Label start of a sequence with NA's - r

This is trivial question, however I don't seem to find neat solution for this. (without excluding NA's first and including them back again). So I'm looking for some ideas without the need of NA's exclusion.
I would like to label the start of a 0 and 1 sequence with 2 and 1 respectively and replace NA's with 0 as well as the remaining sequence of 0's and 1's.
Is the rle function useful here? Base R solution would be welcomed.
Example:
x <- c(rep(NA,10),rep(1,5),rep(NA,5),rep(1,10),rep(NA,3),rep(0,7),rep(NA,15),rep(1,9))
r <- c(0,diff(x)); r[r %in% -1] <- 2
From this sample data:
x
[1] NA NA NA NA NA NA NA NA NA NA 1 1 1 1 1 NA NA NA NA NA 1 1 1 1 1 1 1 1 1 1 NA NA NA 0 0 0 0 0 0 0 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA 1 1 1 1 1 1 1 1 1
Desired output:
[1] 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0


We could use rle to create a grouping variable ('gr') to split the 'x' into a list. Replace the first element that is 0 or 1 with 2 or 1 using match, concatenate with 0s, unlist and then replace the NA elements with 0.
xN <- x
xN[is.na(xN)] <- -999
v1 <- rle(xN)$lengths
gr <- rep(seq_along(v1), v1)
x1 <- unlist(lapply(split(x, gr), function(x)
c(match(x[1],1:0),rep(0,length(x)-1)) ), use.names=FALSE)
x1[is.na(x1)] <- 0
x1
#[1] 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0
#[39] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
Or instead of split, we can use which and diff to replace the values.
x1 <- (!x)+2*(!is.na(x))-1
ind <- which(!is.na(x))
x1[c(ind[c(FALSE,diff(ind)==1)], which(is.na(x)))] <- 0
x1
#[1] 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0
#[39] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
Or we can use rleid from the devel version of data.table as grouping variable. Replace the first element of 0's and 1's with 2 and 1 using match and change the NA values to 0.
library(data.table)#v1.9.5+
DT <- setDT(list(x))
DT[, c(match(V1[1], 1:0), rep(0, .N-1)) ,rleid(V1)][is.na(V1), V1:=0]$V1
#[1] 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0
#[39] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0

