Summarizing data in R - r

I am newbie in R and I am having troubles with summarizing data. I tried to follow tutorials in internet, but unfortunately I had errors all time.
I have a matrix where my response factor is "concentration"
In my experiment I have 3 treatments (a, b and c) and 5 replicates for each treatment. And I get the concentrations of 8 products (PRO1 - PRO8).
TRA PRO1 PRO2 PRO3 PRO4 PRO5 PRO6 PRO7 PRO8
1 a 83 85 59 46 64 8 76 74
2 a 61 71 73 15 87 95 61 9
3 a 78 12 35 23 56 95 67 11
4 a 48 30 75 94 57 15 58 58
5 a 51 92 30 60 22 9 64 5
6 b 46 17 66 79 30 99 3 38
7 b 40 25 11 18 66 25 55 38
8 b 34 94 83 63 30 100 56 31
9 b 3 81 26 73 32 56 4 12
10 b 18 40 13 51 4 44 75 4
11 c 68 28 20 15 13 56 5 82
12 c 50 85 65 85 13 13 34 69
13 c 75 37 11 55 58 69 85 67
14 c 71 30 83 46 87 67 59 70
15 c 10 76 50 20 98 81 57 76
I tried the summaryBy, doBy and these functions, however it did not work for me.
How can I sort my matrix in order to execute these functions and get the mean, sd? My intention is to make a barplot with error bars to see the differences between the treatments for each product.
Thanks


You may try
library(ggplot2)
library(dplyr)
library(tidyr)
gather(df1, Var, Val, -TRA) %>%
group_by(TRA, Var) %>%
summarise(Mean=mean(Val), SD=sd(Val)) %>%
ggplot(., aes(x=TRA, y=Mean, fill=Var))+
geom_bar(position=position_dodge(), stat='identity')+
geom_errorbar(aes(ymin=Mean-SD, ymax=Mean+SD), width=.2,
position=position_dodge(.9))
data
df1 <- structure(list(TRA = c("a", "a", "a", "a", "a", "b", "b", "b",
"b", "b", "c", "c", "c", "c", "c"), PRO1 = c(83L, 61L, 78L, 48L,
51L, 46L, 40L, 34L, 3L, 18L, 68L, 50L, 75L, 71L, 10L), PRO2 = c(85L,
71L, 12L, 30L, 92L, 17L, 25L, 94L, 81L, 40L, 28L, 85L, 37L, 30L,
76L), PRO3 = c(59L, 73L, 35L, 75L, 30L, 66L, 11L, 83L, 26L, 13L,
20L, 65L, 11L, 83L, 50L), PRO4 = c(46L, 15L, 23L, 94L, 60L, 79L,
18L, 63L, 73L, 51L, 15L, 85L, 55L, 46L, 20L), PRO5 = c(64L, 87L,
56L, 57L, 22L, 30L, 66L, 30L, 32L, 4L, 13L, 13L, 58L, 87L, 98L
), PRO6 = c(8L, 95L, 95L, 15L, 9L, 99L, 25L, 100L, 56L, 44L,
56L, 13L, 69L, 67L, 81L), PRO7 = c(76L, 61L, 67L, 58L, 64L, 3L,
55L, 56L, 4L, 75L, 5L, 34L, 85L, 59L, 57L), PRO8 = c(74L, 9L,
11L, 58L, 5L, 38L, 38L, 31L, 12L, 4L, 82L, 69L, 67L, 70L, 76L
)), .Names = c("TRA", "PRO1", "PRO2", "PRO3", "PRO4", "PRO5",
"PRO6", "PRO7", "PRO8"), class = "data.frame", row.names = c("1",
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13",
"14", "15"))

Related

plotting specific columns of a data frame in R

ind
set
inst_0
inst_1
inst_2
Inst_3
inst_4
inst_5
0
1
20
30
50
55
58
60
0
2
34
44
46
67
89
70
0
3
37
89
78
80
90
98
0
4
23
45
67
89
87
89
1
1
34
56
65
78
77
89
1
2
23
32
45
55
66
77
1
3
35
69
88
99
98
57
1
4
23
45
56
78
89
99
2
1
23
34
55
55
77
88
2
2
12
44
55
67
88
90
2
3
12
66
77
91
44
99
2
4
45
55
88
31
56
100
I have a data frame like this above and I would like to make a plot showing this kind of a trend like in the graph below( this is only made for 4 individual in a same set) for the combinations of for example Ind0-set1, Ind1-set1, Ind2-set2...,Ind0-set2,Ind1-set2 and second question is that how to plot multiple line graph separately for each set in one graph?
I am not sure to use ggplot2 or it can be done plot function too.

