Create column in a data.frame with a specific condition - r

I would like to create a column in a data.frame, placing the first time that the year appears in each id.
That is, I have this data:
example <- structure(list(id = structure(c(1, 2, 3, 4, 5), class = "numeric"),
`2007` = c(0, 0, 0, 0, 0), `2008` = c(0, 0, 0, 0, 1), `2009` = c(1,
0, 0, 0, 0), `2010` = c(1, 0, 1, 0, 1), `2011` = c(0, 0,
0, 0, 0), `2012` = c(1, 0, 1, 1, 1), `2013` = c(1, 0, 1,
0, 1), `2014` = c(1, 1, 1, 1, 0), `2015` = c(1, 1, 0, 0,
0), `2016` = c(1, 1, 1, 0, 1)), row.names = c(NA, 5L), class = "data.frame")
And I would like to get the following:
example2 <- structure(list(id = structure(c(1, 2, 3, 4, 5), class = "numeric"),
`2007` = c(0, 0, 0, 0, 0), `2008` = c(0, 0, 0, 0, 1), `2009` = c(1,
0, 0, 0, 0), `2010` = c(1, 0, 1, 0, 1), `2011` = c(0, 0,
0, 0, 0), `2012` = c(1, 0, 1, 1, 1), `2013` = c(1, 0, 1,
0, 1), `2014` = c(1, 1, 1, 1, 0), `2015` = c(1, 1, 0, 0,
0), `2016` = c(1, 1, 1, 0, 1), situation = c(2009, 2014,
2010, 2012, 2008)), row.names = c(NA, 5L), class = "data.frame")
Is it possible to do that ? Every help is welcome. Thanks.

Try this:
#Code
example$situation <- apply(example[,-1],1,function(x) names(x)[min(which(x==1))])
Output:
example
id 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 situation
1 1 0 0 1 1 0 1 1 1 1 1 2009
2 2 0 0 0 0 0 0 0 1 1 1 2014
3 3 0 0 0 1 0 1 1 1 0 1 2010
4 4 0 0 0 0 0 1 0 1 0 0 2012
5 5 0 1 0 1 0 1 1 0 0 1 2008
Or with dplyr and tidyr reshaping and merging:
library(dplyr)
library(tidyr)
#Code
example <- example %>%
left_join(
example %>% pivot_longer(-1) %>%
group_by(id) %>%
summarise(situation=name[min(which(value==1))])
)
Same output.

Related

Create column conditioning the behavior of rows in the dataset

I would like to do something very specific. I have a vast set of data, which, in summary, looks more or less like this, with values 0, 1 and 2:
I need to create a situation variable so that it contains the value 0, 1 and 2.
The value 0 for cases that contain only 0's and 1's in the entire line.
The value 1 for the case where the value 2 appears, but at some point 1 appears before it.
The value 2 for the case where the value 2 appears, but at some point 0 appears before it.
So it's something close to:
structure(list(X1 = c(1, 1, 1, 1, 1, 1, 1, 1, 0, 1), X2 = c(1,
1, 1, 1, 0, 0, 0, 0, 0, 2), X3 = c(0, 1, 1, 1, 1, 0, 0, 1, 0,
0), X4 = c(0, 1, 1, 0, 1, 1, 0, 0, 0, 0), X5 = c(2, 1, 1, 0,
2, 1, 1, 0, 0, 0), X6 = c(2, 1, 1, 0, 2, 1, 1, 0, 0, 0), X7 = c(2,
1, 1, 1, 2, 1, 1, 2, 0, 0), X8 = c(0, 1, 1, 1, 2, 1, 2, 2, 2,
0)), class = "data.frame", row.names = c(NA, 10L))
I wrote a score function and applied it over all the rows of your dataframe.
score <- function(x) {
a <- which(x == 2)
ifelse(length(a) > 0, ifelse(a[1] >=2, 2 - x[a[1] - 1], 1), 0)
}
df <- structure(list(X1 = c(1, 1, 1, 1, 1, 1, 1, 1, 0, 1),
X2 = c(1, 1, 1, 1, 0, 0, 0, 0, 0, 2),
X3 = c(0, 1, 1, 1, 1, 0, 0, 1, 0, 0),
X4 = c(0, 1, 1, 0, 1, 1, 0, 0, 0, 0),
X5 = c(2, 1, 1, 0, 2, 1, 1, 0, 0, 0),
X6 = c(2, 1, 1, 0, 2, 1, 1, 0, 0, 0),
X7 = c(2, 1, 1, 1, 2, 1, 1, 2, 0, 0),
X8 = c(0, 1, 1, 1, 2, 1, 2, 2, 2, 0)),
class = "data.frame", row.names = c(NA, 10L))
df$situation <- sapply(1:nrow(df), function(i) score(as.numeric(df[i,])))
df
Here's a tidyverse approach.
I'll first concatenate all columns together, then use grepl() to look for 12 or 02.
library(tidyverse)
df %>% rowwise() %>%
mutate(concat = paste(c_across(everything()), collapse = "")) %>%
ungroup() %>%
mutate(situation = case_when(
!grepl(2, concat) ~ 0,
grepl("12", concat) ~ 1,
grepl("02", concat) ~ 2
)) %>%
select(-concat)
Output
# A tibble: 10 x 9
X1 X2 X3 X4 X5 X6 X7 X8 situation
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 0 0 2 2 2 0 2
2 1 1 1 1 1 1 1 1 0
3 1 1 1 1 1 1 1 1 0
4 1 1 1 0 0 0 1 1 0
5 1 0 1 1 2 2 2 2 1
6 1 0 0 1 1 1 1 1 0
7 1 0 0 0 1 1 1 2 1
8 1 0 1 0 0 0 2 2 2
9 0 0 0 0 0 0 0 2 2
10 1 2 0 0 0 0 0 0 1
Note that this solution assumes that:
2 will not appear in the first column
1 or 2 in the situation is defined by the number immediately before 2 in your dataset
There will not be a case of 12 and 02 happening in the same row

