I have a dataframe of two columns of T and F.
I want to know
which row is T in the first and F in the second
which row is F in the first and T in the second
which row is F in both
I have very little clues on the matter, con someone shine a light?
You can use case when
library(dplyr)
df = data.frame(x = c("T","T","F","F","F"), y = c("T","F","T","F","X"))
df %>%
mutate(condition = case_when(
x == "T" & y == "T" ~ "Both are T",
x == "T" & y == "F" ~ "First is T fecond is F",
x == "F" & y == "F" ~ "Both are F",
x == "F" & y == "T" ~ "First is F, second is T",
TRUE ~ "Something else"
))
#> x y condition
#> 1 T T Both are T
#> 2 T F First is T fecond is F
#> 3 F T First is F, second is T
#> 4 F F Both are F
#> 5 F X Something else
Created on 2021-08-05 by the reprex package (v2.0.0)
Here is one possible way to solve your problem:
library(dplyr)
df <- data.frame(a = rep(c(T, F, T, F), each=2),
b = rep(c(T, T, F, F), each=2))
# a b
# 1 TRUE TRUE
# 2 TRUE TRUE
# 3 FALSE TRUE
# 4 FALSE TRUE
# 5 TRUE FALSE
# 6 TRUE FALSE
# 7 FALSE FALSE
# 8 FALSE FALSE
df %>%
mutate(newcol = case_when(a & !b ~ "first=T second=F",
!a & b ~ "first=F second=T",
!a & !b ~ "both=F",
TRUE ~ "other"))
# a b newcol
# 1 TRUE TRUE other
# 2 TRUE TRUE other
# 3 FALSE TRUE first=F second=T
# 4 FALSE TRUE first=F second=T
# 5 TRUE FALSE first=T second=F
# 6 TRUE FALSE first=T second=F
# 7 FALSE FALSE both=F
# 8 FALSE FALSE both=F
You can treat [a,b] columns as a 2-bit binary number vector, and a*2+b transfer it from binary to decimal. Thus, 2*a+b+1 is mapped to 1,2,3,4.
Try the base R code below
transform(
df,
newcol = c("both=F", "first=F,second=T", "first=T,second=F", "other")[a * 2 + b + 1]
)
which gives
a b newcol
1 TRUE TRUE other
2 TRUE TRUE other
3 FALSE TRUE first=F,second=T
4 FALSE TRUE first=F,second=T
5 TRUE FALSE first=T,second=F
6 TRUE FALSE first=T,second=F
7 FALSE FALSE both=F
8 FALSE FALSE both=F
Data
df <- data.frame(a = rep(c(T, F, T, F), each=2),
b = rep(c(T, T, F, F), each=2))
Related
Let's say I have a data frame:
data <- data.frame(w = c(1, 2, 3, 4), x = c(F, F, F, F), y = c(T, T, F, T),
z = c(T, F, F, T), z1 = c(12, 4, 5, 15))
data
#> w x y z z1
#> 1 1 FALSE TRUE TRUE 12
#> 2 2 FALSE TRUE FALSE 4
#> 3 3 FALSE FALSE FALSE 5
#> 4 4 FALSE TRUE TRUE 15
Question
How do I filter the rows in which all boolean variables are FALSE? In this case, row 3.
Or in other words, I would like to get a data frame that has at least one TRUE value per row.
Expected output
#> w x y z z1
#> 1 1 FALSE TRUE TRUE 12
#> 2 2 FALSE TRUE FALSE 4
#> 3 4 FALSE TRUE TRUE 15
Attempt
library(tidyverse)
data %>% filter(x == T | y == T | z == T)
#> w x y z z1
#> 1 1 FALSE TRUE TRUE 12
#> 2 2 FALSE TRUE FALSE 4
#> 3 4 FALSE TRUE TRUE 15
Above is a working option, but not scalable at all. Is there a more convenient option using the dplyr's filter() function?
rowSums() is a good option - TRUE is 1, FALSE is 0.
