Complex multiple t tests with data in 2 data.frames in R

Question

I have two data.frames (code below). The row ids in the data.frames match.

 set.seed(12345)
 df1 <- data.frame(id=c(letters),
      g1=c(rep(0,7), rep(1,18), "NA"),
      g2=c(rep("NA", 3), rep(0,23)),
      g3=c(rep(0,5), rep("NA",20),1),
      g4=c(sample(c(0,1), replace=TRUE, size=26)),
      g5=c(sample(c(0,1), replace=TRUE, size=20), "NA", sample(c(0,1), replace=TRUE, size=5)),
      g6=c(rep(1,26)),
      g7=c(sample(c(0,1), replace=TRUE, size=26)),
      g8=c(rep(0,5), rep("NA",21)),
      g9=c(rep("NA",26)),
      g10=c(rep(0,26)))
 df1[,2:11] <- sapply(df1[,2:11], as.numeric)
 df2 <- data.frame(id=c(letters),
      b1=c(runif(7, 3, 7.8), "NA", runif(18, 6, 18)),
      b2=c(runif(7, -1, 4), "NA", runif(18, 0, 5)),
      b3=c(runif(20, 0, 16), "NA", runif(5, -1, 2)),
      b4=c(runif(7, 5, 29), rep("NA", 3), runif(3, -1, 2)),
      b5=c(runif(3, 3, 8), rep("NA",23)),
      b6=c(rep("NA",21), runif(5, 0, 19)),
      b7=c(rep("NA",26)),
      b8=c(runif(26, 1, 9)),
      b9=c(runif(7, -1, 4), "NA", runif(18, -1, 4)),
      b10=c(runif(6, -1, 4), rep("NA", 2), runif(18, 0, 5)),
      b11=c(runif(7, 18, 23), "NA", runif(18, 12, 19)),
      b12=c(runif(7, 0, 4), "NA", runif(18, 0, 4)),
      b13=c(runif(7, 1, 4), "NA", runif(14, -2, 18), rep("NA", 4)),
      b14=c(runif(6, 6, 8), rep("NA", 3), runif(17, 0, 5)),
      b15=c(runif(7, 11, 12), "NA", runif(18, -1, 5)),
      b16=c(runif(7, 3, 4), "NA", runif(18, 12, 21)),
      b17=c(rep("NA", 8), runif(16, 1, 8), rep("NA", 2)))
 df2[,2:18] <- sapply(df2[,2:18], as.numeric)

I would like to use t tests to test if for each "g" column in df1, the 0 and 1 groups have significantly different values in df2.

For example, for b1:

 t.test(df2$b1[1:8], df2$b1[9:26])$p.val
   #[1] 3.846501e-07

I would like to create a new df3 with the results, which looks like this:

 df3 <- data.frame(g=rep("g1", 3),
      b=c("b1", "b2", "b3"),
      mean_0=c(mean(na.omit(df2$b1[1:8])),0,0),
      mean_1=c(mean(na.omit(df2$b1[9:26])),0,0),
      p.val=c(t.test(df2$b1[1:8], df2$b1[9:26])$p.val,1,1),
      p.adjust=c(0,1,1))
 df3 <- df3[order(df3$p.val),]

I do not know how to code this difficult problem. Can anyone help?

How do you want to handle the NA cases and the columns that only have 1 group(only 0 or only 1)? — Dave2e, Aug 09 '22 at 00:58
@Dave2e Thanks! "na.omit". I would not need them in the output. Thank you! :) — Sylvia Rodriguez, Aug 09 '22 at 01:02

Dave2e · Answer 1 · 2022-08-09T02:46:33.037

2

Here is my attempt with a traditional for loops:

#preallocate the result matrix
resultlength <- (ncol(df1)-1) * (ncol(df2)-1)
results <- data.frame(b=vector(mode= "integer", length=resultlength), g=vector(mode= "integer", length=resultlength), p = vector(mode= "numeric", length=resultlength))

#index
i=0  
for (g in (2:ncol(df1)) ) {
   gindex <- g -1
   independent <- df1[, g]
   for(col in (2:ncol(df2)) ) {
      b=col-1
      tdf <- data.frame(dependent=df2[, col], independent )
      tdf <- tdf[complete.cases(tdf), ]
   #   Check to ensure at least a 0 & 1 is present
     if (length(unique(tdf$independent)) == 2) {
         tresult <- tryCatch( t.test(tdf$dependent ~ tdf$independent)$p.value, error = function(e) 1.2)
      }
      else
      {
         tresult <-1.1  #error for lacking 2 groups
      }
      i <- i +1
      results[i,] <- c(b, gindex, tresult)
   }
}

This does have some error checking and correction. Right now it is failing if there is only a single value in one of the groups.

edited Aug 09 '22 at 02:46

answered Aug 09 '22 at 01:44

Dave2e

22,192
18
42
50

Thank you @Dave2e! Yes, you are right, not yet 100% there, but thank you for your answer. Brings me a step closer :) – Sylvia Rodriguez Aug 09 '22 at 01:54
1

@SylviaRodriguez, added a `tryCatch()`, it should now work for all cases. If there is an error with the t.test it should report a p value of 1.1 or 1.2. Please free to adjust accordingly. – Dave2e Aug 09 '22 at 02:48
Thank you very much, Dave. I will try to use this one on my real data. Best answer so far! :) – Sylvia Rodriguez Aug 09 '22 at 12:14

score 1 · Answer 2 · answered Aug 09 '22 at 01:29

One potential solution is to use purrr::map2() to iterate over the different combinations of "g" and "b" columns, e.g.