If you want to do this using ggplot2 then the first step would be to reshape your data to long or tidy format using e.g. tidyr::pivot_longer:
library(tidyr)
library(dplyr)
library(ggplot2)
# Reshape to long
dat <- dat %>%
# Convert all column names to lower case
rename_with(tolower) %>%
pivot_longer(-c(ind, set), names_to = "inst", values_to = "value", names_prefix = "inst_")
After doing so you could create a plot showing all individuals for all sets by using facetting:
ggplot(dat, aes(inst, value, color = factor(ind), group = ind)) +
geom_line() +
geom_point() +
facet_wrap(~set)
Or you could filter your data for your desired combinations to create a plot for e.g. just one set like so:
dat_filtered <- dat[dat$set == 1, ]
ggplot(dat_filtered, aes(inst, value, color = factor(ind), group = ind)) +
geom_line() +
geom_point()
DATA
dat <- data.frame(
ind = c(0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L),
set = c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L),
inst_0 = c(20L, 34L, 37L, 23L, 34L, 23L, 35L, 23L, 23L, 12L, 12L, 45L),
inst_1 = c(30L, 44L, 89L, 45L, 56L, 32L, 69L, 45L, 34L, 44L, 66L, 55L),
inst_2 = c(50L, 46L, 78L, 67L, 65L, 45L, 88L, 56L, 55L, 55L, 77L, 88L),
Inst_3 = c(55L, 67L, 80L, 89L, 78L, 55L, 99L, 78L, 55L, 67L, 91L, 31L),
inst_4 = c(58L, 89L, 90L, 87L, 77L, 66L, 98L, 89L, 77L, 88L, 44L, 56L),
inst_5 = c(60L, 70L, 98L, 89L, 89L, 77L, 57L, 99L, 88L, 90L, 99L, 100L)
)

Replacing values in a data frames based on condition

I want to convert values 0<x<9 & x<50.2 to NA.My data frame has the first five columns which do not have to have replaced, I only want to replace values in column 6 to column 60. I have tried to it in 2 steps as follows but, it also replaced values I dont intend to change
BdsDf[BdsDf > 50.2][6:60] <- NA; BdsDf[BdsDf < 9][6:60] <- NA

Here is one way:
# test data
df = data.frame(lapply(1:60, function(x) {
rnorm(100, 50, 25)
}))
df[,6:60][0 < df[,6:60] & df[,6:60] < 9 & df[,6:60] < 50.2] = NA

With naniar, you can use replace_with_na_at to select the specific columns and add the 2 conditions.
library(dplyr)
library(naniar)
BdsDf %>%
replace_with_na_at(.vars = -1:5,
condition = ~ .x < 9 | .x > 50.2)
Or in base R, you can do the following with the sample dataset. For your data, you would just change 6:8 to 6:60 (like with the commented out portion below).
BdsDf[,6:8][BdsDf[6:8] > 50.2 | BdsDf[6:8] < 9] <- NA
# BdsDf[,6:60][BdsDf[6:60] > 50.2 | BdsDf[6:60] < 9] <- NA
Output
X1 X2 X3 X4 X5 X6 X7 X8
1 75 86 91 16 6 NA NA NA
2 84 68 8 85 19 NA 38 NA
3 18 52 5 17 59 NA 28 NA
4 97 86 45 17 31 NA NA 28
5 41 95 60 80 49 NA 24 NA
6 47 56 65 35 44 18 NA 19
7 29 46 3 22 36 15 NA 10
8 48 50 60 38 47 NA NA 35
9 91 20 50 5 24 40 47 19
10 85 84 15 71 96 NA NA 26
Data
BdsDf <- structure(list(X1 = c(75L, 84L, 18L, 97L, 41L, 47L, 29L, 48L,
91L, 85L), X2 = c(86L, 68L, 52L, 86L, 95L, 56L, 46L, 50L, 20L,
84L), X3 = c(91L, 8L, 5L, 45L, 60L, 65L, 3L, 60L, 50L, 15L),
X4 = c(16L, 85L, 17L, 17L, 80L, 35L, 22L, 38L, 5L, 71L),
X5 = c(6L, 19L, 59L, 31L, 49L, 44L, 36L, 47L, 24L, 96L),
X6 = c(66L, 0L, 84L, 3L, 84L, 18L, 15L, 60L, 40L, 67L), X7 = c(73L,
38L, 28L, 6L, 24L, 91L, 79L, 7L, 47L, 88L), X8 = c(92L, 57L,
66L, 28L, 85L, 19L, 10L, 35L, 19L, 26L)), class = "data.frame", row.names = c(NA,
-10L))

