Iterate over data frame for calculation

user_6122293 0
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I have a data frame with two rows that represent passengers loading onto the bus and passengers leaving the bus:

    A  B  C  D  E  F
In  9  10 6  9  14 10
Out 0  1  2  3  4  3

And I wanted to do a calculation that will result in two more rows of information, where the first is passengers who are on the bus when it arrives at Station A/B/C/etc, and Row 2 is the number of passengers who have left the bus at that station.

The numbers in row 1 should be the same from the previous number in row 2, and row 2 for Station B, for example, is `9 (the number of people who remained on the bus from the previous stop) + BIn (the number of people getting on at the stop) - BOut (the number of people who are getting off the bus at the stop).

The final result should look like:

    A  B  C  D  E  F
In  9  10 6  9  14 10
Out 0  1  2  3  4  3
1   0  9  18 22 28 38
2   9  18 22 28 38 45

How would I iterate through the data frame so that I can obtain these numbers? Is a for loop needed, or is there an easier way to go through this calculation?

dataframe r for-loop
3 Answers

First, I think it makes more sense to have these as columns rather than rows. That way you can take advantage of the vectorized operations in R.

library(data.table)
df <- suppressWarnings(fread('
    A  B  C  D  E  F
In  9  10 6  9  14 10
Out 0  1  2  3  4  3'))
setDT(df) # only required if not starting with a data.table
df
#>        V1     A     B     C     D     E     F
#>    <char> <int> <int> <int> <int> <int> <int>
#> 1:     In     9    10     6     9    14    10
#> 2:    Out     0     1     2     3     4     3

df_tp <- transpose(df, make.names = 'V1', keep.names = 'station')
df_tp
#>    station    In   Out
#>     <char> <int> <int>
#> 1:       A     9     0
#> 2:       B    10     1
#> 3:       C     6     2
#> 4:       D     9     3
#> 5:       E    14     4
#> 6:       F    10     3

Now your last row is the cumulative sum of In minus the cumulative sum of Out. The other one is just a lagged version of that.

df_tp[, net := cumsum(In) - cumsum(Out)]
df_tp[, lagged_net := shift(net, fill = 0)]
df_tp
#>    station    In   Out   net lagged_net
#>     <char> <int> <int> <int>      <int>
#> 1:       A     9     0     9          0
#> 2:       B    10     1    18          9
#> 3:       C     6     2    22         18
#> 4:       D     9     3    28         22
#> 5:       E    14     4    38         28
#> 6:       F    10     3    45         38

Created on 2021-12-07 by the reprex package (v2.0.1)

user_67627550

I think you should take @IceCreamToucan's advice and answer, but if you want to keep the same structure for specific reasons, this inelegant, brute-force for loop will produce your desired output:

df <- data.frame(A = c(9,0),
                 B = c(10,1),
                 C = c(6,2),
                 D = c(9,3),
                 E = c(14, 4),
                 F = c(10, 3))

for (i in 1:ncol(df)){
  if (i == 1){df[3:4,1] <- c(0,df[1,1])}
  else{
    df[3,i] <- df[4,i-1]
    df[4,i] <- sum(df[4,i-1], df[1,i]) - df[2,i]
    }
}
df
#  A  B  C  D  E  F
#1 9 10  6  9 14 10
#2 0  1  2  3  4  3
#3 0  9 18 22 28 38
#4 9 18 22 28 38 45
user_121097880

Or for a tidyverse way to do it:

Load data in the format you shared it:
library(tidyverse)

df <- data.frame(A = c(9,0), 
                 B = c(10,1), 
                 C = c(6,2), 
                 D = c(9,3), 
                 E = c(14,4), 
                 F = c(10,3)) 

> df
  A  B C D  E  F
1 9 10 6 9 14 10
2 0  1 2 3  4  3
Transform to long format:
df <- as_tibble(t(df), rownames = "row_names") %>% 
      rename('In' = V1, 'Out' = V2)

> df

# A tibble: 6 x 3
  row_names    In   Out
  <chr>     <dbl> <dbl>
1 A             9     0
2 B            10     1
3 C             6     2
4 D             9     3
5 E            14     4
6 F            10     3
Add variables you want with cumsum and lag:
df %>% mutate(net = cumsum(In) - cumsum(Out), 
              lag = replace_na(lag(net), 0))

> df

# A tibble: 6 x 5
  row_names    In   Out   net   lag
  <chr>     <dbl> <dbl> <dbl> <dbl>
1 A             9     0     9     0
2 B            10     1    18     9
3 C             6     2    22    18
4 D             9     3    28    22
5 E            14     4    38    28
6 F            10     3    45    38
user_52127420
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