multiply each row of a dataframe by it's vector R
Question
What would be the easiest way (if possible with tidyverse) to multiply each columns x1:x10 with their respective vector. For example: the first row of the new table would be: age = "one", x1 = x1 * 1, x2 = x2 * 2, x3 = x3 * 9, x4 = x4 * 4...etc
x <- data.frame(age = c("one", "two", "three", "four", "five","one", "two", "three", "four", "five"),
replicate(10,sample(0:5,5,rep=TRUE)),
time = c("one", "two", "three", "four", "five","one", "two", "three", "four", "five"),
vector = c("1-2-9-4-5-1-5-6-1-2",
"3-2-3-4-5-2-6-6-1-2",
"1-2-4-4-2-4-5-4-2-1",
"9-2-3-1-5-5-5-3-1-2",
"1-1-3-4-5-1-5-6-3-2"))
age X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 time vector
1 one 4 0 3 5 0 5 0 3 5 4 one 1-2-9-4-5-1-5-6-1-2
2 two 1 5 3 5 1 2 5 0 4 4 two 3-2-3-4-5-2-6-6-1-2
3 three 0 4 5 0 3 0 0 5 2 2 three 1-2-4-4-2-4-5-4-2-1
4 four 0 0 5 5 4 3 2 5 5 1 four 9-2-3-1-5-5-5-3-1-2
5 five 2 1 2 1 4 5 5 2 1 1 five 1-1-3-4-5-1-5-6-3-2
6 one 4 0 3 5 0 5 0 3 5 4 one 1-2-3-4-5-1-5-6-1-2
7 two 1 5 3 5 1 2 5 0 4 4 two 3-2-3-4-5-2-6-6-1-2
8 three 0 4 5 0 3 0 0 5 2 2 three 1-2-4-4-2-4-5-4-2-1
9 four 0 0 5 5 4 3 2 5 5 1 four 9-2-3-1-5-5-5-3-1-2
10 five 2 1 2 1 4 5 5 2 1 1 five 1-1-3-4-5-1-5-6-3-9
I could do it by splitting the last column into multiple columns and then multiply each one but I'm looking for a faster way
thanks
5 answers
solution1
3 ACCPTED 2021-06-12 06:55:36
You may do this in single mutate statement, using dplyr 's powerful cur_data()
set.seed(2021)
x <- data.frame(age = c("one", "two", "three", "four", "five","one", "two", "three", "four", "five"),
replicate(10,sample(0:5,5,rep=TRUE)),
time = c("one", "two", "three", "four", "five","one", "two", "three", "four", "five"),
vector = c("1-2-9-4-5-1-5-6-1-2",
"3-2-3-4-5-2-6-6-1-2",
"1-2-4-4-2-4-5-4-2-1",
"9-2-3-1-5-5-5-3-1-2",
"1-1-3-4-5-1-5-6-3-2"))
library(tidyverse)
x %>% mutate(select(cur_data(), starts_with('X')) * t(map_dfc(strsplit(vector, '-'), as.numeric)))
#> age X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 time vector
#> 1 one 5 10 36 4 25 1 20 12 1 0 one 1-2-9-4-5-1-5-6-1-2
#> 2 two 15 10 0 8 5 4 0 6 4 0 two 3-2-3-4-5-2-6-6-1-2
#> 3 three 1 4 12 12 6 12 25 20 10 5 three 1-2-4-4-2-4-5-4-2-1
#> 4 four 27 10 6 4 20 20 5 15 2 8 four 9-2-3-1-5-5-5-3-1-2
#> 5 five 3 5 9 8 25 5 10 30 3 6 five 1-1-3-4-5-1-5-6-3-2
#> 6 one 5 10 36 4 25 1 20 12 1 0 one 1-2-9-4-5-1-5-6-1-2
#> 7 two 15 10 0 8 5 4 0 6 4 0 two 3-2-3-4-5-2-6-6-1-2
#> 8 three 1 4 12 12 6 12 25 20 10 5 three 1-2-4-4-2-4-5-4-2-1