R- Create function that selects entire row in data frame by column name

This is a question for an R Programming class, but I have been working on it for several hours, over a period of a few days. I have done internet searches and referenced three different books. I have tried very hard to solve it on my own. I am finally asking for help.
I was given a csv, which I read into the program. This is the resulting dataframe, named df:
name hw0 hw1 hw2 hw3 hw4 hw5 hw6 quiz1 quiz2 quiz3 quiz4 quiz5 quiz6 term1
1 20 14 30 100 50 60 36 12 15 30 15 25 25 100
2 A 20 13 30 100 50 60 30 11 15 0 14 25 25 100
3 B 20 14 30 100 50 60 36 8 11 24 8 13 9 95
4 C 20 14 28 100 50 60 36 12 4 25 13 24 14 95
5 D 20 12 30 100 50 0 33 7 15 26 12 22 0 100
6 E 20 14 30 90 30 0 0 10 15 30 15 21 15 100
7 F 20 13 30 100 48 0 36 12 15 30 15 25 23 95
8 G 20 14 26 85 40 42 33 11 15 23 11 17 16 90
9 H 20 0 0 85 50 0 0 0 15 0 0 15 10 85
10 I 20 14 15 0 10 48 30 11 0 27 11 14 16 60
11 J 20 14 29 80 35 0 36 11 13 24 12 14 0 70
12 K 20 14 29 97 50 60 36 4 7 19 11 20 15 100
13 L 20 14 30 100 45 0 36 10 6 26 8 16 7 80
14 M 20 14 30 100 50 60 36 7 15 28 14 25 25 100
15 N 20 11 0 95 20 0 0 8 14 26 7 9 0 95
16 O 20 12 28 97 0 40 0 11 10 27 11 15 15 70
17 P 20 13 0 90 45 0 20 4 13 30 10 20 17 90
18 Q 20 14 30 100 45 0 36 0 12 21 11 14 17 75
term2 term3 exam1 exam2 exam3 final
1 100 100 100 100 95 100
2 100 100 97 97 80 97
3 100 100 83 85 73 73
4 100 100 88 75 56 77
5 100 0 90 87 72 81
6 100 80 92 82 69 79
7 100 100 90 95 87 90
8 100 0 89 79 81 78
9 90 100 62 83 42 75
10 90 72 78 78 66 81
11 0 0 79 77 51 78
12 100 100 79 77 57 81
13 0 100 68 74 76 76
14 100 100 99 98 82 99
15 0 0 70 70 52 61
16 0 0 63 66 0 0
17 100 100 75 72 56 64
18 90 75 72 84 54 63
QUESTION:
checkStudent <- function(df, studentName);
This function extracts a particular student's grades data from a data frame and returns them.
REQUIRED FORMAT:
checkStudent <- function(df, studentName)
{
}
TIPS PROVIDED:
inputs:
df -- a data frame that contains all the grades data
studentName -- name of a student
return:
all the grades for the student whose name is given as studentName
purpose:
extracting a particular student's grades data from a data frame and returning them
PROJECT TESTER- line of code and expected results:
checkStudent(df,"A")
name hw0 hw1 hw2 hw3 hw4 hw5 hw6 quiz1 quiz2 quiz3 quiz4 quiz5
2 A 20 13 30 100 50 60 30 11 15 0 14 25
quiz6 term1 term2 term3 exam1 exam2 exam3 final
2 25 100 100 100 97 97 80 97
I feel like I have been given everything and still can't get it right. I have tried:
checkStudent <- function(df, studentName)
{
name <- studentName
df["name", ]
}
and
checkStudent <- function(df, studentName)
{
subset(df, "name" == studentName, 1:21)
}
and numerous other lines of code, too many to list.
Please help. I am truly stuck.
Again, this needs to be done strictly in R. If it matters, I'm using RStudio. Thank you so much.

