Say I have a df:
df <- data.frame(flag = c(rep(0, 20)),
include = c(rep(1, 20)))
df[c(4,8,16), ]$flag <- 1
df
flag include
1 0 1
2 0 1
3 0 1
4 1 1
5 0 1
6 0 1
7 0 1
8 1 1
9 0 1
10 0 1
11 0 1
12 0 1
13 0 1
14 0 1
15 0 1
16 1 1
17 0 1
18 0 1
19 0 1
20 0 1
What I wish to do is change the include flag to 0 if the row is within +/- two rows of a row where flag == 1. The result would look like:
flag include
1 0 1
2 0 0
3 0 0
4 1 1
5 0 0
6 0 0
7 0 0
8 1 1
9 0 0
10 0 0
11 0 1
12 0 1
13 0 1
14 0 0
15 0 0
16 1 1
17 0 0
18 0 0
19 0 1
20 0 1
I've thought of some 'innovative' (read: inefficient and over complicated) ways to do it but was thinking there must be a simple way I'm overlooking.
Would be nice if the answer was such that I could generalize this to +/- n rows, since I have a lot more data and would be looking to potentially search within +/- 10 rows...
Another option with data.table:
library(data.table)
n = 2
# find the row number where flag is one
flag_one = which(df$flag == 1)
# find the index where include needs to be updated
idx = setdiff(outer(flag_one, -n:n, "+"), flag_one)
# update include in place
setDT(df)[idx[idx >= 1 & idx <= nrow(df)], include := 0][]
# or as #Frank commented the last step with base R would be
# df$include[idx[idx >= 1 & idx <= nrow(df)]] = 0
# flag include
# 1: 0 1
# 2: 0 0
# 3: 0 0
# 4: 1 1
# 5: 0 0
# 6: 0 0
# 7: 0 0
# 8: 1 1
# 9: 0 0
#10: 0 0
#11: 0 1
#12: 0 1
#13: 0 1
#14: 0 0
#15: 0 0
#16: 1 1
#17: 0 0
#18: 0 0
#19: 0 1
#20: 0 1
Put in a function:
update_n <- function(df, n) {
flag_one = which(df$flag == 1)
idx = setdiff(outer(flag_one, -n:n, "+"), flag_one)
df$include[idx[idx >= 1 & idx <= nrow(df)]] = 0
df
}
There must be another simpler way but the first way which I could think of is using sapply and which
df$include[sapply(which(df$flag == 1) , function(x) c(x-2, x-1, x+1, x+2))] <- 0
df
# flag include
#1 0 1
#2 0 0
#3 0 0
#4 1 1
#5 0 0
#6 0 0
#7 0 0
#8 1 1
#9 0 0
#10 0 0
#11 0 1
#12 0 1
#13 0 1
#14 0 0
#15 0 0
#16 1 1
#17 0 0
#18 0 0
#19 0 1
#20 0 1
We first find out all the indices where flag is 1 and then create the required sequence of numbers around each of it and turn that index of include to 0.
For variable n we can do
n = 2
df$include[sapply(which(df$flag == 1),function(x) setdiff(seq(x-n, x+n),x))] <- 0
replace(x = df$include,
list = sapply(1:NROW(df), function(i)
any(df$flag[c(max(1, i-2):max(1, i-1),
min(i+1, NROW(df)):min(i+2, NROW(df)))] == 1)), values = 0)
# [1] 1 0 0 1 0 0 0 1 0 0 1 1 1 0 0 1 0 0 1 1
For n rows,
replace(x = df$include,
list = sapply(1:NROW(df), function(i)
any(df$flag[c(max(1, i-n):max(1, i-1),
min(i+1, NROW(df)):min(i+n, NROW(df)))] == 1)), values = 0)
Another way is to use zoo::rollapply. To determine if a row is within +/- two rows of a row where flag == 1, we check if the maximum flag in a window is 1.