#> 9 four 27 10 6 4 20 20 5 15 2 8 four 9-2-3-1-5-5-5-3-1-2
#> 10 five 3 5 9 8 25 5 10 30 3 6 five 1-1-3-4-5-1-5-6-3-2
or even using across as G.Grothendieck has suggested (that would eliminate use of cur_data()
x %>% mutate(across(starts_with('X')) * t(map_dfc(strsplit(vector, '-'), as.numeric)))
solution2
2 2021-06-12 07:15:29
We can also use the following solution:
library(dplyr)
library(tidyr)
set.seed(123)
x %>%
separate(vector, paste0("Y", seq(1, 10)), "-", convert = TRUE) %>%
mutate(across(starts_with("X"), ~ .x * get(sub("\\X", "\\Y", cur_column()))))
age X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 time Y1 Y2 Y3 Y4 Y5 Y6 Y7 Y8 Y9 Y10
1 one 2 10 45 8 0 0 0 24 3 0 one 1 2 9 4 5 1 5 6 1 2
2 two 15 4 0 8 20 10 12 0 4 2 two 3 2 3 4 5 2 6 6 1 2
3 three 2 8 4 0 4 8 20 0 8 4 three 1 2 4 4 2 4 5 4 2 1
4 four 9 6 6 3 5 15 15 3 2 8 four 9 2 3 1 5 5 5 3 1 2
5 five 1 5 12 0 5 5 5 12 15 6 five 1 1 3 4 5 1 5 6 3 2
6 one 2 10 45 8 0 0 0 24 3 0 one 1 2 9 4 5 1 5 6 1 2
7 two 15 4 0 8 20 10 12 0 4 2 two 3 2 3 4 5 2 6 6 1 2
8 three 2 8 4 0 4 8 20 0 8 4 three 1 2 4 4 2 4 5 4 2 1
9 four 9 6 6 3 5 15 15 3 2 8 four 9 2 3 1 5 5 5 3 1 2
10 five 1 5 12 0 5 5 5 12 15 6 five 1 1 3 4 5 1 5 6 3 2
solution3
1 2021-06-12 04:00:50
Here is a base R approach -
cols <- grep('X\\d+', names(x))
mat <- do.call(rbind, strsplit(x$vector, '-', fixed = TRUE))
x[cols] <- x[cols] * as.numeric(mat)
x
# age X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 time vector
#1 one 2 10 45 8 0 0 0 24 3 0 one 1-2-9-4-5-1-5-6-1-2
#2 two 15 4 0 8 20 10 12 0 4 2 two 3-2-3-4-5-2-6-6-1-2
#3 three 2 8 4 0 4 8 20 0 8 4 three 1-2-4-4-2-4-5-4-2-1
#4 four 9 6 6 3 5 15 15 3 2 8 four 9-2-3-1-5-5-5-3-1-2
#5 five 1 5 12 0 5 5 5 12 15 6 five 1-1-3-4-5-1-5-6-3-2
#6 one 2 10 45 8 0 0 0 24 3 0 one 1-2-9-4-5-1-5-6-1-2
#7 two 15 4 0 8 20 10 12 0 4 2 two 3-2-3-4-5-2-6-6-1-2
#8 three 2 8 4 0 4 8 20 0 8 4 three 1-2-4-4-2-4-5-4-2-1
#9 four 9 6 6 3 5 15 15 3 2 8 four 9-2-3-1-5-5-5-3-1-2
#10 five 1 5 12 0 5 5 5 12 15 6 five 1-1-3-4-5-1-5-6-3-2
data
set.seed(123)
x <- data.frame(age = c("one", "two", "three", "four", "five","one", "two", "three", "four", "five"),
replicate(10,sample(0:5,5,rep=TRUE)),
time = c("one", "two", "three", "four", "five","one", "two", "three", "four", "five"),
vector = c("1-2-9-4-5-1-5-6-1-2",
"3-2-3-4-5-2-6-6-1-2",
"1-2-4-4-2-4-5-4-2-1",
"9-2-3-1-5-5-5-3-1-2",
"1-1-3-4-5-1-5-6-3-2"))
x
# age X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 time vector
#1 one 2 5 5 2 0 0 0 4 3 0 one 1-2-9-4-5-1-5-6-1-2
#2 two 5 2 0 2 4 5 2 0 4 1 two 3-2-3-4-5-2-6-6-1-2