cols = c("x", "y", "z")
## all FALSE
df[rowSums[cols] == 0, ]
## at least 1 TRUE
df[rowSums[cols] >= 1, ]
## etc.
With dplyr, I would use the same idea like this:
df %>%
filter(
rowSums(. %>% select(all_of(cols))) >= 1
)
With dplyr's filter(),
library(dplyr)
filter(data, (x + y + z) > 0 )
w x y z z1
1 1 FALSE TRUE TRUE 12
2 2 FALSE TRUE FALSE 4
3 4 FALSE TRUE TRUE 15
# after #Gregor Thomas's suggestion on using TRUE or FALSE
df[!(apply(!df[, c('x', 'y', 'z')], 1, all)), ]
# without rowSums
df[!(apply(df[, c('x', 'y', 'z')] == FALSE, 1, all)), ]
# with rowSums
df[rowSums(df[, c('x', 'y', 'z')] == FALSE) != 3, ]
# w x y z z1
#1 1 FALSE TRUE TRUE 12
#2 2 FALSE TRUE FALSE 4
#4 4 FALSE TRUE TRUE 15
I have sample vector and some values:
q = c(0.00000000, -0.70218526, -0.60635393, 0.32325554, -0.45921704, -0.57336113, -0.77683717,
-1.76347868, -1.90884891, -0.86157465, -0.72896622, -0.86831735, -0.79357262, -0.65279976,
0.39921356, 0.78018094, 0.75703279, 0.70898895, 1.10155383, 0.88428135, 0.81338108,
0.65611568, 0.89776945, 0.65447442, 0.16289673, 0.19464041, 0.01762445, -0.57663945,
-1.01231868, -0.81204022, -0.99165533, -0.62666993, -1.05661282, -0.78221866, -0.03129549, 1.04051915)
s = -1.59688
i = -0.6373684
z = 0
I need to create a new vector in which boolean values will be filled according to the following conditions:
if q is less than i we fill TRUE until
q becomes more than 0 (that is, z)
or until q becomes less than s.
If Filling has stopped due to the condition of the s value, then you need to wait until the q becomes greater than 0 (that is, z) and
only after that you can start filling TRUE again, otherwise fill in FALSE
As a result, for this sample data, you should get the following result (I filled it in manually):
out <- c(FALSE, TRUE, TRUE, FALSE, FALSE, FALSE, TRUE, FALSE, FALSE, FALSE, FALSE,
FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE,
FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, FALSE)
out
[1] FALSE TRUE TRUE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE
I would like to do it without loops, since they are too slow in R
** FURTHER EDIT in view of OP's request to state logic/strategy**
Actually, your conditions are combination of three conditions. If we create four zones, as I created in the plot above and name the zones as 1 to 4 with
1 where q values are >= z
2 where q values are >= i
3 where q values are >= s
4 where q values are < s
Now, the conditions can be translated as
TRUE when in zone 2 and 3
But if exited once from these TRUE zones, it will become TRUE only if it arrives in zone 3
Moreover, if it has hit zone 4, it can become TRUE only if it arrives/hits zone 1, at least once.
Strategy
To integrate these all, I used tidyverse piped syntax
First divide all values in respective zones (say q1)
As a first condition divide zones 2 & 3 in TRUE and others in `FALSE
As second condition, say c2 , i.e. whether exited from zone 4 and has hit zone 1 or not, mark zone 4 as F and zone 1 as T rest all as NAs.
First value can be NA so replace first value, if NA, with c1
As last condition say c3 i.e. TRUE when arrive in zone 3, mark 3 as TRUE 1 and 4 as FALSE and leave zone 2 as NA to later-on check whether it arrived here from which zone.
First value can be NA so replace first value, if NA, with FALSE
Now only job remains to fill NAs in c2 and c3. Use zoo::na.locf or tidyr::fill which fills all NAs will last available value.