Related

Multiplying multiple columns with each other into a new dataframe in R

I want to multiply many of my binary variables into new columns, so called interactive variables. My dataset is structured like this:
YearCountry <- data.frame( Time = c("2000","2001", "2002", "2003",
"2000","2001", "2002", "2003",
"2000","2001", "2002", "2003"),
AL = c(1,1,1,1,0,0,0,0,0,0,0,0),
FR = c(0,0,0,0,1,1,1,1,0,0,0,0),
UK = c(0,0,0,0,0,0,0,0,1,1,1,1),
Y2000d = c(1,0,0,0,1,0,0,0,1,0,0,0),
Y2001d = c(0,1,0,0,0,1,0,0,0,1,0,0),
Y2002d = c(0,0,1,0,0,0,1,0,0,0,1,0),
Y2003d = c(0,0,0,1,0,0,0,1,0,0,0,1))
YearCountry
Time AL FR UK Y2000d Y2001d Y2002d Y2003d
1 2000 1 0 0 1 0 0 0
2 2001 1 0 0 0 1 0 0
3 2002 1 0 0 0 0 1 0
4 2003 1 0 0 0 0 0 1
5 2000 0 1 0 1 0 0 0
6 2001 0 1 0 0 1 0 0
7 2002 0 1 0 0 0 1 0
8 2003 0 1 0 0 0 0 1
9 2000 0 0 1 1 0 0 0
10 2001 0 0 1 0 1 0 0
11 2002 0 0 1 0 0 1 0
12 2003 0 0 1 0 0 0 1
I need to multiply the binary variable for each of the countries (AL,FR,UK) with each of the binary variables for a given year so that I get #country x #year new variables. In this case I have three countries and four years which gives 12 new variables. My full data contains 105 countries/regions and stretches over twenty years. I therefore need a general formula. I want data that looks like this
Interact <- data.frame(Time = c("2000","2001", "2002", "2003",
"2000","2001", "2002", "2003",
"2000","2001", "2002", "2003"),
Y2000xAL = c(1,0,0,0,0,0,0,0,0,0,0,0),
Y2001xAL = c(0,1,0,0,0,0,0,0,0,0,0,0),
Y2002xAL = c(0,0,1,0,0,0,0,0,0,0,0,0),
Y2003xAL = c(0,0,0,1,0,0,0,0,0,0,0,0),
Y2000xFR = c(0,0,0,0,1,0,0,0,0,0,0,0),
Y2001xFR = c(0,0,0,0,0,1,0,0,0,0,0,0),
Y2002xFR = c(0,0,0,0,0,0,1,0,0,0,0,0),
Y2003xFR = c(0,0,0,0,0,0,0,1,0,0,0,0),
Y2000xUk = c(0,0,0,0,0,0,0,0,1,0,0,0),
Y2001xUK = c(0,0,0,0,0,0,0,0,0,1,0,0),
Y2002xUK = c(0,0,0,0,0,0,0,0,0,0,1,0),
Y2003xUK = c(0,0,0,0,0,0,0,0,0,0,0,1))
Interact
Time Y2000xAL Y2001xAL Y2002xAL Y2003xAL Y2000xFR Y2001xFR Y2002xFR Y2003xFR Y2000xUk Y2001xUK Y2002xUK Y2003xUK
1 2000 1 0 0 0 0 0 0 0 0 0 0 0
2 2001 0 1 0 0 0 0 0 0 0 0 0 0
3 2002 0 0 1 0 0 0 0 0 0 0 0 0
4 2003 0 0 0 1 0 0 0 0 0 0 0 0
5 2000 0 0 0 0 1 0 0 0 0 0 0 0
6 2001 0 0 0 0 0 1 0 0 0 0 0 0
7 2002 0 0 0 0 0 0 1 0 0 0 0 0
8 2003 0 0 0 0 0 0 0 1 0 0 0 0
9 2000 0 0 0 0 0 0 0 0 1 0 0 0
10 2001 0 0 0 0 0 0 0 0 0 1 0 0
11 2002 0 0 0 0 0 0 0 0 0 0 1 0
12 2003 0 0 0 0 0 0 0 0 0 0 0 1

Here's an approach with dplyr::across. We can make the final result into a plain data.frame with purrr:invoke as demonstrated in this answer.
library(dplyr)
library(purrr)
YearCountry %>%
mutate(across(AL:UK, ~ . * select(cur_data(), Y2000d:Y2003d))) %>%
select(-(Y2000d:Y2003d)) %>%
invoke(.f = data.frame) %>%
rename_with(~str_replace(.,"\\.",""))
Time ALY2000d ALY2001d ALY2002d ALY2003d FRY2000d FRY2001d FRY2002d FRY2003d UKY2000d UKY2001d UKY2002d UKY2003d
1 2000 1 0 0 0 0 0 0 0 0 0 0 0
2 2001 0 1 0 0 0 0 0 0 0 0 0 0
3 2002 0 0 1 0 0 0 0 0 0 0 0 0
4 2003 0 0 0 1 0 0 0 0 0 0 0 0
5 2000 0 0 0 0 1 0 0 0 0 0 0 0
6 2001 0 0 0 0 0 1 0 0 0 0 0 0
7 2002 0 0 0 0 0 0 1 0 0 0 0 0
8 2003 0 0 0 0 0 0 0 1 0 0 0 0
9 2000 0 0 0 0 0 0 0 0 1 0 0 0
10 2001 0 0 0 0 0 0 0 0 0 1 0 0
11 2002 0 0 0 0 0 0 0 0 0 0 1 0
12 2003 0 0 0 0 0 0 0 0 0 0 0 1