#3 three 2 4 1 0 2 2 4 0 4 4 three 1-2-4-4-2-4-5-4-2-1
#4 four 1 3 2 3 1 3 3 1 2 4 four 9-2-3-1-5-5-5-3-1-2
#5 five 1 5 4 0 1 5 1 2 5 3 five 1-1-3-4-5-1-5-6-3-2
#6 one 2 5 5 2 0 0 0 4 3 0 one 1-2-9-4-5-1-5-6-1-2
#7 two 5 2 0 2 4 5 2 0 4 1 two 3-2-3-4-5-2-6-6-1-2
#8 three 2 4 1 0 2 2 4 0 4 4 three 1-2-4-4-2-4-5-4-2-1
#9 four 1 3 2 3 1 3 3 1 2 4 four 9-2-3-1-5-5-5-3-1-2
#10 five 1 5 4 0 1 5 1 2 5 3 five 1-1-3-4-5-1-5-6-3-2
solution4
1 2021-06-12 19:00:51
Using base R with read.table
x[startsWith(names(x), "X")] <- read.table(text = x$vector, sep="-",
header = FALSE) * x[startsWith(names(x), "X")]
x
age X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 time vector
1 one 2 10 45 8 0 0 0 24 3 0 one 1-2-9-4-5-1-5-6-1-2
2 two 15 4 0 8 20 10 12 0 4 2 two 3-2-3-4-5-2-6-6-1-2
3 three 2 8 4 0 4 8 20 0 8 4 three 1-2-4-4-2-4-5-4-2-1
4 four 9 6 6 3 5 15 15 3 2 8 four 9-2-3-1-5-5-5-3-1-2
5 five 1 5 12 0 5 5 5 12 15 6 five 1-1-3-4-5-1-5-6-3-2
6 one 2 10 45 8 0 0 0 24 3 0 one 1-2-9-4-5-1-5-6-1-2
7 two 15 4 0 8 20 10 12 0 4 2 two 3-2-3-4-5-2-6-6-1-2
8 three 2 8 4 0 4 8 20 0 8 4 three 1-2-4-4-2-4-5-4-2-1
9 four 9 6 6 3 5 15 15 3 2 8 four 9-2-3-1-5-5-5-3-1-2
10 five 1 5 12 0 5 5 5 12 15 6 five 1-1-3-4-5-1-5-6-3-2
data
x <- structure(list(age = c("one", "two", "three", "four", "five",
"one", "two", "three", "four", "five"), X1 = c(5L, 5L, 1L, 3L,
3L, 5L, 5L, 1L, 3L, 3L), X2 = c(5L, 5L, 2L, 5L, 5L, 5L, 5L, 2L,
5L, 5L), X3 = c(4L, 0L, 3L, 2L, 3L, 4L, 0L, 3L, 2L, 3L), X4 = c(1L,
2L, 3L, 4L, 2L, 1L, 2L, 3L, 4L, 2L), X5 = c(5L, 1L, 3L, 4L, 5L,
5L, 1L, 3L, 4L, 5L), X6 = c(1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L,
5L), X7 = c(4L, 0L, 5L, 1L, 2L, 4L, 0L, 5L, 1L, 2L), X8 = c(2L,
1L, 5L, 5L, 5L, 2L, 1L, 5L, 5L, 5L), X9 = c(1L, 4L, 5L, 2L, 1L,
1L, 4L, 5L, 2L, 1L), X10 = c(0L, 0L, 5L, 4L, 3L, 0L, 0L, 5L,
4L, 3L), time = c("one", "two", "three", "four", "five", "one",
"two", "three", "four", "five"), vector = c("1-2-9-4-5-1-5-6-1-2",
"3-2-3-4-5-2-6-6-1-2", "1-2-4-4-2-4-5-4-2-1", "9-2-3-1-5-5-5-3-1-2",
"1-1-3-4-5-1-5-6-3-2", "1-2-9-4-5-1-5-6-1-2", "3-2-3-4-5-2-6-6-1-2",
"1-2-4-4-2-4-5-4-2-1", "9-2-3-1-5-5-5-3-1-2", "1-1-3-4-5-1-5-6-3-2"
)), class = "data.frame", row.names = c(NA, -10L))
solution5
0 2021-06-12 03:48:06
Here is a {tidyverse} approach using c_across() . The result is a list column which can be extracted using unnest() .
x1 <- x %>%
mutate(vector = str_split(vector, "-")) %>%
rowwise() %>%
mutate(vector = list(as.integer(vector)),
res = list(c_across(X1:X10) * vector)) %>%
ungroup()
x1
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