Your final desired result is combination of all conditions so c1 & c2 & c3
q = c(0.00000000, -0.70218526, -0.60635393, 0.32325554, -0.45921704, -0.57336113, -0.77683717,
-1.76347868, -1.90884891, -0.86157465, -0.72896622, -0.86831735, -0.79357262, -0.65279976,
0.39921356, 0.78018094, 0.75703279, 0.70898895, 1.10155383, 0.88428135, 0.81338108,
0.65611568, 0.89776945, 0.65447442, 0.16289673, 0.19464041, 0.01762445, -0.57663945,
-1.01231868, -0.81204022, -0.99165533, -0.62666993, -1.05661282, -0.78221866, -0.03129549, 1.04051915)
s = -1.59688
i = -0.6373684
z = 0
library(tidyverse)
q %>% as.data.frame() %>% setNames('q') %>%
mutate(q1 = case_when(q >= z ~ 1,
q >= i ~ 2,
q >= s ~ 3,
TRUE ~ 4),
c1 = q1 %in% c(2,3),
c2 = case_when(q1 == 4 ~ F,
q1 == 1 ~ T,
TRUE ~ NA),
c2 = ifelse(row_number() == 1 & is.na(c2), c1, c2),
c3 = case_when(q1 %in% c(1,4) ~ F,
q1 == 3 ~ T,
TRUE ~ NA),
c3 = ifelse(row_number() ==1 & is.na(c3), F, c3)) %>%
fill(c2, c3) %>%
transmute(output = c1 & c2 & c3) %>% pull(output)
#> [1] FALSE TRUE TRUE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE
#> [13] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
#> [25] FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE
Created on 2021-06-02 by the reprex package (v2.0.0)
OLD ANSWER
#data given
q = c(0.00000000, -0.70218526, -0.60635393, 0.32325554, -0.45921704, -0.57336113, -0.77683717,
-1.76347868, -1.90884891, -0.86157465, -0.72896622, -0.86831735, -0.79357262, -0.65279976,
0.39921356, 0.78018094, 0.75703279, 0.70898895, 1.10155383, 0.88428135, 0.81338108,
0.65611568, 0.89776945, 0.65447442, 0.16289673, 0.19464041, 0.01762445, -0.57663945,
-1.01231868, -0.81204022, -0.99165533, -0.62666993, -1.05661282, -0.78221866, -0.03129549, 1.04051915)
s = -1.59688
i = -0.6373684
z = 0
#loading libraries
library(dplyr)
library(tidyr)
#creating zones
q1 <- dplyr::case_when(q >= z ~ 1,
q >= i ~ 2,
q >= s ~ 3,
TRUE ~ 4)
#first condition
c1 <- dplyr::case_when(q1 %in% c(2,3) ~ T,
TRUE ~ F)
#second condition (third in above statements)
c2 <- dplyr::case_when(q1 == 4 ~ F,
q1 == 1 ~ T,
TRUE ~ NA)
c2[1] <- ifelse(is.na(c2[1]), c1[1], c2[1])
c2 <- tidyr::fill(data.frame(id = 1:length(q), c2 = c2), c2)$c2
#third condition
c3 <- dplyr::case_when(q1 == 3 ~ T,
q1 %in% c(1,4) ~ F,
TRUE ~ NA)
c3[1] <- ifelse(is.na(c3[1]), F, c3[1])
c3 <- tidyr::fill(data.frame(id = 1:length(q), c3 = c3), c3)$c3
#creating output
output <- (c1 & c2 & c3)
> output
[1] FALSE TRUE TRUE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[21] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE
#check it with your given `out`
> which((c1 & c2 & c3) == out)
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
#OR
> which((c1 & c2 & c3) != out)
integer(0)
UPDATE If you want to use baseR only, use these expressions/codes for c2 and c3
#second condition
c2 <- case_when(q1 == 4 ~ F,
q1 == 1 ~ T,
TRUE ~ NA)
c2 <- c2[!cumsum(!is.na(c2)) | !is.na(c2)][cumsum(!cumsum(!is.na(c2)) | !is.na(c2))]
#third condition
c3 <- case_when(q1 == 3 ~ T,
q1 %in% c(1,4) ~ F,
TRUE ~ NA)
c3 <- c3[!cumsum(!is.na(c3)) | !is.na(c3)][cumsum(!cumsum(!is.na(c3)) | !is.na(c3))]
for new data
q <- c(-0.01563733, -0.05829460, -0.05884189, -0.08954093, -0.13268677, -0.31748724, -0.40060792, -0.08515156, -0.14303489, -0.24525535, -0.93842637, -0.77738228, -1.29502715, -0.89000932, -1.49038656, -1.64953167, -1.67114179, -1.47482366, -0.85874778, -1.01021450, -0.90078260, -1.24313333, -0.99053914, -1.11684140, -1.34073045, -1.36406163, -1.25163185, -1.42429376, -1.48127185, -1.79040671, -2.26811789, -1.82124304, -1.85208201, -1.76394637, -1.63173292)
i = -0.489
s = -1.032
z = 0
#after running the above code
> output
[1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE FALSE FALSE FALSE FALSE
[17] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[33] FALSE FALSE FALSE
and chart
for a new vector with random values
set.seed(202)
q <- runif(35, -2, 2)
I am trying to create three new columns with values depending on a particular order of three logical type columns.