Create new columns using across() and if_else()

I have survey data that has a binary 1, 0 (indicating peak or off-peak) variable with the related peak or off-peak numbers in two separate columns.
structure(list(q9_jul_2019 = c(1, 0, 1, 0, 1, 0), q9_aug_2019 = c(1,
0, 1, 0, 1, 0), q9_sep_2019 = c(1, 0, 1, 0, 1, 0), q9_oct_2019 = c(0,
0, 1, 0, 1, 0), q9_nov_2019 = c(0, 0, 1, 0, 1, 0), q9_dec_2019 = c(0,
0, 1, 0, 0, 0), q9_jan_2020 = c(0, 0, 1, 0, 0, 0), q9_feb_2020 = c(0,
1, 0, 1, 0, 0), q9_mar_2020 = c(1, 1, 0, 1, 0, 0), q9_apr_2020 = c(1,
1, 1, 1, 0, 1), q9_may_2020 = c(0, 1, 0, 0, 0, 0), q9_jun_2020 = c(0,
0, 0, 0, 0, 0), q15 = c(1, 10, 30, 0, 2, 0), q22 = c(0, 10, 6,
0, 0, 0)), row.names = c(NA, 6L), class = "data.frame")
I have created new monthly columns that have the associated visitation numbers in that column but I'm sure there must be a neater way to do it using across(). I haven't been able to make it work though, so at the moment I'm stuck at the following:
survey <- survey %>%
mutate(visitation_jul_19 = if_else(q9_jul_2019 == 1, q15, q22),
visitation_aug_19 = if_else(q9_aug_2019 == 1, q15, q22),
visitation_sep_19 = if_else(q9_sep_2019 == 1, q15, q22),
visitation_oct_19 = if_else(q9_oct_2019 == 1, q15, q22),
visitation_nov_19 = if_else(q9_nov_2019 == 1, q15, q22),
visitation_dec_19 = if_else(q9_dec_2019 == 1, q15, q22),
visitation_jan_20 = if_else(q9_jan_2020 == 1, q15, q22),
visitation_feb_20 = if_else(q9_feb_2020 == 1, q15, q22),
visitation_mar_20 = if_else(q9_mar_2020 == 1, q15, q22),
visitation_apr_20 = if_else(q9_apr_2020 == 1, q15, q22),
visitation_may_20 = if_else(q9_may_2020 == 1, q15, q22),
visitation_jun_20 = if_else(q9_jun_2020 == 1, q15, q22))
You may try
library(dplyr)
survey %>%
mutate(across(q9_jul_2019:q9_jun_2020, ~ ifelse(.x == 1, q15, q22)))
q9_jul_2019 q9_aug_2019 q9_sep_2019 q9_oct_2019 q9_nov_2019 q9_dec_2019 q9_jan_2020 q9_feb_2020 q9_mar_2020 q9_apr_2020
1 1 1 1 0 0 0 0 0 1 1
2 10 10 10 10 10 10 10 10 10 10
3 30 30 30 30 30 30 30 6 6 30
4 0 0 0 0 0 0 0 0 0 0
5 2 2 2 2 2 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0
q9_may_2020 q9_jun_2020 q15 q22
1 0 0 1 0
2 10 10 10 10
3 6 6 30 6
4 0 0 0 0
5 0 0 2 0
6 0 0 0 0