1) model.matrix We split the names by the number of characters in them (the countries have 2 characters in their names and the years have 6) and paste pluses in each. (Alternately use Plus(grep("^..$", nms, value = TRUE)) to get the country names and use that in place of spl["2"] and similarly Plus(grep("^Y....d$", nms, value = TRUE)) in place of spl["6"].)
c(`2` = "AL+FR+UK", `6` = "Y2000d+Y2001d+Y2002d+Y2003d")
and from that the formula:
~(AL + FR + UK):(Y2000d + Y2001d + Y2002d + Y2003d) + 0
and then compute its model matrix.
The formula could also be expanded to one accepted by lm by modifying the sprintf format so we might not even need to create the model matrix. For example, if we had a response vector R then we could write: s <- sprintf("R ~ (%s)*(%s)", spl["2"], spl["4"]); fo <- formula(s); lm(fo, YearCountry) to include all variables and the interactions of countries and year as well as an intercept.
Plus <- function(x) paste(x, collapse = "+")
nms <- names(YearCountry)[-1]
spl <- sapply(split(nms, nchar(nms)), Plus)
s <- sprintf("~ (%s):(%s)+0", spl["2"], spl["6"])
fo <- formula(s)
model.matrix(fo, YearCountry)
giving this matrix:
AL:Y2000d AL:Y2001d AL:Y2002d AL:Y2003d FR:Y2000d FR:Y2001d FR:Y2002d FR:Y2003d UK:Y2000d UK:Y2001d UK:Y2002d UK:Y2003d
1 1 0 0 0 0 0 0 0 0 0 0 0
2 0 1 0 0 0 0 0 0 0 0 0 0
3 0 0 1 0 0 0 0 0 0 0 0 0
4 0 0 0 1 0 0 0 0 0 0 0 0
5 0 0 0 0 1 0 0 0 0 0 0 0
6 0 0 0 0 0 1 0 0 0 0 0 0
7 0 0 0 0 0 0 1 0 0 0 0 0
8 0 0 0 0 0 0 0 1 0 0 0 0
9 0 0 0 0 0 0 0 0 1 0 0 0
10 0 0 0 0 0 0 0 0 0 1 0 0
11 0 0 0 0 0 0 0 0 0 0 1 0
12 0 0 0 0 0 0 0 0 0 0 0 1
attr(,"assign")
[1] 1 2 3 4 5 6 7 8 9 10 11 12
Alternately we can write it compactly like this:
Plus <- function(x) paste(x, collapse = "+")
nms <- names(YearCountry)
s <- sprintf("~ (%s):(%s)+0", Plus(nms[2:4]), Plus(nms[5:8]))
fo <- formula(s)
model.matrix(fo, YearCountry)
2) eList Another approach is to use list comprehensions. With the eList package we can do this:
library(eList)
DF(for(i in YearCountry[2:4]) for(j in YearCountry[5:8]) i*j)
giving this data frame. Use as.matrix(...) on it if you want a matrix.
AL.Y2000d AL.Y2001d AL.Y2002d AL.Y2003d FR.Y2000d FR.Y2001d FR.Y2002d FR.Y2003d UK.Y2000d UK.Y2001d UK.Y2002d UK.Y2003d
1 1 0 0 0 0 0 0 0 0 0 0 0
2 0 1 0 0 0 0 0 0 0 0 0 0
3 0 0 1 0 0 0 0 0 0 0 0 0
4 0 0 0 1 0 0 0 0 0 0 0 0
5 0 0 0 0 1 0 0 0 0 0 0 0
6 0 0 0 0 0 1 0 0 0 0 0 0
7 0 0 0 0 0 0 1 0 0 0 0 0
8 0 0 0 0 0 0 0 1 0 0 0 0
9 0 0 0 0 0 0 0 0 1 0 0 0
10 0 0 0 0 0 0 0 0 0 1 0 0
11 0 0 0 0 0 0 0 0 0 0 1 0
12 0 0 0 0 0 0 0 0 0 0 0 1
3) listcompr listcompr is another list comprehension package. Note that the development version of this package is needed in order to use bycol=. Replace gen.named.matrix with gen.named.data.frame if you want a data frame.
# devtools::github_github("patrickroocks/listcompr")
library(listcompr)
nms <- names(YearCountry)
gen.named.matrix("{nms[i]}.{nms[j]}", YearCountry[[i]] * YearCountry[[j]],
i = 2:4, j = 5:8, bycol = TRUE)

How can I convert matrix to different matrices in R?