eg I have this:
a b c
1 TRUE TRUE TRUE
2 TRUE FALSE TRUE
3 TRUE FALSE TRUE
And depending if going across the row the values are TRUE, TRUE, TRUE as in row 1, then create three new columns with the values 1,1,1 but if the order is TRUE,FALSE,TRUE as in row 2 and 3 then the values would be 2,3,3. Just to note, a value of TRUE does not = 1 but rather a value I define depending on all three logical values (A total of 8 possible combinations each defined by three separate numbers). So I get something like this:
a b c d e f
1 TRUE TRUE TRUE 5 5 2
2 TRUE FALSE TRUE 2 3 3
3 TRUE FALSE TRUE 2 3 3
If someone could point me in the right direction to do this as efficiently as possible it would be greatly appreciated as I am relatively new to R.
If there is no logic in getting values for the columns and you need to add conditions individually for each combination you can use if/else.
df[c('d', 'e', 'f')] <- t(apply(df, 1, function(x) {
if (x[1] && x[2] && x[3]) c(5, 5, 2)
else if (x[1] && !x[2] && x[3]) c(2, 3, 3)
#add more conditions
#....
}))
df
# a b c d e f
#1 TRUE TRUE TRUE 5 5 2
#2 TRUE FALSE TRUE 2 3 3
#3 TRUE FALSE TRUE 2 3 3
Here's a dplyr solution using case_when. On the left side of the ~ you define your conditions, and on the right side of the ~ you assign a value for when those conditions are met. If a condition is not met (i.e. all FALSE values), you will return NA.
df %>%
mutate(d =
case_when(
a == TRUE & b == TRUE & c == TRUE ~ 5,
a == TRUE & b == FALSE & c == TRUE ~ 2
),
e =
case_when(
a == TRUE & b == TRUE & c == TRUE ~ 5,
a == TRUE & b == FALSE & c == TRUE ~ 3
),
f =
case_when(
a == TRUE & b == TRUE & c == TRUE ~ 2,
a == TRUE & b == FALSE & c == TRUE ~ 3
))
Which gives you:
a b c d e f
<lgl> <lgl> <lgl> <dbl> <dbl> <dbl>
1 TRUE TRUE TRUE 5 5 2
2 TRUE FALSE TRUE 2 3 3
3 TRUE FALSE TRUE 2 3 3
Data:
df <- tribble(
~a, ~b, ~c,
TRUE, TRUE, TRUE,
TRUE, FALSE, TRUE,
TRUE, FALSE, TRUE
)
I'm having a hard time understand how R is treating the AND and OR operators when I'm using filter from dplyr.
Here's an example to illustrate:
library(dplyr)
xy <- data.frame(x=1:6, y=c("a", "b"), z= c(rep("d",3), rep("g",3)))
> xy
x y z
1 1 a d
2 2 b d
3 3 a d
4 4 b g
5 5 a g
6 6 b g
Using filter I want to eliminate all rows where x==1 and z==d. This would lead me to believe I want to use the AND operator: &
> filter(xy, x != 1 & z != "d")
x y z
1 4 b g
2 5 a g
3 6 b g
But this removes all rows that have either x==1 or z==d. What's more confusing, is that when I use the OR operator, | I get the desired result:
> filter(xy, x != 1 | z != "d")
x y z
1 2 b d
2 3 a d
3 4 b g
4 5 a g
5 6 b g
Also, this does work, however not as desirable for if I were stringing together == and != in the same conditional evaluation.
> filter(xy, !(x == 1 & z == "d"))
x y z
1 2 b d
2 3 a d
3 4 b g
4 5 a g
5 6 b g
Can someone explain what I'm missing?