Obtaining a previous value from a column and differencing in R

I have a bit of a question about obtaining a previous value from one column and putting it into another column. I have a data.table object (which can easily be converted into an xts object) as follows:
The dput output is as follows:
structure(list(Time = structure(c(1122855314, 1122855315, 1122855316,
1122855317, 1122855318, 1122855319, 1122855320, 1122955811, 1122955812,
1122955813, 1122955814, 1123027212, 1123027213, 1123027214, 1123027215,
1123027216, 1123027217), class = c("POSIXct", "POSIXt"), tzone = "Australia/Melbourne"),
`Inventory_{t}` = c(0, 2, 2, 2, 5, 8, 3, 7, 6, 6, 1, 0, 1,
1, 3, 3, 3), `Inventory_{t-1}` = c(0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0), `Delta Inventory_{t-1}` = c(0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), .Names = c("Time",
"Inventory_{t}", "Inventory_{t-1}", "Delta Inventory_{t-1}"), row.names = c(NA,
-17L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x00000000028b0788>)
I would like to 'fill in' the "Inventory_{t-1}" such that it takes the value which was in "Inventory_{t}" one second earlier and puts it into that cell. Similarly, for "Delta Inventory_{t-1}" I want it to be equal to Delta Inventory_{t-1} = Inventory_{t-1} - Inventory_{t-2}
I should also note that at the start of each new day, the initial values for "Inventory_{t-1}" and "Delta Inventory_{t-1}" must be 0.
With this information, I would like to get a new data.table/xts object which looks like this:
structure(list(Time = structure(c(1122855314, 1122855315, 1122855316,
1122855317, 1122855318, 1122855319, 1122855320, 1122955811, 1122955812,
1122955813, 1122955814, 1123027212, 1123027213, 1123027214, 1123027215,
1123027216, 1123027217), class = c("POSIXct", "POSIXt"), tzone = "Australia/Melbourne"),
`Inventory_{t}` = c(0, 2, 2, 2, 5, 8, 3, 7, 6, 6, 1, 0, 1,
1, 3, 3, 3), `Inventory_{t-1}` = c(0, 0, 2, 2, 2, 5, 8, 0,
7, 6, 6, 0, 0, 1, 1, 3, 3), `Delta Inventory_{t-1}` = c(0,
0, 2, 0, 0, 3, 3, 0, 7, -1, 0, 0, 0, 1, 0, 2, 0)), .Names = c("Time",
"Inventory_{t}", "Inventory_{t-1}", "Delta Inventory_{t-1}"), row.names = c(NA,
-17L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x00000000028b0788>)
The things is, this issue is very straightforward for me to solve if I use loops, but since I have so much data I was hoping for a much faster way to do this, so if anyone can help me out with this I'd really appreciate it, thanks in advance.
This can be solved using the shift() function. The OP has requested to restart the calculation anew every day. This is accomplished by the by = parameter:
z[, `:=`(`Inventory_{t-1}` = shift(`Inventory_{t}`, fill = 0),
`Delta Inventory_{t-1}` = shift(`Inventory_{t}`, fill = 0) -
shift(`Inventory_{t}`, n = 2L, fill = 0)), by = .(Day = as.Date(Time))][]
Time Inventory_{t} Inventory_{t-1} Delta Inventory_{t-1}
1: 2005-08-01 10:15:14 0 0 0
2: 2005-08-01 10:15:15 2 0 0
3: 2005-08-01 10:15:16 2 2 2
4: 2005-08-01 10:15:17 2 2 0
5: 2005-08-01 10:15:18 5 2 0
6: 2005-08-01 10:15:19 8 5 3
7: 2005-08-01 10:15:20 3 8 3
8: 2005-08-02 14:10:11 7 0 0
9: 2005-08-02 14:10:12 6 7 7
10: 2005-08-02 14:10:13 6 6 -1
11: 2005-08-02 14:10:14 1 6 0
12: 2005-08-03 10:00:12 0 0 0
13: 2005-08-03 10:00:13 1 0 0
14: 2005-08-03 10:00:14 1 1 1
15: 2005-08-03 10:00:15 3 1 0
16: 2005-08-03 10:00:16 3 3 2
17: 2005-08-03 10:00:17 3 3 0

