Treshold values row-wise in a dataframe - r

Consider an example data frame:
A B C v
5 4 2 3
7 1 3 5
1 2 1 1
I want to set all elements of a row to 1 if the element is bigger or equal than v, and 0 otherwise. The example data frame would result in the following:
A B C v
1 1 0 3
1 0 0 5
1 1 1 1
How can I do this efficiently? The number of columns will be much higher, and I would like a solution that does not require me to specify the names of the columns individually, and will apply it to all of them (except v) instead.
My solution with a for loop is way too slow.


We can create a logical matrix and coerce to binary
df1[-4] <- +(df1[-4] >= df1$v)

Related

Use if-else function on data frame with multiple values

I have a data frame that contains multiple values in each spot, like this:
ID<-c(1,1,1,2,2,2,2,3,3,4,4,4,5,6,6)
W<-c(29,72,32,33,34,44,42,78,32,42,18,26,10,34,39)
df1<-data.frame(ID, W)
df<-ddply(df1, .(ID), summarize,
X=paste(unique(W),collapse=","))
ID X
1 1 29,72,32
2 2 33,34,44,42
3 3 78,32
4 4 42,18,26
5 5 10
6 6 34,39
I am trying to generate another column using an if-else function so that every ID that has an X value greater than 70 will show a 1, and all others will show a 0, like this:
ID X Y
1 1 29,72,32 1
2 2 33,34,44,42 0
3 3 78,32 1
4 4 42,18,26 0
5 5 10 0
6 6 34,39 0
This is the code that I tried:
df$Y <- ifelse(df$X>=70, 1, 0)
But it doesn't work; it only seems to put the first value of each spot through the function:
ID X Y
1 1 29,72,32 0
2 2 33,34,44,42 0
3 3 78,32 1
4 4 42,18,26 0
5 5 10 0
6 6 34,39 0
It worked fine on my one column that has only one value per spot. Is there a way to get to the if-else function to evaluate every value in each spot and assign a 1 if any of them fit the statement?
Thank you, I'm sorry that I do not know a lot of R vocabulary yet.

As 'X' is a string, we can split the 'X' at the , to create a list of vectors, loop over the list with map check if there are any numeric converted values are greater than 70
library(dplyr)
library(purrr)
df %>%
mutate(Y = map_int(strsplit(X, ","), ~ +(any(as.numeric(.x) > 70))))

How do I identifying the first zero in a group of ordered columns?

I'm trying to format a dataset for use in some survival analysis models. Each row is a school, and the time-varying columns are the total number of students enrolled in the school that year. Say the data frame looks like this (there are time invariate columns as well).
Name total.89 total.90 total.91 total.92
a 8 6 4 0
b 1 2 4 9
c 7 9 0 0
d 2 0 0 0
I'd like to create a new column indicating when the school "died," i.e., the first column in which a zero appears. Ultimately I'd like to have this column be "years since 1989" and can re-name columns accordingly.
A more general version of the question, for a series of time ordered columns, how do I identify the first column in which a given value occurs?

Here's a base R approach to get a column with the first zero (x = 0) or NA if there isn't one:
data$died <- apply(data[, -1], 1, match, x = 0)
data
# Name total.89 total.90 total.91 total.92 died
# 1 a 8 6 4 0 4
# 2 b 1 2 4 9 NA
# 3 c 7 9 0 0 3
# 4 d 2 0 0 0 2

Here is an option using max.col with rowSums
df1$died <- max.col(!df1[-1], "first") * NA^!rowSums(!df1[-1])
df1$died
#[1] 4 NA 3 2

how to convert a list with different length of lists to a dataframe in r

I have a list containing three different length of vectors with unique elements for each vector.
data <- list(ARG=letters[1:8],BRZ=c("a","b","c","f","h","g","l","m","n"),US=c("u","b","c","e","h","f","q","a","n","t"))
I would like to convert this list to a data frame by mergering them together, the result is expected as below or similar output, Thank you for helping this.
ID ARG BRZ US
a 1 1 1
b 1 1 1
c 1 1 1
d 1
e 1 1
f 1 1 1
g 1 1
h 1 1 1
l 1
m 1
n 1 1
q 1
t 1
u 1

We use mtabulate and transpose the output
library(qdapTools)
t(mtabulate(data))
Or if we are using base R, then stack into a data.frame with 2 columns and apply the table
table(stack(data))
Assuming that there are no duplicates for each entry. If there are duplicates, then we may need a logical vector coerced to binary
+(table(stack(data)) >0)

How to tidy up a character column?