This is a question of boolean algebra. The logical expression !(x == 1 & z == d) is equivalent to x != 1 | z != d, just the same as -(x + y) is equivalent to -x - y. Eliminating the bracket, you change all == to != and all & to | and vice versa. This leads to the fact that
!(x == 1 & z == "d")
is NOT the same as
x != 1 & z != "d"
but rather
x != 1 | z != "d"
A couple tips that won't fit in a comment:
If you're having trouble understanding how something is working in R, I'd highly recommend running each individual piece of the operation. With dplyr, it's easy to keep track on intermediate steps and display them all:
mutate(xy,
A = x != 1,
B = z != 'd',
A_and_B = A & B,
A_or_B = A | B
)
# x y z A B A_and_B A_or_B
# 1 1 a d FALSE FALSE FALSE FALSE
# 2 2 b d TRUE FALSE FALSE TRUE
# 3 3 a d TRUE FALSE FALSE TRUE
# 4 4 b g TRUE TRUE TRUE TRUE
# 5 5 a g TRUE TRUE TRUE TRUE
# 6 6 b g TRUE TRUE TRUE TRUE
I think that if you look at the definition of each column its values will make perfect sense. Then, after going one step at a time, hopefully the results will make sense too.
As others have stated in various ways, you're setting yourself up for a hard time from the start with
Using filter I want to eliminate all rows where x==1 and z==d
Don't think of filter as eliminating rows, think of it as keeping rows. If you mentally invert your goal to "keep all rows where..." you'll set yourself up for a more direct translation of words to code.
The result of filter is the rows where the specified condition is true.
Take for example x != 1 & z != "d". What are the rows where this condition is true? The output you got. The other rows were removed, because the condition was not true for those rows.
In this example, your real intention was to eliminate rows where x == 1 and z == "d".
In other words, you want to keep the rows where the condition x == 1 and z == "d" is false.
Putting that into code becomes filter(xy, !(x == 1 and z == "d")).
It's ironic that this looks much like your intention, and very different from what you actually tried to write.
If you forget this logic of filter,
you can remind yourself with a simpler experiment, filter(xy, TRUE) which will return all rows, and filter(xy, FALSE) which will return none.
# x != 1 & z != "d" evaluates to a single TRUE/FALSE vector which subsets the data
# note how & and | behave in isolation:
TRUE & TRUE # T AND T = T
## [1] TRUE
TRUE & FALSE # T AND F = F
## [1] FALSE
FALSE & FALSE # F AND F = F
## [1] FALSE
TRUE | TRUE # T OR T = T
## [1] TRUE
TRUE | FALSE # T OR F = T
## [1] TRUE
FALSE | FALSE # F OR F = F
## [1] FALSE
# Apply over vectors
(x1 <- xy$x != 1)
## [1] FALSE TRUE TRUE TRUE TRUE TRUE
(z1 <- xy$z != "d")
## [1] FALSE FALSE FALSE TRUE TRUE TRUE
x1 & z1 # you get last 3 rows
## [1] FALSE FALSE FALSE TRUE TRUE TRUE
x1 | z1 # you get all but 1st row (which contains 1 and d)
## [1] FALSE TRUE TRUE TRUE TRUE TRUE
I have the following data frame
df <- data.frame(A1 = c("A","A","A","A","A","A","A","A","A","A","B","B","B","B","B","B","B","B","B","B"),
B2 = c("C","D","C","D","C","D","C","D","C","D","C","D","C","D","C","D","C","D","C","D"),
C3 = c("E","F","E","F","E","F","E","F","E","F","E","F","E","F","E","F","E","F","E","F"),
D4=c(1,12,5,41,45,4,5,6,12,7,3,4,6,8,12,4,12,1,6,7))
and I would like to subset all the rows for which the first 3 column match the vector c("A","C","E")
I have tried to use which but it does not work
vct <- c("A","C","E")
df[which(df[1:3] == vct)]
You can probably use paste (or interaction):
vct <- c("A","C","E")
do.call(paste, df[1:3]) %in% paste(vct, collapse = " ")
# [1] TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE FALSE
# [12] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
df[do.call(paste, df[1:3]) %in% paste(c("A", "C", "E"), collapse = " "), ]
# A1 B2 C3 D4
# 1 A C E 1
# 3 A C E 5
# 5 A C E 45
# 7 A C E 5
# 9 A C E 12
## with "interaction"
df[interaction(df[1:3], drop=TRUE) %in% paste(vct, collapse = "."), ]
You can also do something like this:
df[with(df, A1 == "A" & B2 == "C" & C3 == "E"), ]