How to frequency of consecutive rows with the same number for several columns

I have a dataset as follows:
structure(list(chr = c(1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1,
1, 0, 0, 1, 1, 1, 1), leftPos = c(240000, 1080000, 1200000, 1320000,
1440000, 1800000, 2400000, 2520000, 3120000, 3360000, 3480000,
3600000, 3720000, 4200000, 4560000, 4920000, 5040000, 5160000,
5280000, 6e+06), chr.1 = c(1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1,
1, 0, 0, 1, 1, 1, 1, 1), leftPos.1 = c(240000, 1080000, 1200000,
1320000, 1440000, 1800000, 2400000, 2520000, 3120000, 3360000,
3480000, 3600000, 3720000, 4200000, 4560000, 4920000, 5040000,
5160000, 5280000, 6e+06), ASample = c(0,
0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0), Sample1 = c(0,
1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1), Sample2 = c(0,
1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1), Sample3 = c(0,
1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1), Sample4 = c(0,
0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1), Sample5 = c(0,
1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1), Sample6 = c(0,
1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1), Sample7 = c(0,
0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1), Sample8 = c(0,
1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1), Sample9 = c(0,
0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1), Sample10 = c(0,
1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1), Sample11 = c(0,
1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1), Sample12 = c(0,
1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1), Sample13 = c(0,
0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0), Sample14 = c(0,
1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1), Sample15 = c(0,
1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1)), .Names = c("chr",
"leftPos", "chr.1", "leftPos.1", "Sample1",
"Sample2",
"Sample3", "Sample4",
"Sample5", "Sample6",
"Sample7", "Sample8",
"Sample9", "Sample10",
"Sample11", "Sample12",
"Sample13", "Sample14",
"Sample15"), row.names = c(NA,
20L), class = "data.frame")
I need to count the number of rows per column where there is more than one of the same 1 or -1
I would like to be able to count the number of consecutive rows for each column, grouped by chr, that have three consecutive 1 or -1 within a chromosome (column called chr).
The ideal output would be something like (not taken from the dput data above)
chr numberOfConsecutive1s FreqSample1 FreqSample2 FreqSample3 etc
1 2 3 2 14
1 3 5 2 2
1 4 5 0 6
1 5 4 3 5
1 6 3 0 3
1 7 7 5 7
1 8 5 0 2
1 9 54 2 6
1 10 34 77 7
2 2 6 4 2
2 3 23 34 34
2 4 5 37 2
2 5 55 24 22
2 6 2 0 11
2 7 3 14 5
2 8 2 5 77
2 9 5 23 34
2 10 5 11 34
3 1 32 0 2
So far I have tried the following which simply converts non consecutive 1s into 0s so I am left with the consecutive 1s only. I don't know how to count them up as per the desired output.
dx<-DAT_list2res
f0 <- function( colNr, dx )
{
col <- dx[,colNr]
n1 <- which( col == 1 ) # The `1`-rows.
d0 <- which( diff(col) == 0 ) # Consecutive entries are equal.
dc0 <- which( diff(dx[,1]) == 0 ) # Same chromosome.
m <- intersect( n1-1, intersect( d0, dc0 ) )
return ( setdiff( 1:nrow(dx), union(m,m+1) ) )
}
g <- function( dx )
{
for ( i in 3:ncol(dx) ) { dx[f0(i,dx),i] <- 0 }
return ( dx )
}
dx<-g(dx)
EDIT
I also tried this as suggested by bramtayl:
result =
consecFreq %>%
select(-chr) %>%
gather(variable, chr, 5:190) %>%
group_by(variable) %>%
mutate(ID =
chr %>%
lag %>%
`!=`(chr) %>%
plyr::mapvalues(NA, FALSE) %>%
cumsum) %>%
count(variable, chr, ID) %>%
rename(numberOfConsecutive1s = n) %>%
count(variable, chr, numberOfConsecutive1s) %>%
spread(variable, n, fill = 0)
but it gives me an 'index out of bounds' error. If I ignore the spread line I get an odd output as well so I'm not sure this is the answer
REVISED
Based on clarifications, this approach uses the rle function for each chromosome to find runs of consecutive 1's or -1's and then table to count the number of runs for each value. This gives NA for samples which have no counts for a particular value so the last line of the code converts the NA's to 0's if this is helpful. Finally there seems to be a problem with your structure input in that Cytospongex10_SLX.9395.FastSeqK.fq.gz.res is missing from the .Names section of structure. This causes all the column names to be shifted and the last column name to be NA which can cause problems in the execution.
The code below assigns the correct names to the input data (in data.frame df) and then calculates the frequencies as described above.
colnames(data) <- c("chr",
"leftPos", "chr.1", "leftPos.1", "Cytospongex10_SLX.9395.FastSeqK.fq.gz.res", "Sample1",
"Sample2",
"Sample3", "Sample4",
"Sample5", "Sample6",
"Sample7", "Sample8",
"Sample9", "Sample10",
"Sample11", "Sample12",
"Sample13", "Sample14",
"Sample15")
chr_labels <- sort(unique(data$chr))
sampl_freqs <- data.frame(chr=1, numberOfConsecutive1s=1, count=0)
for( sampl in colnames(data)[-(1:5)]) {
freqs <- data.frame()
for( chr in chr_labels ) {
runs <- rle(data[data$chr == chr,sampl])
freqs_chr <- data.frame(chr=chr, table(runs$length[runs$values %in% c(-1,1)], dnn = "numberOfConsecutive1s") )
freqs <- rbind(freqs, freqs_chr)
}
sampl_freqs <- merge.data.frame(sampl_freqs, freqs, by = c("chr","numberOfConsecutive1s"), all=TRUE)
colnames(sampl_freqs) <- c(head(colnames(sampl_freqs),-1),paste("Freq",sampl,sep=""))
}
# clean up from sampl_freqs definition
sampl_freqs <- sampl_freqs[,-3]
# To convert NA's to 0
sampl_freqs <- data.frame(sampl_freqs[,1:2], sapply(sampl_freqs[,-(1:2)], function(x) ifelse(is.na(x), 0, x)))
Similar to above, but uses dplyr
library(reshape2)
library(dplyr)
df <- melt(data[,-(2:5)], id.vars="chr", variable.name="sample")
sampl_freqs <- df %>% group_by(sample, chr ) %>%
do(data.frame(unclass(rle(.$value))) %>%
filter(values %in% c(-1,1)) ) %>%
group_by(sample, chr, lengths) %>%
summarize(Freq = n() ) %>%
dcast( chr + lengths ~ sample, value.var = "Freq" )
sampl_freqs <- with(sampl_freqs,data.frame( chr, numberOfConsecutive1s = lengths ,
sapply(sampl_freqs[,-(1:2)], function(x) ifelse(is.na(x), 0, x))))
I think you want something like this:
library(dplyr)
library(tidyr)
min_chunk_length = 1
result =
data %>%
rename(chromosome = chr) %>%
select(chromosome, Sample1:Sample15) %>%
gather(sample, value, Sample1:Sample15) %>%
group_by(chromosome, sample) %>%
mutate(non_zero = value %in% c(1, -1),
chunk_ID =
non_zero %>%
lag %>%
`!=`(non_zero) %>%
plyr::mapvalues(NA, FALSE) %>%
cumsum) %>%
filter(non_zero = TRUE) %>%
group_by(chromosome, sample, chunk_ID) %>%
mutate(length_of_chunk = n()) %>%
filter(length_of_chunk > min_chunk_length) %>%
count(chromosome, sample) %>%
spread(sample, n, fill = 0)