What I have:
test_df <- data.frame(isolate=c(1,2,3,4,1,2,3,4,5),label=c(1,1,1,1,2,2,2,2,2),alignment=c("--at","at--","--at","--at","a--","acg","a--","a--", "agg"))
> test_df
isolate label alignment
1 1 1 --at
2 2 1 at--
3 3 1 --at
4 4 1 --at
5 1 2 a--
6 2 2 acg
7 3 2 a--
8 4 2 a--
9 5 2 agg
What I want:
I'd like to explode the alignment field into two columns, position and character:
> test_df
isolate label aln_pos aln_char
1 1 1 1 -
2 1 1 2 -
3 1 1 3 a
4 1 1 4 t
...
Not all alignments are the same length, but all alignments with the same label have the same length.
What I've tried:
I was thinking I could use separate to first make each position have its own column, then use gather turn those columns into key value pairs. However, I haven't been able to get the separate part right.

Since you mentioned tidyr::gather, you could try this:
test_df <- data.frame(isolate=c(1,2,3,4,1,2,3,4,5),
label=c(1,1,1,1,2,2,2,2,2),
alignment=c("--at","at--","--at","--at","a--","acg","a--","a--", "agg"),
stringsAsFactors = FALSE)
library(tidyverse)
test_df %>%
mutate(alignment = strsplit(alignment,"")) %>%
unnest(alignment)

In base R, you can use indexing along with creation of a list with strsplit like this.
# make variable a character vector
test_df$alignment <- as.character(test_df$alignment)
# get list of individual characters
myList <- strsplit(test_df$alignment, split="")
then build the data.frame
# construct data.frame
final_df <- cbind(test_df[rep(seq_len(nrow(test_df)), lengths(myList)),
c("isolate", "label")],
aln_pos=sequence(lengths(myList)),
aln_char=unlist(myList))
Here, we take the first two columns of the original data.frame and repeat the rows using rep with a vector input in its second argument telling it how many times to repeat the corresponding value in its first argument. The number of times is calculated with lengths. The second argument of cbind is a call to sequence taking the same lengths output. this produces counts from 1 to the corresponding length. The third argument is the unlisted character values.
this returns
head(final_df, 10)
isolate label aln_pos aln_char
1 1 1 1 -
1.1 1 1 2 -
1.2 1 1 3 a
1.3 1 1 4 t
2 2 1 1 a
2.1 2 1 2 t
2.2 2 1 3 -
2.3 2 1 4 -
3 3 1 1 -
3.1 3 1 2 -

Frequency of Characters in Strings as columns in data frame using R

I have a data frame initial of the following format
> head(initial)
Strings
1 A,A,B,C
2 A,B,C
3 A,A,A,A,A,B
4 A,A,B,C
5 A,B,C
6 A,A,A,A,A,B
and the data frame I want is final
> head(final)
Strings A B C
1 A,A,B,C 2 1 1
2 A,B,C 1 1 1
3 A,A,A,A,A,B 5 1 0
4 A,A,B,C 2 1 1
5 A,B,C 1 1 1
6 A,A,A,A,A,B 5 1 0
to generate the data frames the following codes can be used to keep the number of rows high
initial<-data.frame(Strings=rep(c("A,A,B,C","A,B,C","A,A,A,A,A,B"),100))
final<-data.frame(Strings=rep(c("A,A,B,C","A,B,C","A,A,A,A,A,B"),100),A=rep(c(2,1,5),100),B=rep(c(1,1,1),100),C=rep(c(1,1,0),100))
What is the fastest way I can achieve this? Any help will be greatly appreciated

We can use base R methods for this task. We split the 'Strings' column (strsplit(...)), set the names of the output list with the sequence of rows, stack to convert to data.frame with key/value columns, get the frequency with table, convert to 'data.frame' and cbind with the original dataset.
cbind(df1, as.data.frame.matrix(
table(
stack(
setNames(
strsplit(as.character(df1$Strings),','), 1:nrow(df1))
)[2:1])))
# Strings A B C D
#1 A,B,C,D 1 1 1 1
#2 A,B,B,D,D,D 1 2 0 3
#3 A,A,A,A,B,C,D,D 4 1 1 2
or we can use mtabulate after splitting the column.
library(qdapTools)
cbind(df1, mtabulate(strsplit(as.character(df1$Strings), ',')))
# Strings A B C D
#1 A,B,C,D 1 1 1 1
#2 A,B,B,D,D,D 1 2 0 3
#3 A,A,A,A,B,C,D,D 4 1 1 2
Update
For the new dataset 'initial', the second method works. If we need to use the first method with the correct order, convert to factor class with levels specified as the unique elements of 'ind'.
df1 <- stack(setNames(strsplit(as.character(initial$Strings), ','),
seq_len(nrow(initial))))
df1$ind <- factor(df1$ind, levels=unique(df1$ind))
cbind(initial, as.data.frame.matrix(table(df1[2:1])))

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