library(tidyverse)
library(broom)

set.seed(12345)
df1 <- data.frame(id=c(letters),
                  g1=c(rep(0,7), rep(1,18), "NA"),
                  g2=c(rep("NA", 3), rep(0,23)),
                  g3=c(rep(0,5), rep("NA",20),1),
                  g4=c(sample(c(0,1), replace=TRUE, size=26)),
                  g5=c(sample(c(0,1), replace=TRUE, size=20), "NA", sample(c(0,1), replace=TRUE, size=5)),
                  g6=c(rep(1,26)),
                  g7=c(sample(c(0,1), replace=TRUE, size=26)),
                  g8=c(rep(0,5), rep("NA",21)),
                  g9=c(rep("NA",26)),
                  g10=c(rep(0,26)))
df1[,2:11] <- sapply(df1[,2:11], as.numeric)

df2 <- data.frame(id=c(letters),
                  b1=c(runif(7, 3, 7.8), "NA", runif(18, 6, 18)),
                  b2=c(runif(7, -1, 4), "NA", runif(18, 0, 5)),
                  b3=c(runif(20, 0, 16), "NA", runif(5, -1, 2)),
                  b4=c(runif(7, 5, 29), rep("NA", 3), runif(3, -1, 2)),
                  b5=c(runif(3, 3, 8), rep("NA",23)),
                  b6=c(rep("NA",21), runif(5, 0, 19)),
                  b7=c(rep("NA",26)),
                  b8=c(runif(26, 1, 9)),
                  b9=c(runif(7, -1, 4), "NA", runif(18, -1, 4)),
                  b10=c(runif(6, -1, 4), rep("NA", 2), runif(18, 0, 5)),
                  b11=c(runif(7, 18, 23), "NA", runif(18, 12, 19)),
                  b12=c(runif(7, 0, 4), "NA", runif(18, 0, 4)),
                  b13=c(runif(7, 1, 4), "NA", runif(14, -2, 18), rep("NA", 4)),
                  b14=c(runif(6, 6, 8), rep("NA", 3), runif(17, 0, 5)),
                  b15=c(runif(7, 11, 12), "NA", runif(18, -1, 5)),
                  b16=c(runif(7, 3, 4), "NA", runif(18, 12, 21)),
                  b17=c(rep("NA", 8), runif(16, 1, 8), rep("NA", 2)))
df2[,2:18] <- sapply(df2[,2:18], as.numeric)

# In your example you use "t.test(df2$b1[1:8], df2$b1[9:26])$p.val"
# but these columns don't line up with the values in df1;
# you should instead use "t.test(df2$b1[1:7], df2$b1[8:25])$p.val", e.g.

t.test(df2$b1[1:7], df2$b1[8:25])$p.val
#> [1] 8.505498e-07
t.test(df2[df1$g1 == 0,]$b1, df2[df1$g1 == 1,]$b1)$p.val
#> [1] 8.505498e-07

df3 <- data.frame(g=rep("g1", 3),
                  b=c("b1", "b2", "b3"),
                  mean_0=c(mean(na.omit(df2$b1[1:8])),0,0),
                  mean_1=c(mean(na.omit(df2$b1[9:26])),0,0),
                  p.val=c(t.test(df2$b1[1:8], df2$b1[9:26])$p.val,1,1),
                  p.adjust=c(0,1,1))
df3 <- df3[order(df3$p.val),]
df3
#>    g  b   mean_0   mean_1        p.val p.adjust
#> 1 g1 b1 4.870685 12.10767 3.846501e-07        0
#> 2 g1 b2 0.000000  0.00000 1.000000e+00        1
#> 3 g1 b3 0.000000  0.00000 1.000000e+00        1

df4 <- map2(.x = rep(c("g1", "g4", "g5", "g7"), each = 4),
     .y = rep(c("b1", "b2", "b3", "b4"), times = 4),
     .f = ~tidy(t.test(df2[df1[[.x]] == 0,][[.y]],
                       df2[df1[[.x]] == 1,][[.y]])) %>%
       mutate(g = .x,
              b = .y) %>%
       select(g,
              b,
              "mean_0" = estimate1,
              "mean_1" = estimate2,
              p.value))
result <- bind_rows(df4)
result
#> # A tibble: 16 × 5
#>    g     b     mean_0 mean_1     p.value
#>    <chr> <chr>  <dbl>  <dbl>       <dbl>
#>  1 g1    b1      4.87  12.0  0.000000851
#>  2 g1    b2      2.41   2.74 0.584      
#>  3 g1    b3      7.82   7.53 0.907      
#>  4 g1    b4     19.1   11.5  0.0611     
#>  5 g4    b1     10.7    9.85 0.683      
#>  6 g4    b2      2.59   2.78 0.798      
#>  7 g4    b3      7.84   7.11 0.782      
#>  8 g4    b4     12.8   13.9  0.822      
#>  9 g5    b1     10.1    9.81 0.883      
#> 10 g5    b2      2.48   2.81 0.614      
#> 11 g5    b3      8.05   7.06 0.743      
#> 12 g5    b4     10.8   14.6  0.471      
#> 13 g7    b1      9.52  11.3  0.325      
#> 14 g7    b2      2.57   3.04 0.396      
#> 15 g7    b3      8.20   5.51 0.318      
#> 16 g7    b4     16.5    8.31 0.0923

^{Created on 2022-08-09 by the reprex package (v2.0.1)}

NB. This only works for columns where the t.test() doesn't return an error. In this example I 'handpicked' the "g" and "b" columns that aren't all zeros / all ones / all NAs so the example would complete, so you may need to alter your approach for your actual data.

Thank you, Jared! Yes, this works. You are right, I will still have to find a solution for the NAs. That's part of my problem. Thanks for your answer! — Sylvia Rodriguez, Aug 09 '22 at 01:37

Complex multiple t tests with data in 2 data.frames in R

2 Answers2