You're really close.
Variables in R should never be encapsulated in quotes, but always are free standing. Additionally your code is just printing the row, it is not returning it.
Here's a slightly modify version of your first attempt, without the quotes.
checkStudent <- function(df, studentName)
{
name <- studentName
return(df[name, ])
}
Edit: Oops, I realized your rows aren't named as the students.
You'll need to make it more like this:
checkStudent <- function(df, studentName)
{
my_row <- which(df$name == studentName)
return(df[my_row, ])
}

Try with logical subsetting:
checkStudent <- function(x,y) x[x['name']==y,]
Test:
checkStudent(df,"A")
# name hw0 hw1 hw2 hw3 hw4 hw5 hw6 quiz1 quiz2 quiz3 quiz4 quiz5 quiz6 term1 term2 term3 exam1 exam2 exam3 final
#1 A 20 13 30 100 50 60 30 11 15 0 14 25 25 100 100 100 97 97 80 97
data:
df <- structure(list(name = structure(1:17, .Label = c("A", "B", "C",
"D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P",
"Q"), class = "factor"), hw0 = c(20L, 20L, 20L, 20L, 20L, 20L,
20L, 20L, 20L, 20L, 20L, 20L, 20L, 20L, 20L, 20L, 20L), hw1 = c(13L,
14L, 14L, 12L, 14L, 13L, 14L, 0L, 14L, 14L, 14L, 14L, 14L, 11L,
12L, 13L, 14L), hw2 = c(30L, 30L, 28L, 30L, 30L, 30L, 26L, 0L,
15L, 29L, 29L, 30L, 30L, 0L, 28L, 0L, 30L), hw3 = c(100L, 100L,
100L, 100L, 90L, 100L, 85L, 85L, 0L, 80L, 97L, 100L, 100L, 95L,
97L, 90L, 100L), hw4 = c(50L, 50L, 50L, 50L, 30L, 48L, 40L, 50L,
10L, 35L, 50L, 45L, 50L, 20L, 0L, 45L, 45L), hw5 = c(60L, 60L,
60L, 0L, 0L, 0L, 42L, 0L, 48L, 0L, 60L, 0L, 60L, 0L, 40L, 0L,
0L), hw6 = c(30L, 36L, 36L, 33L, 0L, 36L, 33L, 0L, 30L, 36L,
36L, 36L, 36L, 0L, 0L, 20L, 36L), quiz1 = c(11L, 8L, 12L, 7L,
10L, 12L, 11L, 0L, 11L, 11L, 4L, 10L, 7L, 8L, 11L, 4L, 0L), quiz2 = c(15L,
11L, 4L, 15L, 15L, 15L, 15L, 15L, 0L, 13L, 7L, 6L, 15L, 14L,
10L, 13L, 12L), quiz3 = c(0L, 24L, 25L, 26L, 30L, 30L, 23L, 0L,
27L, 24L, 19L, 26L, 28L, 26L, 27L, 30L, 21L), quiz4 = c(14L,
8L, 13L, 12L, 15L, 15L, 11L, 0L, 11L, 12L, 11L, 8L, 14L, 7L,
11L, 10L, 11L), quiz5 = c(25L, 13L, 24L, 22L, 21L, 25L, 17L,
15L, 14L, 14L, 20L, 16L, 25L, 9L, 15L, 20L, 14L), quiz6 = c(25L,
9L, 14L, 0L, 15L, 23L, 16L, 10L, 16L, 0L, 15L, 7L, 25L, 0L, 15L,
17L, 17L), term1 = c(100L, 95L, 95L, 100L, 100L, 95L, 90L, 85L,
60L, 70L, 100L, 80L, 100L, 95L, 70L, 90L, 75L), term2 = c(100L,
100L, 100L, 100L, 100L, 100L, 100L, 90L, 90L, 0L, 100L, 0L, 100L,
0L, 0L, 100L, 90L), term3 = c(100L, 100L, 100L, 0L, 80L, 100L,
0L, 100L, 72L, 0L, 100L, 100L, 100L, 0L, 0L, 100L, 75L), exam1 = c(97L,
83L, 88L, 90L, 92L, 90L, 89L, 62L, 78L, 79L, 79L, 68L, 99L, 70L,
63L, 75L, 72L), exam2 = c(97L, 85L, 75L, 87L, 82L, 95L, 79L,
83L, 78L, 77L, 77L, 74L, 98L, 70L, 66L, 72L, 84L), exam3 = c(80L,
73L, 56L, 72L, 69L, 87L, 81L, 42L, 66L, 51L, 57L, 76L, 82L, 52L,
0L, 56L, 54L), final = c(97L, 73L, 77L, 81L, 79L, 90L, 78L, 75L,
81L, 78L, 81L, 76L, 99L, 61L, 0L, 64L, 63L)), .Names = c("name",
"hw0", "hw1", "hw2", "hw3", "hw4", "hw5", "hw6", "quiz1", "quiz2",
"quiz3", "quiz4", "quiz5", "quiz6", "term1", "term2", "term3",
"exam1", "exam2", "exam3", "final"), row.names = c(NA, -17L), class = "data.frame")