Here I have some codes generating a matrix like this:
N = 200
T = 10
mu_0 <- matrix(diag(1, T))
dim(mu_0) <- c(T,T)
mu_t_0 <- matrix(rep(t(mu_0), N), ncol = T, byrow = TRUE)
And generally the result looks like this
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1 1 0 0 0 0 0 0 0 0 0
2 0 1 0 0 0 0 0 0 0 0
3 0 0 1 0 0 0 0 0 0 0
4 0 0 0 1 0 0 0 0 0 0
5 0 0 0 0 1 0 0 0 0 0
6 0 0 0 0 0 1 0 0 0 0
7 0 0 0 0 0 0 1 0 0 0
8 0 0 0 0 0 0 0 1 0 0
9 0 0 0 0 0 0 0 0 1 0
10 0 0 0 0 0 0 0 0 0 1
11 1 0 0 0 0 0 0 0 0 0
12 0 1 0 0 0 0 0 0 0 0
13 0 0 1 0 0 0 0 0 0 0
14 0 0 0 1 0 0 0 0 0 0
15 0 0 0 0 1 0 0 0 0 0
16 0 0 0 0 0 1 0 0 0 0
17 0 0 0 0 0 0 1 0 0 0
18 0 0 0 0 0 0 0 1 0 0
19 0 0 0 0 0 0 0 0 1 0
20 0 0 0 0 0 0 0 0 0 1
...
Now for later calculation I want to split this large matrix into different small matrices like this:
Matrix One:
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
...
Matrix Two:
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
...
I tried the split function but I cannot get what I want. Are there any solutions?

You can use asplit to split an array or matrix by its margin.
x <- asplit(mu_t_0, 2)
str(x)
#List of 10
# $ : num [1:2000(1d)] 1 0 0 0 0 0 0 0 0 0 ...
# $ : num [1:2000(1d)] 0 1 0 0 0 0 0 0 0 0 ...
# $ : num [1:2000(1d)] 0 0 1 0 0 0 0 0 0 0 ...
# $ : num [1:2000(1d)] 0 0 0 1 0 0 0 0 0 0 ...
# $ : num [1:2000(1d)] 0 0 0 0 1 0 0 0 0 0 ...
# $ : num [1:2000(1d)] 0 0 0 0 0 1 0 0 0 0 ...
# $ : num [1:2000(1d)] 0 0 0 0 0 0 1 0 0 0 ...
# $ : num [1:2000(1d)] 0 0 0 0 0 0 0 1 0 0 ...
# $ : num [1:2000(1d)] 0 0 0 0 0 0 0 0 1 0 ...
# $ : num [1:2000(1d)] 0 0 0 0 0 0 0 0 0 1 ...
# - attr(*, "dim")= int 10

This one puts all the columns into a list.
res <- lapply(1:ncol(mu_t_0), function(i) mu_t_0[, i, drop=F])
head(res[[1]])
# [,1]
# [1,] 1
# [2,] 0
# [3,] 0
# [4,] 0
# [5,] 0
# [6,] 0
head(res[[2]])
# [,1]
# [1,] 0
# [2,] 1
# [3,] 0
# [4,] 0
# [5,] 0
# [6,] 0
To extract the single one column matrices use list2env; the objects in the list need names beforehand.
names(res) <- paste0("m.", 1:length(res))
list2env(res, env=.GlobalEnv)
ls()
# [1] "m.1" "m.10" "m.2" "m.3" "m.4" "m.5" "m.6" "m.7" "m.8" "m.9" "mu_0" "mu_t_0"
# [13] "N" "res" "T"

If you want individual vectors named in the global environment (i.e. object V1, V2, and etc.):
invisible(mapply(assign, names(as.data.frame(mu_t_0)), as.data.frame(mu_t_0), MoreArgs=list(envir = globalenv())))
Though I'm really interested in why the OP doesn't want to just work with a single matrix...

