recoding based on two condtions in r - r

I have an example dataset looks like:
data <- as.data.frame(c("A","B","C","X1_theta","X2_theta","AB_theta","BC_theta","CD_theta"))
colnames(data) <- "category"
> data
category
1 A
2 B
3 C
4 X1_theta
5 X2_theta
6 AB_theta
7 BC_theta
8 CD_theta
I am trying to generate a logical variable when the category (variable) contains "theta" in it. However, I would like to assign the logical value as "FALSE" when cell values contain "X1" and "X2".
Here is what I did:
data$logic <- str_detect(data$category, "theta")
> data
category logic
1 A FALSE
2 B FALSE
3 C FALSE
4 X1_theta TRUE
5 X2_theta TRUE
6 AB_theta TRUE
7 BC_theta TRUE
8 CD_theta TRUE
here, all cells value that have "theta" have the logical value of "TRUE".
Then, I wrote this below to just assign "FALSE" when the cell value has "X" in it.
data$logic <- ifelse(grepl("X", data$category), "FALSE", "TRUE")
> data
category logic
1 A TRUE
2 B TRUE
3 C TRUE
4 X1_theta FALSE
5 X2_theta FALSE
6 AB_theta TRUE
7 BC_theta TRUE
8 CD_theta TRUE
But this, of course, overwrote the previous application
What I would like to get is to combine two conditions:
> data
category logic
1 A FALSE
2 B FALSE
3 C FALSE
4 X1_theta FALSE
5 X2_theta FALSE
6 AB_theta TRUE
7 BC_theta TRUE
8 CD_theta TRUE
Any thoughts?
Thanks

We can create the 'logic', by detecting substring 'theta' at the end and not having 'X' ([^X]) as the starting (^) character
libary(dplyr)
library(stringr)
library(tidyr)
data %>%
mutate(logic = str_detect(category, "^[^X].*theta$"))
If we need to split the column into separate columns based on the conditions
data %>%
mutate(logic = str_detect(category, "^[^X].*theta$"),
category = case_when(logic ~ str_replace(category, "_", ","),
TRUE ~ as.character(category))) %>%
separate(category, into = c("split1", "split2"), sep= ",", remove = FALSE)
# category split1 split2 logic
#1 A A <NA> FALSE
#2 B B <NA> FALSE
#3 C C <NA> FALSE
#4 X1_theta X1_theta <NA> FALSE
#5 X2_theta X2_theta <NA> FALSE
#6 AB,theta AB theta TRUE
#7 BC,theta BC theta TRUE
#8 CD,theta CD theta TRUE
Or in base R
data$logic <- with(data, grepl("^[^X].*theta$", category))
Another option is to have two grepl condition statements
data$logic <- with(data, grepl("theta$", category) & !grepl("^X\\d+", category))
data$logic
#[1] FALSE FALSE FALSE FALSE FALSE TRUE TRUE TRUE

Not the cleanest in the world (since it adds 2 unnecessary cols) but it gets the job done:
data <- as.data.frame(c("A","B","C","X1_theta","X2_theta","AB_theta","BC_theta","CD_theta"))
colnames(data) <- "category"
data$logic1 <- ifelse(grepl('X',data$category), FALSE, TRUE)
data$logic2 <- ifelse(grepl('theta',data$category),TRUE, FALSE)
data$logic <- ifelse((data$logic1 == TRUE & data$logic2 == TRUE), TRUE, FALSE)
print(data)
I think you can also remove the logic1 and logic2 cols if you want but I usually don't bother (I'm a messy coder haha).
Hope this helped!
EDIT: akrun's grepl solution does what I'm doing way more cleanly (as in, it doesn't require the extra cols). I definitely recommend that approach!

Related

How to match multiple columns without merge?