Adding several box plots in one

I have a dataset where I have three different groups of individuals, let´s call them Green, Red, and Blue. Then I have data covering 92 proteins in their blood, from which I have readings for each individual in each group.
I would like to get a good overview of the variances and means for each protein for each group. Which means that I would like to make a multiple box plot graph.
I would like to have the different proteins on the x-axis, and three box plots (preferably in different colors) (one for each group) above every protein, with numeric protein weight on the y-axis.
How do I do this?
I am currently working with a data frame where the groups are divided by the rows, and the different protein readings is in each column.
Tried to add a picture, but apparently you need reputation-points…
I´ve heard that you can use the melt command in reshape2, but I need guidance in how to use it.
Please, simplify the answers. I´m not very experienced when it comes to R.

Look, I realize things are frustrating when you are first getting started, but you're going to have to ask specific and targeted questions for people to be willing and able to help you out in a structured way.
Having said that, let's walk through a structured example. I am only going to use 9 proteins here, but you should get the idea.
library(ggplot2)
library(reshape2)
# Setup a data frame, since the question did not provide one...
df <- structure(list(Individual = 1:12,
Group = structure(c(2L, 1L, 3L, 2L, 1L, 3L, 2L, 1L, 3L, 2L, 1L, 3L),
.Label = c("Blue", "Green", "Red"), class = "factor"),
Protein_1 = c(82L, 23L, 19L, 100L, 33L, 86L, 32L, 41L, 39L, 59L, 93L, 99L),
Protein_2 = c(86L, 50L, 86L, 90L, 37L, 20L, 26L, 38L, 87L, 81L, 23L, 49L),
Protein_3 = c(81L, 31L, 5L, 10L, 79L, 40L, 27L, 73L, 64L, 30L, 87L, 64L),
Protein_4 = c(52L, 15L, 25L, 12L, 63L, 52L, 60L, 33L, 27L, 32L, 53L, 93L),
Protein_5 = c(19L, 75L, 25L, 14L, 33L, 60L, 73L, 13L, 92L, 92L, 91L, 12L),
Protein_6 = c(33L, 49L, 29L, 58L, 51L, 12L, 61L, 48L, 71L, 18L, 84L, 31L),
Protein_7 = c(84L, 57L, 28L, 99L, 47L, 54L, 72L, 97L, 73L, 46L, 68L, 37L),
Protein_8 = c(15L, 16L, 46L, 95L, 57L, 86L, 30L, 83L, 45L, 12L, 49L, 82L),
Protein_9 = c(84L, 91L, 33L, 10L, 91L, 91L, 4L, 88L, 42L, 82L, 76L, 95L)),
.Names = c("Individual", "Group", "Protein_1", "Protein_2", "Protein_3",
"Protein_4", "Protein_5", "Protein_6", "Protein_7", "Protein_8", "Protein_9"),
class = "data.frame", row.names = c(NA, -12L))
head(df)
# Individual Group Protein_1 Protein_2 Protein_3 Protein_4 Protein_5 Protein_6 Protein_7 Protein_8 Protein_9
# 1 1 Green 82 86 81 52 19 33 84 15 84
# 2 2 Blue 23 50 31 15 75 49 57 16 91
# 3 3 Red 19 86 5 25 25 29 28 46 33
# 4 4 Green 100 90 10 12 14 58 99 95 10
# 5 5 Blue 33 37 79 63 33 51 47 57 91
# 6 6 Red 86 20 40 52 60 12 54 86 91
?melt
df.melted <- melt(df, id.vars = c("Individual", "Group"))
head(df.melted)
# Individual Group variable value
# 1 1 Green Protein_1 82
# 2 2 Blue Protein_1 23
# 3 3 Red Protein_1 19
# 4 4 Green Protein_1 100
# 5 5 Blue Protein_1 33
# 6 6 Red Protein_1 86
# First Protein
# Notice I am using subset()
ggplot(data = subset(df.melted, variable == "Protein_1"),
aes(x = Group, y = value)) + geom_boxplot(aes(fill = Group))
# Second Protein
ggplot(data = subset(df.melted, variable == "Protein_2"),
aes(x = Group, y = value)) + geom_boxplot(aes(fill = Group))
# and so on...
# You could also use facets
ggplot(data = df.melted, aes(x = Group, y = value)) +
geom_boxplot(aes(fill = Group)) +
facet_wrap(~ variable)
And yes, I realize that the color groupings do not align with the colors of the plot...I will leave that as an exercise... You have to be willing to tinker, explore, and fail many times.

