如何计算每行每列的周期总和并将其除以 R 中的总最小值

Question

I have a calculation where I accumulate the sum of each flower in the table below for each year. However, I would like to use the same calculation to also divide the accumulated values by a total minimum value.

(Data Frame)

 dat <- data.frame(
  stringsAsFactors = FALSE,
  flower = c("lily", "rose", "daisy"),
  x1902 = c(23L, 50L, 30L),
  x1950 = c(23L, 110L, 37L),
  x2010 = c( 23L, 115L, 47L),
  x2012 = c( 31L, 131L, 49L),
  x2021 = c( 36L, 131L, 49L),
  total_min = c(5L, 5L, 2L)

The code that I used to do the accumulation as follows:

dat[, 2:6] <- t(apply(dat[,2:6], 1, cumsum))

So now I need to divide the accumulated amount by the total min for each flower type for each year. I tried the code as follows but I am not getting the correct results:

dat[,2:6] <- t(apply(dat[,2:6], 1, cumsum)/dat$total_min)

(Table 1 - DataFrame)

flower	x1902	x1950	x2010	x2012	x2021	total_min
lily	23	0	0	8	5	5
rose	50	60	5	16	0	5
daisy	30	7	10	2	0	2

Calculating the sum for each flower in each year. The end result gives me:

(Table 2 - Accumulated results )

flower	x1902	x1950	x2010	x2012	x2021
lily	23	23	23	31	36
rose	50	110	115	131	131
daisy	30	37	47	49	49

The final result should look like Table 3

(Table 3 - expected results)

flower	x1902	x1950	x2010	x2012	x2021
lily	4.6	4.6	4.6	6.2	7.2
rose	10	22	23	26.2	26.2
daisy	15	18.5	23.5	24.5	24.5

Answer 1

The problems with the code in the question are:

• the final ) is in the wrong place
• although not wrong it would be less error prone to create a new version of the data with a different name.
• also it would be a bit more robust if we extract the columns whose names start with x rather than hard coding 2:6.

1) Using dat0 shown in the Note at the end we have:

xnms <- startsWith(names(dat0), "x")
dat2 <- replace(dat0, xnms, t(apply(dat0[xnms], 1, cumsum)) / dat0$total_min)
dat2
##   flower x1902 x1950 x2010 x2012 x2021 total_min
## 1   lily   4.6   4.6   4.6   6.2   7.2         5
## 2   rose  10.0  22.0  23.0  26.2  26.2         5
## 3  daisy  15.0  18.5  23.5  24.5  24.5         2

2) Another approach is Reduce :

replace(dat0, xnms, 
  as.data.frame(Reduce("+", dat0[xnms], acc = TRUE)) / dat0$total_min)

Note

dat0 <-
structure(list(flower = c("lily", "rose", "daisy"), x1902 = c(23L, 
50L, 30L), x1950 = c(0L, 60L, 7L), x2010 = c(0L, 5L, 10L), x2012 = c(8L, 
16L, 2L), x2021 = c(5L, 0L, 0L), total_min = c(5L, 5L, 2L)),
class = "data.frame", row.names = c(NA, -3L))

dat0
##   flower x1902 x1950 x2010 x2012 x2021 total_min
## 1   lily    23     0     0     8     5         5
## 2   rose    50    60     5    16     0         5
## 3  daisy    30     7    10     2     0         2