I have those two df's:
ID1 <- c("TRZ00897", "AAR9832", "NZU44447683209", "sxc89898989M", "RSU765th89", "FFF")
Date1 <- c("2022-08-21","2022-03-22","2022-09-24", "2022-09-21", "2022-09-22", "2022-09-22")
Data1 <- data.frame(ID1,Date1)
ID <- c("RSU765th89", "NZU44447683209", "AAR9832", "TRZ00897","ERD895655", "FFF", "IUHG0" )
Date <- c("2022-09-22","2022-09-21", "2022-03-22", "2022-08-21", "2022-09-21", "2022-09-22", "2022-09-22" )
Data2 <- data.frame(ID,Date)
I tried to get exact matches. An exact match is if ID and Date are the same in both df's, for example: "TRZ00897" "2022-08-21" is an exact match, because it is present in both df's
With the following line of code:
match(Data1$ID1, Data2$ID) == match(Data1$Date1, Data2$Date)
the output is:
TRUE TRUE NA NA TRUE FALSE
Obviously the last one should not be FALSE because "FFF" "2022-09-22" is in both df. The reason why it is FALSE is, that the Date"2022-09-22" occurred already in Data2 at index position 1.
match(Data1$ID1, Data2$ID)
4 3 2 NA 1 6
match(Data1$Date1, Data2$Date)
4 3 NA 2 1 1
So at the end, there is index position 6 and 1 which is not equal --> FALSE
How can I change this? Which function should I use to get the correct answer.
Note, I don't need to merge or join etc. I'm really looking for a function that can detect those patterns.
Combine the columns then match:
match(paste(Data1$ID1, Data1$Date1), paste(Data2$ID, Data2$Date))
# [1] 4 3 NA NA 1 6
To get logical outut use %in%:
paste(Data1$ID1, Data1$Date1) %in% paste(Data2$ID, Data2$Date)
# [1] TRUE TRUE FALSE FALSE TRUE TRUE
Try match with asplit (since you have different column names for two dataframes, I have to manually remove the names using unname, which can be avoided if both of them have the same names)
> match(asplit(unname(Data1), 1), asplit(unname(Data2), 1))
[1] 4 3 NA NA 1 6
Another option that is memory-expensive option is using interaction
> match(interaction(Data1), interaction(Data2))
[1] 4 3 NA NA 1 6
With mapply and %in%:
apply(mapply(`%in%`, Data1, Data2), 1, all)
[1] TRUE TRUE FALSE FALSE TRUE TRUE
rowSums(mapply(`%in%`, Data1, Data2)) == ncol(Data1)
Edit; for a subset of columns:
idx <- c(1, 2)
apply(mapply(`%in%`, Data1[idx], Data2[idx]), 1, all)
#[1] TRUE TRUE FALSE FALSE TRUE TRUE

R: How to make extra rows from a column?

I have a data-set of human hands, where currently a single person is defined as a single observation. I want to reshape dataframe to have hands as individual observations. I tried something with "dplyr" package and "gather" function but had no success at all.
So from this, where each person is on one row :
id Gender Age Present_R Present_L Dominant
1 F 2 TRUE TRUE R
2 F 5 TRUE FALSE L
3 M 8 FALSE FALSE R
to this, where each hand is on one row:
id Gender Age Hand Present Dominant
1 F 2 R TRUE TRUE
2 F 2 L TRUE FALSE
3 F 5 R TRUE FALSE
4 F 5 L FALSE TRUE
5 M 8 R FALSE TRUE
6 M 8 L FALSE FALSE
Note that hand dominance becomes logical.
We can gather into 'long' format, arrange by 'id', then create the 'Dominant' by unlisting the 'Present' columns, 'Hand' by removing the substring of the 'Hand' column
library(tidyverse)
gather(df1, Hand, Present, Present_R:Present_L) %>%
arrange(id) %>%
mutate(Dominant = unlist(df1[c("Present_L", "Present_R")]),
id = row_number(),
Hand = str_remove(Hand, ".*_"))
# id Gender Age Dominant Hand Present
#1 1 F 2 TRUE R TRUE
#2 2 F 2 FALSE L TRUE
#3 3 F 5 FALSE R TRUE
#4 4 F 5 TRUE L FALSE
#5 5 M 8 TRUE R FALSE
#6 6 M 8 FALSE L FALSE
Based on the OP' comments, it seems like we need to compare the 'Dominant' with the 'Hand'
gather(df1, Hand, Present, Present_R:Present_L) %>%
arrange(id) %>%
mutate(id = row_number(),
Hand = str_remove(Hand, ".*_"),
Dominant = Dominant == Hand)
# id Gender Age Dominant Hand Present
#1 1 F 2 TRUE R TRUE
#2 2 F 2 FALSE L TRUE
#3 3 F 5 FALSE R TRUE
#4 4 F 5 TRUE L FALSE
#5 5 M 8 TRUE R FALSE
#6 6 M 8 FALSE L FALSE
With a small data frame (i.e., few variables, regardless of the number of cases), "hand-coding" may be the easiest approach:
with(df, data.frame(id = c(id,id), Gender=c(Gender,Gender), Age=c(Age, Age),
Hand = c(rep("R", nrow(df)), rep("L", nrow(df))),
Present = c(Present_R, Present_L),
Dominant = c(Dominant=="R", Dominant=="L")
))