Find the largest product in a dataframe

I have a dataframe and I need to find the largest product in the data frame in the same direction(up, down, left, right, or diagonally).
The dataframe is as follows:
structure(list(V1 = c(8L, 49L, 81L, 52L, 22L, 24L, 32L, 67L,
24L, 21L, 78L, 16L, 86L, 19L, 4L, 88L, 4L, 20L, 20L, 1L), V2 = c(2L,
49L, 49L, 70L, 31L, 47L, 98L, 26L, 55L, 36L, 17L, 39L, 56L, 80L,
52L, 36L, 42L, 69L, 73L, 70L), V3 = c(22L, 99L, 31L, 95L, 16L,
32L, 81L, 20L, 58L, 23L, 53L, 5L, 0L, 81L, 8L, 68L, 16L, 36L,
35L, 54L), V4 = c(97L, 40L, 73L, 23L, 71L, 60L, 28L, 68L, 5L,
9L, 28L, 42L, 48L, 68L, 83L, 87L, 73L, 41L, 29L, 71L), V5 = c(38L,
17L, 55L, 4L, 51L, 99L, 64L, 2L, 66L, 75L, 22L, 96L, 35L, 5L,
97L, 57L, 38L, 72L, 78L, 83L), V6 = c(15L, 81L, 79L, 60L, 67L,
3L, 23L, 62L, 73L, 0L, 75L, 35L, 71L, 94L, 35L, 62L, 25L, 30L,
31L, 51L), V7 = c(0L, 18L, 14L, 11L, 63L, 45L, 67L, 12L, 99L,
76L, 31L, 31L, 89L, 47L, 99L, 20L, 39L, 23L, 90L, 54L), V8 = c(40L,
57L, 29L, 42L, 89L, 2L, 10L, 20L, 26L, 44L, 67L, 47L, 7L, 69L,
16L, 72L, 11L, 88L, 1L, 69L), V9 = c(0L, 60L, 93L, 69L, 41L,
44L, 26L, 95L, 97L, 20L, 15L, 55L, 5L, 28L, 7L, 3L, 24L, 34L,
74L, 16L), V10 = c(75L, 87L, 71L, 24L, 92L, 75L, 38L, 63L, 17L,
45L, 94L, 58L, 44L, 73L, 97L, 46L, 94L, 62L, 31L, 92L), V11 = c(4L,
17L, 40L, 68L, 36L, 33L, 40L, 94L, 78L, 35L, 3L, 88L, 44L, 92L,
57L, 33L, 72L, 99L, 49L, 33L), V12 = c(5L, 40L, 67L, 56L, 54L,
53L, 67L, 39L, 78L, 14L, 80L, 24L, 37L, 13L, 32L, 67L, 18L, 69L,
71L, 48L), V13 = c(7L, 98L, 53L, 1L, 22L, 78L, 59L, 63L, 96L,
0L, 4L, 0L, 44L, 86L, 16L, 46L, 8L, 82L, 48L, 61L), V14 = c(78L,
43L, 88L, 32L, 40L, 36L, 54L, 8L, 83L, 61L, 62L, 17L, 60L, 52L,
26L, 55L, 46L, 67L, 86L, 43L), V15 = c(52L, 69L, 30L, 56L, 40L,
84L, 70L, 40L, 14L, 33L, 16L, 54L, 21L, 17L, 26L, 12L, 29L, 59L,
81L, 52L), V16 = c(12L, 48L, 3L, 71L, 28L, 20L, 66L, 91L, 88L,
97L, 14L, 24L, 58L, 77L, 79L, 32L, 32L, 85L, 16L, 1L), V17 = c(50L,
4L, 49L, 37L, 66L, 35L, 18L, 66L, 34L, 34L, 9L, 36L, 51L, 4L,
33L, 63L, 40L, 74L, 23L, 89L), V18 = c(77L, 56L, 13L, 2L, 33L,
17L, 38L, 49L, 89L, 31L, 53L, 29L, 54L, 89L, 27L, 93L, 62L, 4L,
57L, 19L), V19 = c(91L, 62L, 36L, 36L, 13L, 12L, 64L, 94L, 63L,
33L, 56L, 85L, 17L, 55L, 98L, 53L, 76L, 36L, 5L, 67L), V20 = c(8L,
0L, 65L, 91L, 80L, 50L, 70L, 21L, 72L, 95L, 92L, 57L, 58L, 40L,
66L, 69L, 36L, 16L, 54L, 48L)), .Names = c("V1", "V2", "V3",
"V4", "V5", "V6", "V7", "V8", "V9", "V10", "V11", "V12", "V13",
"V14", "V15", "V16", "V17", "V18", "V19", "V20"), class = "data.frame", row.names = c(NA,
-20L))
The dataframe looks like this:
> newjd
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20
1 8 2 22 97 38 15 0 40 0 75 4 5 7 78 52 12 50 77 91 8
2 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 4 56 62 0
3 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 3 49 13 36 65
4 52 70 95 23 4 60 11 42 69 24 68 56 1 32 56 71 37 2 36 91
5 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
6 24 47 32 60 99 3 45 2 44 75 33 53 78 36 84 20 35 17 12 50
7 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
8 67 26 20 68 2 62 12 20 95 63 94 39 63 8 40 91 66 49 94 21
9 24 55 58 5 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
10 21 36 23 9 75 0 76 44 20 45 35 14 0 61 33 97 34 31 33 95
11 78 17 53 28 22 75 31 67 15 94 3 80 4 62 16 14 9 53 56 92
12 16 39 5 42 96 35 31 47 55 58 88 24 0 17 54 24 36 29 85 57
13 86 56 0 48 35 71 89 7 5 44 44 37 44 60 21 58 51 54 17 58
14 19 80 81 68 5 94 47 69 28 73 92 13 86 52 17 77 4 89 55 40
15 4 52 8 83 97 35 99 16 7 97 57 32 16 26 26 79 33 27 98 66
16 88 36 68 87 57 62 20 72 3 46 33 67 46 55 12 32 63 93 53 69
17 4 42 16 73 38 25 39 11 24 94 72 18 8 46 29 32 40 62 76 36
18 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 4 36 16
19 20 73 35 29 78 31 90 1 74 31 49 71 48 86 81 16 23 57 5 54
20 1 70 54 71 83 51 54 69 16 92 33 48 61 43 52 1 89 19 67 48
Any suggestion is highly appreciated.

