Styjun

First I create a pair of example dataframes:

df = data.frame("sample1" = runif(10), "sample2" = runif(10), "sample3" = runif(10), "sample4" = runif(10))
traits = data.frame("var1" = c(rep("group1", 2), rep("group2", 2)))
rownames(traits) = colnames(df)

If I create a formula as a text string, I can plug it right into lm()

> row = t(df[1,])
> ModString = "row ~ traits$var1"
> Mod = lm(as.formula(ModString))
> Mod

Call:
lm(formula = as.formula(ModString))

Coefficients:
 (Intercept) traits$var1group2 
 0.7799 0.1788 

But if I try to do the same thing with parLapply, I get an error indicating that the "traits" argument was not working as expected:

> num_cores <- detectCores() - 1
> cl <- makeCluster(num_cores)
> results <- parLapply(cl = cl, seq(1:10), function(i, df, traits)
+ row = df[i,]
+ ModString = "vector ~ traits$factor1"
+ Mod = lm(ModString)
+ return(Mod)
+ , df = df, traits = traits)
Error in checkForRemoteErrors(val) : 
 9 nodes produced errors; first error: object 'traits' not found

But what is strange is that the "traits" argument IS making it into the parLapply as I am using, it seems to be something about the way lm() works that is the problem. I can input and return "traits" just fine:

> cl <- makeCluster(num_cores)
> results <- parLapply(cl = cl, seq(1:10), function(i, df, traits)
+ row = df[i,]
+ traits2 = traits
+ ModString = "vector ~ traits$factor1"
+ return(list(traits2, row, ModString))
+ , df = df, traits = traits)
> results
[[1]]
[[1]][[1]]
 var1
sample1 group1
sample2 group1
sample3 group2
sample4 group2

[[1]][[2]]
 sample1 sample2 sample3 sample4
1 0.6941108 0.8656177 0.9807334 0.936609

[[1]][[3]]
[1] "vector ~ traits$factor1"


[[2]]
[[2]][[1]]
 var1
sample1 group1
sample2 group1
sample3 group2
sample4 group2

[[2]][[2]]
 sample1 sample2 sample3 sample4
2 0.1007983 0.5599374 0.0208095 0.8082196

[[2]][[3]]
[1] "vector ~ traits$factor1"


[[3]]
[[3]][[1]]
 var1
sample1 group1
sample2 group1
sample3 group2
sample4 group2

[[3]][[2]]
 sample1 sample2 sample3 sample4
3 0.9633059 0.7564143 0.913617 0.4179525

[[3]][[3]]
[1] "vector ~ traits$factor1"


[[4]]
[[4]][[1]]
 var1
sample1 group1
sample2 group1
sample3 group2
sample4 group2

[[4]][[2]]
 sample1 sample2 sample3 sample4
4 0.06625104 0.390351 0.511572 0.8386714

[[4]][[3]]
[1] "vector ~ traits$factor1"


[[5]]
[[5]][[1]]
 var1
sample1 group1
sample2 group1
sample3 group2
sample4 group2

[[5]][[2]]
 sample1 sample2 sample3 sample4
5 0.6135228 0.4926991 0.08513074 0.105647

[[5]][[3]]
[1] "vector ~ traits$factor1"


[[6]]
[[6]][[1]]
 var1
sample1 group1
sample2 group1
sample3 group2
sample4 group2

[[6]][[2]]
 sample1 sample2 sample3 sample4
6 0.7121677 0.6554129 0.6409468 0.4906039

[[6]][[3]]
[1] "vector ~ traits$factor1"


[[7]]
[[7]][[1]]
 var1
sample1 group1
sample2 group1
sample3 group2
sample4 group2

[[7]][[2]]
 sample1 sample2 sample3 sample4
7 0.4651641 0.546514 0.4039608 0.1758802

[[7]][[3]]
[1] "vector ~ traits$factor1"


[[8]]
[[8]][[1]]
 var1
sample1 group1
sample2 group1
sample3 group2
sample4 group2

[[8]][[2]]
 sample1 sample2 sample3 sample4
8 0.5121237 0.4950444 0.9662431 0.6851582

[[8]][[3]]
[1] "vector ~ traits$factor1"


[[9]]
[[9]][[1]]
 var1
sample1 group1
sample2 group1
sample3 group2
sample4 group2

[[9]][[2]]
 sample1 sample2 sample3 sample4
9 0.2486208 0.135422 0.2128657 0.7332921

[[9]][[3]]
[1] "vector ~ traits$factor1"


[[10]]
[[10]][[1]]
 var1
sample1 group1
sample2 group1
sample3 group2
sample4 group2

[[10]][[2]]
 sample1 sample2 sample3 sample4
10 0.06203028 0.7916495 0.3528376 0.2259685

[[10]][[3]]
[1] "vector ~ traits$factor1"

What embarrassingly trivial detail am I missing here?

I would do it this way; note the radically different data organization:

library(dplyr)
library(tidyr)
library(tibble)
library(parallel)

#You seem to have rows of data that should be columns,
# this puts things in a form more suitable for work in R
df_new <- df %>% 
 mutate(row = 1:n()) %>% 
 gather(key = sample,value = val,sample1:sample4) %>% 
 arrange(row,sample)

#Data in rownames is not terribly useful
traits_new <- rownames_to_column(traits,"sample")

#Now we can put it all in *one* data frame
df_new <- left_join(df_new,
 traits_new,
 by = "sample")

#...and split it into a list representing each of the df's you
# want a lm() fit on
df_new_split <- split(df_new,df_new$row)

#Wrapper for lm with the only formula we need
fit_lm <- function(x)
 lm(val ~ var1,data = x)


num_cores <- detectCores() - 1
cl <- makeCluster(num_cores)

results <- parLapply(cl = cl,df_new_split,fit_lm)

OK I feel really silly but I'm going to leave the question up because it's a great example of how easy it is to get confused when copy-pasting and editing multiple versions of code. I did not consistently use as.formula in my parLapply, and also forgot to change variable name vector to row and to transpose it.

So. The following works just dandy:

require(parallel)
num_cores <- detectCores() - 1
cl <- makeCluster(num_cores)
results <- parLapply(cl = cl, seq(1:10), function(i, df, traits)
 row = t(df[i,])
 ModString = "row ~ traits[,"var1"]"
 Mod = lm(as.formula(ModString))
 return(Mod)
, df = df, traits = traits)