I have a dataset (df) which looks like:
df
X24_TT X35_FTT X55_FTT X70_FTT X85_TT X86_FTT X90_FTT X96_FTT
0 1 0 0 0 1 0 1
1 0 0 1 1 1 0 1
1 0 0 0 0 1 0 1
0 1 0 0 0 0 1 1
0 0 1 1 0 1 0 0
1 0 0 0 0 0 0 0
I want to keep only those columns whose column sums are greater than 20th percentile i.e.
a) Calculate colsum for each column in "df" i.e an array of colsums
b) Find 20th percentile value of that array of colsums
c) Keep columns whose colsum > 20th percentile value
If we calculate the 20th percentile of colsum of "df" it comes to be 1.
Therefore in a new dataset(nm), only those columns appears whose column sum are greater than 1 i.e.
nm
X24_TT X35_FTT X70_FTT X86_FTT X96_FTT
0 1 0 1 1
1 0 1 1 1
1 0 0 1 1
0 1 0 0 1
0 0 1 1 0
1 0 0 0 0
Please note that number of columns are huge therefore provide dynamic solutions
Related
I want to create a new column based on some conditions imposed on several columns. For example, here is an example dataset:
a <- data.frame(x=c(1,0,1,0,0), y=c(0,0,0,0,0), z=c(1,1,0,0,0))
a
x y z
1 1 0 1
2 0 0 1
3 1 0 0
4 0 0 0
5 0 0 0
Specifically, if for any particular row 1 is present, then the new column returns 1. If all are 0, then the new column returns 0. So the dataset with the new column will be
x y z w
1 1 0 1 1
2 0 0 1 1
3 1 0 0 1
4 0 0 0 0
5 0 0 0 0
My initial thought was to use %in% but couldn't get the result I want. Thank you for your help!
If your data frame consists of binary values, e.g., only 0 and 1, you can try the code below with rowSums
a$w <- +(rowSums(a)>0)
such that
> a
x y z w
1 1 0 1 1
2 0 0 1 1
3 1 0 0 1
4 0 0 0 0
5 0 0 0 0
We can use rowMaxs from matrixStats
library(matrixStats)
a$w <- rowMaxs(as.matrix(a))
a$w
#[1] 1 1 1 0 0
You can find max of each row :
a$w <- do.call(pmax, a)
a
# x y z w
#1 1 0 1 1
#2 0 0 1 1
#3 1 0 0 1
#4 0 0 0 0
#5 0 0 0 0
which can also be done with apply :
a$w <- apply(a, 1, max)
I have a dataset which looks this this:
A B X50_TT_1.0 X50_TT_1.1 X60_DD_2.0 X60_DD_2.1 X100_L2V_7.0 X100_L2V_7.1
3 1 1 0 0 1 1 0
6 3 0 1 0 1 0 1
2 3 1 0 0 1 1 0
10 5 0 1 1 0 1 0
0 0 1 0 1 0 0 1
I want to have new data frame (df) which only contains columns which ends with 1.1, 2.1 i.e.
df
X50_TT_1.1 X60_DD_2.1 X100_L2V_7.1
0 1 0
1 1 1
0 1 0
1 0 0
0 0 1
As here I only shows few columns but actually it contains more than 100 columns. Therefore, kindly provide the solution which can be applicable to as many columns dataset consists.
Thanks in advance.
I guess the pattern is, that the column ends on ".1" may you need to adapt it at that point.
My data I am using
original_data
A B X50_TT_1.0 X50_TT_1.1 X60_DD_2.0 X60_DD_2.1 X100_L2V_7.0 X100_L2V_7.1
1 3 1 1 0 0 1 1 0
Actually this is for everything ending with "1"
df <- original_data[which(grepl(".1$", names(original_data)))]
For ending with ".1" you have to use:
df <- original_data[which(grepl("\\.1$", names(original_data)))]
For original_data both gave me the same result:
df
X50_TT_1.1 X60_DD_2.1 X100_L2V_7.1
1 0 1 0
I'm looking for a function in R which can do the permutation. For example, I have a vector with five 1 and ten 0 like this:
> status=c(rep(1,5),rep(0,10))
> status
[1] 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0
Now I'd like to randomly permute the position of these numbers but keep the same number of 0 and 1 in vector and to get new series of number, for example to get something like this:
1 1 0 1 0 1 0 0 0 0 0 1 0 0 0
or
1 0 0 0 0 0 0 1 1 0 0 1 0 1 0
I found the function sample() can help us to sample, but the number of 1 and 0 is not the same each time. Do you know how can I do this with R? Thanks in advance.