Filling a table with additional columns if they don't exist

I've the following difficult problem. Here short example of my data. Assume that I've two data sets (my real example has something about 20). The data frames result as a list computed by a self written function with lapply. So, I put the data frames in my example in a list, too. Then I "rbind" them to compute a frequency table.
df1 <- data.frame(rev(seq(12:0)), paste0("a=",sample(0:12, 13, replace=T)))
colnames(df1) <- c("k", "a")
df2 <- data.frame(rev(seq(12:0)), paste0("a=",sample(0:12, 13, replace=T)))
colnames(df2) <- c("k", "a")
list_df <- list(df1,df2)
df_combine<- plyr::ldply(list_df, rbind)
freq_foo <- table(df_combine$k,df_combine$a)
I get a frequency table of the following form.
a=0 a=11 a=12 a=2 a=5 a=6 a=7 a=8 a=3 a=9
1 1 0 0 0 0 0 0 1 0 0
2 1 0 0 0 0 0 0 0 0 1
3 1 0 0 0 0 1 0 0 0 0
4 0 0 0 1 0 1 0 0 0 0
5 0 0 0 1 1 0 0 0 0 0
6 0 0 0 0 0 0 1 0 0 1
7 0 1 1 0 0 0 0 0 0 0
8 1 0 0 0 0 1 0 0 0 0
9 0 0 0 0 0 0 2 0 0 0
10 0 0 1 0 1 0 0 0 0 0
11 1 1 0 0 0 0 0 0 0 0
12 0 0 0 0 0 0 1 0 1 0
13 1 0 1 0 0 0 0 0 0 0
I want to extend and manipulate my table in the following way:
First the table should go over a range of a=0 to a=15. So if there is a missing column, it should be added. And 2nd) I want to order the columns from 0 to 15.
For the first problem I tried
if(freq_foo$paste0("a=",0:15) == F){freq_foo$paste("a=",0:15) <- 0}
but this should work only for data frames and not for tables. Also. i've no idea how to order the columns with an ascending order. The data type isnt important to me because I just want to use the output for further calculations. So, it can also be a data frame instead of a table.

#convert freq_foo table to dataframe
df <- as.data.frame.matrix(freq_foo)
#add all zeros column for missing column name in 0:15 series
df[, paste0("a=", c(0:15)[!(c(0:15) %in% as.numeric(gsub(".*=(\\d+)", "\\1", names(df))))])] <- 0
#order columns from 0 to 15
df <- df[, order(as.numeric(gsub(".*=(\\d+)", "\\1", names(df))))]
Output is:
a=0 a=1 a=2 a=3 a=4 a=5 a=6 a=7 a=8 a=9 a=10 a=11 a=12 a=13 a=14 a=15
1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0
2 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0
3 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0
4 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0
5 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0
6 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
7 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0
8 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0
10 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
11 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
12 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0
13 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0
(Edit: Updated code after getting a requirement clarification from OP)

incorrect number of dimensions even if the size of the array is the same

I have been working on this program for many days, and decide to rewrite it today....
But this problem keeps bothering me.
I thought the csm[1,] and Prank[1,] has the same dimension.
Who can help me with this problem?
Prank<-read.csv("result.csv")
nrP<-nrow(Prank)
ncP<-ncol(Prank)
csm<-matrix(0,nrP*3,ncP)
ccsm<-matrix(0,nrP*3,ncP)
nrC<-nrow(csm)
ncC<-ncol(csm)
nrP
[1] 30
ncP
[1] 144
nrC
[1] 90
ncC
[1] 144
Prank[1,]
P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P13 P14 P15 P16 P17 P18 P19 P20 P21 P22 P23 P24 P25 P26 P27 P28 P29 P30 P31 P32
1 4 2 3 1 4 2 3 1 4 2 3 1 3 1 4 2 4 2 3 1 4 1 3 2 4 1 3 2 4 2 3 1
P33 P34 P35 P36 P37 P38 P39 P40 P41 P42 P43 P44 P45 P46 P47 P48 P49 P50 P51 P52 P53 P54 P55 P56 P57 P58 P59 P60 P61
1 4 1 3 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
P62 P63 P64 P65 P66 P67 P68 P69 P70 P71 P72 P73 P74 P75 P76 P77 P78 P79 P80 P81 P82 P83 P84 P85 P86 P87 P88 P89 P90
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
P91 P92 P93 P94 P95 P96 P97 P98 P99 P100 P101 P102 P103 P104 P105 P106 P107 P108 P109 P110 P111 P112 P113 P114 P115
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
P116 P117 P118 P119 P120 P121 P122 P123 P124 P125 P126 P127 P128 P129 P130 P131 P132 P133 P134 P135 P136 P137 P138
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
P139 P140 P141 P142 P143 P144
1 0 0 0 0 0 0
csm[1,]
[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[59] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[117] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
csm[1,]<-Prank[1,]
csm[1,]
Error in csm[1, ] : incorrect number of dimensions