Create unique groups row wise based on logical vector in data.frame

I think there must be a solution on SO for this but I've been searching around solutions with almost what I want, but not quite. Looking for a tidyverse solution, if possible.
I have a data.frame, say newdf:
newdf <- data.frame(inside.city = c(TRUE, TRUE, TRUE, FALSE, FALSE, TRUE, FALSE, FALSE))
newdf
inside.city
1 TRUE
2 TRUE
3 TRUE
4 FALSE
5 FALSE
6 TRUE
7 FALSE
8 FALSE
Every time someone "leaves the city" (inside.city == FALSE), I want to give their trip a unique group number, so that the resulting data.frame looks like this:
inside.city group
1 TRUE NA
2 TRUE NA
3 TRUE NA
4 FALSE 1
5 FALSE 1
6 TRUE NA
7 FALSE 2
8 FALSE 2
Assume the data are already ordered by the date.
How can I do this efficiently?
Here's a way using mutate(). I just transform the column twice to simplify things
library(dplyr)
newdf %>% mutate(group=cumsum(!inside.city & lag(inside.city, default=TRUE)),
group=ifelse(inside.city, NA, group))
Basically you just increment when you see a FALSE after a TRUE and then set the TRUE values to NA.

Collapse consecutive rows in a data frame

I have this example data.frame:
df <- data.frame(a = c(1,2,3,5,7,8),b=c(2,3,4,6,8,9))
And I'd like to collapse all rows i whose b column value is equal to a column value at their subsequent row (i+1) such that in the collapsed row they their a column will be that of row i and their b column will be that of row i+1. This has to be done as long as there are no consecutive rows that meet this condition.
For the example df rows 1-3 are to be collapsed, row 4 left as is, and then rows 5-6 collapsed, giving:
res.df <- data.frame(a = c(1,5,7), b = c(4,6,9))
This isn't overly pretty, but it is vectorised comparing a cutdown version of df$a to df$b.
grps <- rev(cumsum(rev(c(tail(df$a,-1) != head(df$b,-1),TRUE))))
#[1] 3 3 3 2 1 1
cbind(df["a"], b=ave(df$b,grps,FUN=max) )[!duplicated(grps),]
# a b
#1 1 4
#4 5 6
#5 7 9
Breaking it down probably helps explain the first part:
tail(df$a,-1) != head(df$b,-1)
#[1] FALSE FALSE TRUE TRUE FALSE
c(tail(df$a,-1) != head(df$b,-1),TRUE)
#[1] FALSE FALSE TRUE TRUE FALSE TRUE
rev(c(tail(df$a,-1) != head(df$b,-1),TRUE))
#[1] TRUE FALSE TRUE TRUE FALSE FALSE
cumsum(rev(c(tail(df$a,-1) != head(df$b,-1),TRUE)))
#[1] 1 1 2 3 3 3

how to find if all elements in a subset of a data.frame row are TRUE

I have a data.frame with a block of columns that are logicals, e.g.
> tmp <- data.frame(a=c(13, 23, 52),
+ b=c(TRUE,FALSE,TRUE),
+ c=c(TRUE,TRUE,FALSE),
+ d=c(TRUE,TRUE,TRUE))
> tmp
a b c d
1 13 TRUE TRUE TRUE
2 23 FALSE TRUE TRUE
3 52 TRUE FALSE TRUE
I'd like to compute a summary column (say: e) that is a logical AND over the whole range of logical columns. In other words, for a given row, if all b:d are TRUE, then e would be TRUE; if any b:d are FALSE, then e would be FALSE.
My expected result is:
> tmp
a b c d e
1 13 TRUE TRUE TRUE TRUE
2 23 FALSE TRUE TRUE FALSE
3 52 TRUE FALSE TRUE FALSE
I want to indicate the range of columns by indices, as I have a bunch of columns, and the names are cumbersome. The following code works, but i'd rather use a vectorized approach to improve performance.
> tmp$e <- NA
> for(i in 1:nrow(tmp)){
+ tmp[i,"e"] <- all(tmp[i,2:(ncol(tmp)-1)]==TRUE)
+ }
> tmp
a b c d e
1 13 TRUE TRUE TRUE TRUE
2 23 FALSE TRUE TRUE FALSE
3 52 TRUE FALSE TRUE FALSE
Any way to do this without using a for loop to step through the rows of the data.frame?
You can use rowSums to loop over rows... and some fancy footwork to make it quasi-automated:
# identify the logical columns
boolCols <- sapply(tmp, is.logical)
# sum each row of the logical columns and
# compare to the total number of logical columns
tmp$e <- rowSums(tmp[,boolCols]) == sum(boolCols)
By using rowSums in ifelse statement, in one go it can be acheived:
tmp$e <- ifelse(rowSums(tmp[,2:4] == T) == 3, T, F)

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