Calculate correlation with cor only for numerical columns

Question

I have a dataframe and would like to calculate the correlation  with Spearman  data is categorical and ranked  but only for a subset of columns  I tried with all  but R s cor   function only accepts numerical data  x must be numeric  says the error message   even if Spearman is used   One brute approach is to delete the non-numerical columns from the dataframe  This is not as elegant  for speed I still don t want to calculate correlations between all columns    I hope there is a way to simply say  calculate correlations for columns x  y  z   Column references could by number or by name  I suppose the flexible way to provide them would be through a vector   Any suggestions are appreciated

User · Accepted Answer

if you have a dataframe where some columns are numeric and some are other (character or factor) and you only want to do the correlations for the numeric columns, you could do the following:

set.seed(10)

x = as.data.frame(matrix(rnorm(100), ncol = 10))
x$L1 = letters[1:10]
x$L2 = letters[11:20]

cor(x)

Error in cor(x) : 'x' must be numeric

but

cor(x[sapply(x, is.numeric)])

             V1         V2          V3          V4          V5          V6          V7
V1   1.00000000  0.3025766 -0.22473884 -0.72468776  0.18890578  0.14466161  0.05325308
V2   0.30257657  1.0000000 -0.27871430 -0.29075170  0.16095258  0.10538468 -0.15008158
V3  -0.22473884 -0.2787143  1.00000000 -0.22644156  0.07276013 -0.35725182 -0.05859479
V4  -0.72468776 -0.2907517 -0.22644156  1.00000000 -0.19305921  0.16948333 -0.01025698
V5   0.18890578  0.1609526  0.07276013 -0.19305921  1.00000000  0.07339531 -0.31837954
V6   0.14466161  0.1053847 -0.35725182  0.16948333  0.07339531  1.00000000  0.02514081
V7   0.05325308 -0.1500816 -0.05859479 -0.01025698 -0.31837954  0.02514081  1.00000000
V8   0.44705527  0.1698571  0.39970105 -0.42461411  0.63951574  0.23065830 -0.28967977
V9   0.21006372 -0.4418132 -0.18623823 -0.25272860  0.15921890  0.36182579 -0.18437981
V10  0.02326108  0.4618036 -0.25205899 -0.05117037  0.02408278  0.47630138 -0.38592733
              V8           V9         V10
V1   0.447055266  0.210063724  0.02326108
V2   0.169857120 -0.441813231  0.46180357
V3   0.399701054 -0.186238233 -0.25205899
V4  -0.424614107 -0.252728595 -0.05117037
V5   0.639515737  0.159218895  0.02408278
V6   0.230658298  0.361825786  0.47630138
V7  -0.289679766 -0.184379813 -0.38592733
V8   1.000000000  0.001023392  0.11436143
V9   0.001023392  1.000000000  0.15301699
V10  0.114361431  0.153016985  1.00000000

User · Answer

I found an easier way by looking at the R script generated by Rattle  It looks like below   correlations  lt - cor mydata  c 1 3 5 87 89 90 94 98    use  pairwise   method  spearman

User · Answer

Another option would be to just use the excellent corrr package https   github com drsimonj corrr and do  require corrr  require dplyr   myData   gt       select x y z    gt      or do negative or range selections here    correlate     gt      rearrange     gt      rearrange by correlations    shave     Shave off the upper triangle for a cleaner result   Steps 3 and 4 are entirely optional and are just included to demonstrate the usefulness of the package

User · Answer

For numerical data you have the solution  But it is categorical data  you said  Then life gets a bit more complicated     Well  first   The amount of association between two categorical variables is not measured with a Spearman rank correlation  but with a Chi-square test for example   Which is logic actually  Ranking means there is some order in your data  Now tell me which is larger  yellow or red  I know  sometimes R does perform a spearman rank correlation on categorical data  If I code yellow 1 and red 2  R would consider red larger than yellow   So  forget about Spearman for categorical data  I ll demonstrate the chisq-test and how to choose columns using combn    But you would benefit from a bit more time with Agresti s book   http   www amazon com Categorical-Analysis-Wiley-Probability-Statistics dp 0471360937  set seed 1234  X  lt - rep c  A   B   20  Y  lt - sample c  C   D   40 replace T   table X Y  chisq test table X Y  correct F    I don t use Yates continuity correction   Let s make a matrix with tons of columns  Data  lt - as data frame            matrix              sample letters 1 3  2000 replace T               ncol 25                          You want to select which columns to use columns  lt - c 3 7 11 24  vars  lt - names Data  columns     say you need to know which ones are associated with each other  out  lt -  apply  combn columns 2  2 function x             chisq test table Data  x 1   Data  x 2    correct F  p value             out  lt - cbind as data frame t combn vars 2    out    Then you should get     gt  out    V1  V2       out 1  V3  V7 0 8116733 2  V3 V11 0 1096903 3  V3 V24 0 1653670 4  V7 V11 0 3629871 5  V7 V24 0 4947797 6 V11 V24 0 7259321   Where V1 and V2 indicate between which variables it goes  and  out  gives the p-value for  association  Here all variables are independent  Which you would expect  as I created the data at random

[r] Calculate correlation with cor(), only for numerical columns

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