We can use sample
sample(status)
#[1] 1 0 0 1 0 0 1 0 0 0 0 1 0 1 0
sample(status)
#[1] 0 0 0 0 1 1 0 0 1 1 0 0 0 1 0
If we use sample to return the entire vector, it will do the permutation and give the frequency count same for each of the unique elements
colSums(replicate(5, sample(status)))
#[1] 5 5 5 5 5
i.e. we get 5 one's in each of the sampling. So, the remaining 0's would be 10.
I have newly started to learn R, so my question may be utterly ridiculous. I have a data frame
data<- data.frame('number'=1:11, 'col1'=sample(10:20),'col2'=sample(10:20),'col3'=sample(10:20),'col4'=sample(10:20),'col5'=sample(10:20), 'date'= c('12-12-2014','12-11-2014','12-10-2014','12-09-2014', '12-08-2014','12-07-2014','12-06-2014','12-05-2014','12-04-2014', '12-04-2014', '12-03-2014') )
The number column is an 'id' column and the last column is a date.
I want to count the number of times that each number occurs across (not per column, but the whole data frame containing data) the columns 2:6 and when they occurred.
I am stuck on the first part having tried the following using data.table:
count <- function(){
i = 1
DT <-data.table(data[2:6])
for (i in 10:20){
DT[, .N, by =i]
i = i + 1
}
}
which gives an error that I don't begin to understand
Error in `[.data.table`(DT, , .N, by = i) :
The items in the 'by' or 'keyby' list are length (1). Each must be same length as rows in x or number of rows returned by i (11)
Can someone help, please. Also with the second part that I have not even attempted yet i.e. associating a date or a row number with each occurrence of a number
Perhaps you may want this
library(reshape2)
table(melt(data[,-1], id.var='date')[,-2])
# value
#date 10 11 12 13 14 15 16 17 18 19 20
# 12-03-2014 0 0 1 0 0 1 0 0 1 2 0
# 12-04-2014 2 0 0 2 2 0 1 0 1 1 1
# 12-05-2014 0 0 0 0 0 0 1 1 2 0 1
# 12-06-2014 1 1 0 0 0 1 0 1 0 0 1
# 12-07-2014 0 1 0 1 0 1 1 1 0 0 0
# 12-08-2014 1 1 0 0 1 0 0 1 1 0 0
# 12-09-2014 0 0 2 0 1 2 0 0 0 0 0
# 12-10-2014 0 0 1 1 0 0 1 0 0 1 1
# 12-11-2014 0 1 1 0 0 0 1 0 0 1 1
# 12-12-2014 1 1 0 1 1 0 0 1 0 0 0
Or if you need a data.table solution (from #Arun's comments)
library(data.table)
dcast.data.table(melt(setDT(data),
id="date", measure=2:6), date ~ value)
I would like to generate a random binary combination (row order) in my dataframe df:
bin
2
2
2
2
3
2
3
2
In this example I intend to generate 6 times 0 (the same number of 2) and two times 1 (the same number of 3). I expect something like that:
bin
0
0
1
0
0
1
0
0
Any ideas? Thank you
So given a vector bin
bin<-c(2,2,2,2,3,2,3,2)
You would like to create a new vector that contains the same number of 0's as the number of 2's in bin, and the same number of 1's as the number of 3's in bin. Assuming that's correct, then
sample(rep(0:1, table(bin)))
Should do the trick. Here are the results of running that command several times:
# 0 0 0 0 1 1 0 0
# 0 0 0 1 0 0 1 0
# 0 0 0 1 0 0 1 0
# 0 0 1 0 1 0 0 0