Sum Two Columns in R

sum two columns in R

The sum function will add all numbers together to produce a single number, not a vector (well, at least not a vector of length greater than 1).

It looks as though at least one of your columns is a factor. You could convert them into numeric vectors by checking this

head(as.numeric(data$col1))  # make sure this gives you the right output

And if that looks right, do

data$col1 <- as.numeric(data$col1)
data$col2 <- as.numeric(data$col2)

You might have to convert them into characters first. In which case do

data$col1 <- as.numeric(as.character(data$col1))
data$col2 <- as.numeric(as.character(data$col2))

It's hard to tell which you should do without being able to see your data.

Once the columns are numeric, you just have to do

data$col3 <- data$col1 + data$col2

summing multiple columns in an R data-frame quickly

Here's an alternative approach using tidyverse:

library(tidyverse)

# input columns of interest
cols = c("mpg", "cyl", "disp", "hp", "drat")

mtcars %>% 
  group_by(id = row_number()) %>%  # for each row
  nest(cols) %>%                   # nest selected columns
  mutate(SUM = map_dbl(data, sum)) # calculate the sum of those columns

# # A tibble: 32 x 3
#      id data               SUM
#   <int> <list>           <dbl>
# 1     1 <tibble [1 x 5]>  301.
# 2     2 <tibble [1 x 5]>  301.
# 3     3 <tibble [1 x 5]>  232.
# 4     4 <tibble [1 x 5]>  398.
# 5     5 <tibble [1 x 5]>  565.
# 6     6 <tibble [1 x 5]>  357.
# 7     7 <tibble [1 x 5]>  631.
# 8     8 <tibble [1 x 5]>  241.
# 9     9 <tibble [1 x 5]>  267.
# 10    10 <tibble [1 x 5]>  320.
# # ... with 22 more rows

The output here is a data frame containing the row id (id), the data used at each row (data) and the calculated sum (SUM).

You can get a vector of the calculated SUM if you add ... %>% pull(SUM).

How to sum multiple columns in two data frames in r

Here's a base R option :

tmp <- cbind(df1, df2)
data.frame(sapply(split.default(tmp, names(tmp)), rowSums))

#  V1 V2 V3 V4 V5
#1  4  8  5  5  4
#2  6 10  7  7  0

data

df1 < -structure(list(V1 = 2:3, V2 = 4:5, V3 = c(5L, 7L)), 
class = "data.frame", row.names = c(NA, -2L))

df2 <- structure(list(V1 = 2:3, V5 = c(4L, 0L), V2 = 4:5, V4 = c(5L, 
7L)), class = "data.frame", row.names = c(NA, -2L))

Conditional cumulative sum from two columns

You were essentially in the right direction. Since you provide an .init value to accumulate, the resulting vector is of size n+1, with the first value being .init. You have to remove the first value to get a vector that fit to your column size.

Then, if you want NAs on the remaining values, here's a way to do it. Also, since the "starting row" is the third, .init has to be set to 8.

df %>%
  mutate(test = 
           ifelse(source == "B", accumulate(add, .init = 8, ~.x + .y)[-1], NA))

# A tibble: 6 x 4
  source value   add  test
  <chr>  <dbl> <dbl> <dbl>
1 A          5     1    NA
2 A         10     1    NA
3 B         NA     1    11
4 B         NA     2    13
5 B         NA     3    16
6 C         20     4    NA

join and sum columns together R

Using plyr and dplyr you can do this:

df %>% 
  rowwise() %>% 
  mutate(f_new=sum(f, f2, na.rm = T))

# A tibble: 6 x 5
#   ca     f    f2    f3   f_new
#  <fct> <dbl> <dbl> <dbl> <dbl>
#1   a     3    NA     3     3
#2   b     4     5     0     9
#3   a     0     6     6     6
#4   c    NA     1     3     1
#5   b     3     9     0    12
#6   b     4     7     8    11

This method will retain and NA values

R: data.table group and sum two columns

From your comments it seems like you are getting the desired result but in scientific notation. Try rounding to 3 decimals if you want to see it as you say:

DT[, lapply(.SD, function(x) round(sum(x), 3)), by = c("PARK", "WTG")]

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