The problem is that Prank[1, ] is a data.frame (i.e. a list) so when you try to assign it to the first row of csm, it has the unexpected side-effect of converting csm to a list. At that point, doing csm[1, ] does not make any sense (a list has a single dimension) hence the error.
A solution is to unlist Prank[1, ] before assigning:
csm[1,] <- unlist(Prank[1,])

read.csv() returns a data.frame, and unless all of the columns of Prank are numeric, the assignment
csm[1,]<-Prank[1,]
will cause csm to be coerced to a list because Prank[1,] is not a numeric vector. You will want to make sure that Prank[1,] is a numeric vector (i.e. is.numeric(Prank[1,])).
Revised suggestion: take a look at data.frame (head(Prank)) and it may be obvious that one or more columns are not numeric. To inspect the classes of each field in prank, you can use
lapply(Prank,class)
or
sapply(Prank,class)
If all the fields in Prank are integer or numeric, you can coerce them all to numeric via
Prank[] <- lapply(Prank,as.numeric)
If not all the fields are numeric, you will want to coerce the problem fields to numeric, or
or remove the offending fields from Prank (e. g. Prank$ProblemField <- NULL) before the assignment.

For loop in a data table with dependencies in R?

I want to update the following datatable:
DT1: (This datatable columns values I need to edit based on my input)
BIC BCC1 BCC2 BCC6 BCC8 BCC9 BCC10 BCC11
990081899A 0 1 0 0 0 0 0
9900023620 0 1 1 0 0 0 0
9900427160 0 1 0 0 0 0 0
990064457TA 1 1 0 0 0 0 0
990066595A 0 0 0 1 0 0 1
990088248A 0 0 0 1 0 0 1
990088882C1 0 0 0 1 0 0 1
990088882C2 0 0 0 1 1 0 0
990088882C3 0 0 0 1 1 0 1
990088882C4 0 0 0 0 1 0 1
990088882C5 0 0 0 0 1 0 1
I want to loop through DT1 column names except first column to check if my input and any of the column names are equal. If they are equal, then check through all the rows of that column if the value is equal to 1. If yes, then set some of the other column values of that row equals to 0.
I am doing this now:
>Hierarchy <- function(Dt, cc, Hier){
Dt_cols<-setdiff(names(Dt), "HIC")
Dt_rows<-1:nrow(Dt)
for(j in 1:length(Dt_cols)){
if(Dt_cols[j] == cc){
for(i in 1:length(Dt_rows)){
if(eval(parse(text = paste("Dt[",i,",",eval(Dt_cols[j]),"]"))) == 1){
for(k in 1:length(Hier)){
if(Dt_cols[j] == Hier[k]){
hierVar<-as.character(eval(Dt_cols[j]));
Dt[i,hierVar]<- 0
}}}}}}return(Dt)}
If I am passing following arguments to this function:
>Hierarchy(DT1,"BCC8", c("BCC9","BCC10","BCC11","BCC12"))
Result should be:
BIC BCC1 BCC2 BCC6 BCC8 BCC9 BCC10 BCC11
990081899A 0 1 0 0 0 0 0
9900023620 0 1 1 0 0 0 0
9900427160 0 1 0 0 0 0 0
990064457TA 1 1 0 0 0 0 0
990066595A 0 0 0 1 0 0 0
990088248A 0 0 0 1 0 0 0
990088882C1 0 0 0 1 0 0 0
990088882C2 0 0 0 1 0 0 0
990088882C3 0 0 0 1 0 0 0
990088882C4 0 0 0 0 1 0 0
990088882C5 0 0 0 0 1 0 0
But with this function is not working properly. I am not able to find another way or correct way of doing. Any suggestions are appreciated. Thanks!

Resources