使用循环计算列的累积和

Question

I have a dataframe with gene expression data by lane (column). What I would like to do is write a loop that takes the sum of each row but progressively adds in another column each time. So each time I loop through I add another column to my dataframe that contains the sums of each row plus another column to the end of the dataframe. In the example below I did this using the apply() function by hand but this is very inefficient and not feasible for a large data set. I messed around with the cumsum() function but couldn't seem to get it to work for this. Very possible I missed something obvious but any guidance would be great!

#Example dataframe

c1 <- c('G1', 'G2', 'G3')
c2 <- c(5, 3, 1)
c3 <- c(3, 7, 1)
c4 <- c(6, 3, 4)
c5 <- c(6, 4, 3)
df <- data.frame(c1, c2, c3, c4, c5)

#Cal cumulative sums
sum.2.3 <- apply(df[,2:3],1,sum)
sum.2.4 <- apply(df[,2:4],1,sum)
sum.2.5 <- apply(df[,2:5],1,sum)

df <- cbind(df, sum.2.3, sum.2.4, sum.2.5)

Answer 1

If the problem is the loop, you use apply inside it.

Code

start_col <- 2

end_col <- ncol(df)

for(i in (start_col+1):end_col){
  
  var_name <- paste("sum",start_col,i,sep = ".")
  
  df[,var_name] <- apply(df[,start_col:i],1,sum)
  
}

Output

  c1 c2 c3 c4 c5 sum.2.3 sum.2.4 sum.2.5
1 G1  5  3  6  6       8      14      20
2 G2  3  7  3  4      10      13      17
3 G3  1  1  4  3       2       6       9

Answer 2

You can use Reduce()

Reduce(`+`, df[-1], accumulate = TRUE)[-1]

[[1]]
[1]  8 10  2

[[2]]
[1] 14 13  6

[[3]]
[1] 20 17  9

Assign into the data frame:

df[paste0("sum.2.", 3:5)] <-  Reduce(`+`, df[-1], accumulate = TRUE)[-1]

Gives:

  c1 c2 c3 c4 c5 sum.2.3 sum.2.4 sum.2.5
1 G1  5  3  6  6       8      14      20
2 G2  3  7  3  4      10      13      17
3 G3  1  1  4  3       2       6       9

Answer 3

No loop needed.

df <- data.frame(
    c1 = c('G1', 'G2', 'G3'),
    c2 = c(5, 3, 1),
    c3 = c(3, 7, 1),
    c4 = c(6, 3, 4),
    c5 = c(6, 4, 3))

cbind(df, setNames(as.data.frame(t(apply(df[,-1], 1, cumsum))[,-1]), paste0("sum.2.", 3:5)))

#>   c1 c2 c3 c4 c5 sum.2.3 sum.2.4 sum.2.5
#> 1 G1  5  3  6  6       8      14      20
#> 2 G2  3  7  3  4      10      13      17
#> 3 G3  1  1  4  3       2       6       9

Answer 4

Using rowCumsums from matrixStats

library(matrixStats)
df[paste0("sum.2.", 3:5)] <- rowCumsums(as.matrix(df[2:5]))[,-1]

-output

> df
  c1 c2 c3 c4 c5 sum.2.3 sum.2.4 sum.2.5
1 G1  5  3  6  6       8      14      20
2 G2  3  7  3  4      10      13      17
3 G3  1  1  4  3       2       6       9

Answer 5

You can use both the mutate function from the dplyr package and the rowSums base function.

library(dplyr)

c1 <- c('G1', 'G2', 'G3')
c2 <- c(5, 3, 1)
c3 <- c(3, 7, 1)
c4 <- c(6, 3, 4)
c5 <- c(6, 4, 3)
df <- data.frame(c1, c2, c3, c4, c5)

df <- df %>% 
  dplyr::mutate(sum.2.3 = rowSums(across(c2:c3)),
                sum.2.4 = rowSums(across(c2:c4)),
                sum.2.5 = rowSums(across(c2:c5)))

Result

  c1 c2 c3 c4 c5 sum.2.3 sum.2.4 sum.2.5
1 G1  5  3  6  6       8      14      20
2 G2  3  7  3  4      10      13      17
3 G3  1  1  4  3       2       6       9