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Alejandro Carrera
Asked: 2020-04-25 17:08:50 +0800 CST 2020-04-25 17:08:50 +0800 CST

How to add across rows?

  • 772

I am trying to create a function, let's say "generic" to add rows in a database.

My idea is that the function "recognizes" the numeric elements in the database and adds by rows.

Suppose I have the data of the following image:

Screenshot: enter image description here

The function would take from the second column and add 2+3+3+6+4+2+0 and so on.

The code I have written so far is as follows:

suma_renglones < - function (x) {
  y = integer(nrow(x))
  for (i in 1: nrow(x)) {
    for (j in 1: ncol(x)) {
      if (is.numeric(x[i, j])) {
        y[i] = y[i] + (x[i, j])
      }
    }
    y[i]
  }
}

However, when applying the function, the column designated to store the result of the sum shows only NULL results. Any help or guidance will be greatly appreciated.

PS: I know there are already solutions about it elsewhere as shown in this example , however my intention is to make a "own" function.

r

3 Answers

  1. Best Answer

    Let's see, you say that you are adding in a column of the original matrix, but you are not doing that, you are simply creating a vector with the sums y, also remember that your routine cannot modify an object passed by parameter because it is outside its "scope ", what can be done, is to return a new matrix with the additional column and lastly "step" on the original object

    Reformulating your code a bit

    suma_renglones <- function (x) {
    y=integer(nrow(x))
    for(i in 1:nrow(x)){
        for (j in 1:ncol(x)) {
            if(is.numeric(x[i,j])) 
            {
                y[i] = y[i]+(x[i,j])
            }
        }
    }
    x["suma"] <- y
    return(x)
}
d = suma_renglones(d)
d

    Return

            A        B       C N1 N2 N3 N4 N5 N6 N7 N8 N9 N10 suma
1  Apples  Bananas Peaches  9  2  3  4  5  6  7  8  9  10   63
2 Oranges Kumquats   Plums  1  1  3  5  5  5  7  6  6  12   51
3   Pears    Kiwis  Grapes  2  4  6  8  5  6  7  6  9   1   54

    What are we doing?

    • At the end of the cycles I have a vector ythat contains the sum of each row
    • I add to the matrix xa new column "sum" with the sums
    • Return xsince the function cannot modify the original array, being out of its "scope"
    • 1

  2. Here I have a solution using dplyrand tidyrwithout loops, so it should be faster:

    I build an example with the first 2 lines

    df <- data.frame(
  x = c("employee_1", "employee_2"),
  day_1 = c(2, 19),
  day_2 = c(3, 23),
  day_3 = c(3, 18))

         x day_1 day_2 day_3
employee_1     2     3     3
employee_2    19    23    18

    First I have to "transpose" the dataframe, which I do with gatherand spreadto calculate the sum of each column, and then I have to transpose the result again and do a inner_joinwith the original table. At last, what I do is change the name of the columns and selectorder the result so that the sum remains as the last column.

    library(dplyr)
library(tidyr)

df %>%
  gather(var, val, 2:ncol(df)) %>%
  spread(x, val) %>% 
  select(-1) %>% 
  summarise_each(funs(sum)) %>% 
  gather() %>% 
  inner_join(df, by = c("key" = "x")) %>% 
  rename(suma = value,
         x = key) %>% 
  select(-suma, everything())

         x day_1 day_2 day_3 suma
employee_1     2     3     3    8
employee_2    19    23    18   60
    • 1

  3. Dear Patrick,

    The solution I designed was this:

        suma_renglones <- function (x) {
y= integer(nrow(x))
for (i in 1: nrow(x)) {
for (j in 1: ncol(x)) {
if(is.numeric(x[[i,j]])
{
y[i] <- y[[i]] + x[[i,j]]
}
}
cat (y[i], file= "temp", "\n")
}
temp <- cbind(y)
x <- cbind(x, temp)
}

    Comparing it to your code, I think I didn't do too badly in the end, right? However, your solution is clearly much cleaner and more efficient. I really appreciate the time you took to help me. Greetings again :)

    • 0