Code multiple levels as 2 factor labels

I have a data frame with some columns:
that I want to transform into a factor,
in which the different levels are coded as -2, -1, 0, 1, 2, 3, 4
for which I want the levels to be labeled as 0 or 1 following this convention:
-2 = 1
-1 = 1
0 = 0
1 = 1
2 = 1
3 = 1
4 = 0
I have the following code:
#Convert to factor
dat[idx] <- lapply(dat[idx], factor, levels = -2:4, labels = c(1, 1, 0, 1, 1, 1, 0))
#Drop unused factor levels
dat <- droplevels(dat)
This works, but it gives me the following warning:
In `levels<-`(`*tmp*`, value = if (nl == nL) as.character(labels) else paste0(labels, :
duplicated levels in factors are deprecated
I tried the following code (per Ananda Mahto's suggestion) but no luck:
levels(dat[idx]) <- list(`0` = c(0, 4), `1` = c(-2, -1, 1, 2, 3))
I figured there has to be a better way to do this, any suggestions?
My data looks like this:
structure(list(Timestamp = structure(c(1380945601, 1380945603,
1380945605, 1380945607, 1380945609, 1380945611, 1380945613, 1380945615,
1380945617, 1380945619), class = c("POSIXct", "POSIXt"), tzone = ""),
FCB2C01 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), RCB2C01 = c(0,
0, 0, 0, 0, 0, 0, 0, 0, 0), FCB2C02 = c(1, 1, 1, 1, 1, 1,
1, 1, 1, 1), RCB2C02 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), FCB2C03 = c(0,
0, 0, 0, 0, 0, 0, 0, 0, 0), RCB2C03 = c(0, 0, 0, 0, 0, 0,
0, 0, 0, 0), FCB2C04 = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1), RCB2C04 = c(0,
0, 0, 0, 0, 0, 0, 0, 0, 0), FCB2C05 = c(1, 1, 1, 1, 1, 1,
1, 1, 1, 1), RCB2C05 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), FCB2C06 = c(1,
1, 1, 1, 1, 1, 1, 1, 1, 1), RCB2C06 = c(0, 0, 0, 0, 0, 0,
0, 0, 0, 0), FCB2C07 = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1), RCB2C07 = c(0,
0, 0, 0, 0, 0, 0, 0, 0, 0), FCB2C08 = c(1, 1, 1, 1, 1, 1,
1, 1, 1, 1), RCB2C08 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), FCB2C09 = c(1,
1, 1, 1, 1, 1, 1, 1, 1, 1), RCB2C09 = c(0, 0, 0, 0, 0, 0,
0, 0, 0, 0), FCB2C10 = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1), RCB2C10 = c(0,
0, 0, 0, 0, 0, 0, 0, 0, 0)), .Names = c("Timestamp", "FCB2C01",
"RCB2C01", "FCB2C02", "RCB2C02", "FCB2C03", "RCB2C03", "FCB2C04",
"RCB2C04", "FCB2C05", "RCB2C05", "FCB2C06", "RCB2C06", "FCB2C07",
"RCB2C07", "FCB2C08", "RCB2C08", "FCB2C09", "RCB2C09", "FCB2C10",
"RCB2C10"), row.names = c(NA, 10L), class = "data.frame")
And the column index:
idx <- seq(2,21,2)
If I correctly understand what you want to do, the "right" way would be to use the levels function to specify your levels. Compare the following:
set.seed(1)
x <- sample(-2:4, 10, replace = TRUE)
YourApproach <- factor(x, levels = -2:4, labels = c(1, 1, 0, 1, 1, 1, 0))
# Warning message:
# In `levels<-`(`*tmp*`, value = if (nl == nL) as.character(labels) else paste0(labels, :
# duplicated levels in factors are deprecated
YourApproach
# [1] 1 0 1 0 1 0 0 1 1 1
# Levels: 1 1 0 1 1 1 0
xFac <- factor(x, levels = -2:4)
levels(xFac) <- list(`0` = c(0, 4), `1` = c(-2, -1, 1, 2, 3))
xFac
# [1] 1 0 1 0 1 0 0 1 1 1
# Levels: 0 1
Note the difference in the "Levels" in each of those. This also means that the underlying numeric representation is going to be different:
> as.numeric(YourApproach)
[1] 2 3 5 7 2 7 7 5 5 1
> as.numeric(xFac)
[1] 2 1 2 1 2 1 1 2 2 2

Resources