

edited Mar 15, 2023 at 17:08

answered Mar 15, 2023 at 16:08

akrun

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3 Comments

Bellis Over a year ago

Unfortunately, it only outputs a matrix with NAs and the specific columns that I want to examine for correlation are not being examined at the moment :D

akrun Over a year ago

@Bellis with your data, I don't get NAs (updated with output)

Bellis Over a year ago

Ah ok, now it works, thanks. And I probably expressed myself misleadingly, I don't want the correlations between e.g. Ca and N because they are from different groups. I'm looking for the within group correlation, e.g. N and P. I'm very sorry and thank you very much for your effort!

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