Save the output of a for loop in R

Question

Suppose I have a binomial distribution where n=12, p=0.2. I split this sample into 4 chunks(groups), each chunk has group size 3. Then I remove the output whose sum is equal to 0. For the remaining outputs, what I'm trying to do is combining all remaining outputs into a new vector. Here's my code

set.seed(123)
sample1=rbinom(12,1,0.2)
chuck2=function(x,n)split(x,cut(seq_along(x),n,labels=FALSE))
chunk=chuck2(sample1,4)
for (i in 1:4){
  aa=chunk[[i]]
  if (sum(aa)!=0){
    a.no0=aa
    print(a.no0)
  }
}

And here's the output:

[1] 1 1 0
[1] 0 1 0
[1] 0 1 0

I want to combine these three outputs into a new vector like:

[1] 1 1 0 0 1 0 0 1 0

but I have no idea how it works, any hints please?

The chunk2 function seems unnecessary - it largely ends up making a matrix. See my solution for something that is easier to follow. — Cole
– Cole, Commented Jun 10, 2019 at 0:25

heds1 · Accepted Answer · 2019-06-09 20:55:58Z

2

set.seed(123)
sample1=rbinom(12,1,0.2)
chuck2=function(x,n)split(x,cut(seq_along(x),n,labels=FALSE))
chunk=chuck2(sample1,4)  

int_vector <- c()

for (i in 1:4){
    aa=chunk[[i]]
    if (sum(aa)!=0){
        a.no0=aa
        int_vector <- c(int_vector, a.no0)
    }
}

int_vector
# [1] 1 1 0 0 1 0 0 1 0

edited Jun 9, 2019 at 20:55

answered Jun 9, 2019 at 20:20

heds1

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4 Comments

Jim O. Over a year ago

simple and straightforward. I like your answer better than mine : )

heds1 Over a year ago

Thanks =) making a list and then unlisting it was my first thought too. But it seems like there's always a better way!

Jim O. Over a year ago

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NelsonGon Over a year ago

Better to rename vector to avoid potential clashes with vector(the function). IDE's generally deal with this, and apparently R but it is in my opinion better to avoid such names altogether.

Jim O. · Accepted Answer · 2019-06-09 20:19:58Z

0

Create a list() and assign it a variable name. Next, you add that variable inside the loop, then append the looping values in the list.

new_vector <- list()

for (i in 1:4){
  aa=chunk[[i]]
  if (sum(aa)!=0){
    a.no0=aa
    new_vector <- append(new_vector, a.no0)
  }
}
new_vector

This will return:

[[1]]
[1] 1

[[2]]
[1] 1

[[3]]
[1] 0

[[4]]
[1] 0

[[5]]
[1] 1

[[6]]
[1] 0

[[7]]
[1] 0

[[8]]
[1] 1

[[9]]
[1] 0

But I think you want a flattened vector:

as.vector(unlist(new_vector))

[1] 1 1 0 0 1 0 0 1 0

answered Jun 9, 2019 at 20:19

Jim O.

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Comments

B Williams · Accepted Answer · 2019-06-09 20:25:09Z

0

Doesn't directly address your issue, but this can be accomplished without a for-loop:

library(dplyr)
set.seed(123)
sample1 <- rbinom(12, 1, 0.2)

as.data.frame(matrix(sample1, ncol = 3, byrow = TRUE)) %>% 
  mutate(test = rowSums(.), id = 1:n()) %>% 
  filter(test > 0) %>% 
  dplyr::select(-test) %>% 
  gather(key, value, -id) %>% 
  arrange(id, key) %>% 
  .$value

answered Jun 9, 2019 at 20:25

B Williams

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Comments

abcxyz · Accepted Answer · 2019-06-09 21:16:18Z

0

Two versions without for loop.

data:

set.seed(123)
sample1 <- rbinom(12, 1, 0.2)

base-R functional version:

split.sample1 <- split(sample1,cut(seq_along(sample1),4,labels=FALSE))
sumf <- function(x) if(sum(x) == 0) NULL else x
result <- unlist(lapply(split.sample1,sumf),use.names=F)

> result
[1] 1 1 0 0 1 0 0 1 0

modern use of pipe %>% operator version:

library(magrittr) # for %>% operator
grp.indx <- cut(seq_along(sample1),4,labels=FALSE)
split.sample1 <- sample1 %>% split(grp.indx)
result <- split.sample1 %>% lapply(sumf) %>% unlist(use.names=F)

> result
[1] 1 1 0 0 1 0 0 1 0

edited Jun 9, 2019 at 21:16

answered Jun 9, 2019 at 21:11

abcxyz

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Comments

Cole · Accepted Answer · 2019-06-10 00:20:53Z

It seems like your function makes a pseudo matrix as a list. This instead directly makes a matrix from sample1 and then outputs a vector where rowSums are greater than 0.

set.seed(123)
sample1 = rbinom(12, 1, 0.2)

chunk_mat = matrix(sample1, ncol = 3, byrow = T)

as.vector(t(chunk_mat[which(rowSums(chunk_mat) != 0), ]))

Here are benchmarks - I have the chuck2 in the global environment but each function still has to generate the chunk dataframe / matrix / list so that they're apples to apples.

Unit: microseconds
            expr      min        lq       mean    median        uq       max neval
     cole_matrix   19.902   26.2515   38.60094   43.3505   47.4505    56.801   100
 heds_int_vector 4965.201 5101.9010 5616.53893 5251.8510 5490.9010 23417.401   100
 bwilliams_dplyr 5278.602 5506.4010 5847.55298 5665.7010 5821.5515  9413.801   100
      Simon_base  128.501  138.0010  196.46697  185.6005  203.1515  2481.101   100
  Simon_magrittr  366.601  392.5005  453.74806  455.1510  492.0010   739.501   100

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