m <- as.matrix(df)
ll <- c(lapply(1:20, function(X) row(m)==X), ## Rows
lapply(1:20, function(X) col(m)==X), ## Columns
lapply(-19:19, function(X) (col(m)+X)==row(m)), ## NW-SE diagonals
lapply(-19:19, function(X) (col(m)+X)==(21-row(m)))) ## SW-NE diagonals
## Calculate product of each row, column, and diagonal
pp <- sapply(ll, function(X) prod(m[X]))
max(pp)
# [1] 1.824798e+35
m[ll[[which.max(pp)]]]
# [1] 75 87 71 24 92 75 38 63 17 45 94 58 44 73 97 46 94 62 31 92
Or, more generally:
maxProdElements <- function(m) {
mm <- nrow(m)
nn <- ncol(m)
ll <- c(lapply(seq_len(mm), function(X) which(row(m)==X)),
lapply(seq_len(nn), function(X) which(col(m)==X)),
lapply((-mm+1):(nn-1), function(X) which(col(m)==(row(m)+X))),
lapply((-mm+1):(nn-1), function(X) which((nn+1-col(m))==(row(m)+X))))
pp <- sapply(ll, function(X) prod(m[X]))
ll[[which.max(pp)]]
}
## Try it with a 2 by 5 matrix
M <- matrix(1:10, ncol=5)
M[maxProdElements(M)]
# [1] 2 4 6 8 10
prod(M[maxProdElements(M)])
# [1] 3840

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