We need rollapply rather than rollmax because we need to specify partial = T.
is_within_flag_window <- function(flag, n) {
zoo::rollapply(flag, width = (2 * n) + 1, partial = T, FUN = max) == 1
}
df %>%
mutate(include = ifelse(flag == 1, 1,
ifelse(is_within_flag_window(flag, 2), 0,
1)))
Use which and outer.
df$include[outer(which(df$flag==1), -2:2, `+`)] <- 0
If flag=1 within one or two positions of each other then restore the ones overwritten at position 0. Note this step is critical in case the "flag" overlaps in a particular range.
df$include[which(df$flag==1)] <- 1
flag include
1 0 1
2 0 0
3 0 0
4 1 1
5 0 0
6 0 0
7 0 0
8 1 1
9 0 0
10 0 0
11 0 1
12 0 1
13 0 1
14 0 0
15 0 0
16 1 1
17 0 0
18 0 0
19 0 1
20 0 1
If flag = 1 within one or two rows of the beginning or end of the dataset, R will throw errors. Use this:
## assign i for convenience/readability
i <- pmax(1, pmin(nrow(df), outer(which(df$flag==1), -2:2, `+`)))
df$include[i] <- 0
Restore 1s as before
Related
I want to create a new column based on some conditions imposed on several columns. For example, here is an example dataset:
a <- data.frame(x=c(1,0,1,0,0), y=c(0,0,0,0,0), z=c(1,1,0,0,0))
a
x y z
1 1 0 1
2 0 0 1
3 1 0 0
4 0 0 0
5 0 0 0
Specifically, if for any particular row 1 is present, then the new column returns 1. If all are 0, then the new column returns 0. So the dataset with the new column will be
x y z w
1 1 0 1 1
2 0 0 1 1
3 1 0 0 1
4 0 0 0 0
5 0 0 0 0
My initial thought was to use %in% but couldn't get the result I want. Thank you for your help!
If your data frame consists of binary values, e.g., only 0 and 1, you can try the code below with rowSums
a$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'm looking for a better way to achieve what the code below does with a for loop. The goal is to create a dataframe (or matrix) where each row is a possible n-length sequence of 1s and 0s, followed by an n+1th column which contains a number corresponding to one of the previous columns that contains a 0.
So in the n == 3 case for example, we want to include a row like this:
1 0 0 2
but not this:
1 0 0 1
Here's the code I have now (assuming n == 3 for simplicity):
library(tidyverse)
df <- expand.grid(x = 0:1, y = 0:1, z = 0:1, target = 1:3, keep = FALSE)
for (row in 1:nrow(df)) {
df$keep[row] <- df[row, df$target[row]] == 0
}
df <- df %>%
filter(keep == TRUE) %>%
select(-keep)
head(df)
# x y z target
# 1 0 0 0 1
# 2 0 1 0 1
# 3 0 0 1 1
# 4 0 1 1 1
# 5 0 0 0 2
# 6 1 0 0 2
# 7 0 0 1 2
# 8 1 0 1 2
# 9 0 0 0 3
# 10 1 0 0 3
# 11 0 1 0 3
# 12 1 1 0 3
Seems like there has to be a better way to do this, especially with dplyr. But I can't figure out how to use the value of target to specify the column to filter on.
Using base R, we can create a row/column index to filter values from the dataframe and keep rows where the extracted value is 0.
df[df[cbind(seq_len(nrow(df)), df$target)] == 0, ]
# x y z target
#1 0 0 0 1
#3 0 1 0 1
#5 0 0 1 1
#7 0 1 1 1
#9 0 0 0 2
#10 1 0 0 2
#13 0 0 1 2
#14 1 0 1 2
#17 0 0 0 3
#18 1 0 0 3
#19 0 1 0 3
#20 1 1 0 3
data
df <- expand.grid(x = 0:1, y = 0:1, z = 0:1, target = 1:3)
Problem
I'm trying to create a new column (b) based on values from a previous column (a). Column a is binary, consisting of either 0's or 1's. If there are three or more 1's in a row in column a, then keep them in column b. I'm close to the desired output, but when there are two 1's in a row, the ifelse grabs the second value because it's meeting the first condition.
Desired Output–Column b
df <- data.frame(a = c(1,1,1,0,0,1,0,1,1,0,1,1,1,0,1,1,0,1,1,1,1),
b = c(1,1,1,0,0,0,0,0,0,0,1,1,1,0,0,0,0,1,1,1,1))
df
a b
1 1 1
2 1 1
3 1 1
4 0 0
5 0 0
6 1 0
7 0 0
8 1 0 #
9 1 0 #
10 0 0
11 1 1
12 1 1
13 1 1
14 0 0
15 1 0 #
16 1 0 #
17 0 0
18 1 1
19 1 1
20 1 1
21 1 1
Failed Attempt...s
require(dplyr)
df_fail <- df %>% mutate(b=ifelse((lag(df$a) + df$a) > 1 |(df$a + lead(df$a) + lead(df$a,2)) >= 3, df$a,NA))
df_fail
a b
1 1 1
2 1 1
3 1 1
4 0 0
5 0 0
6 1 0
7 0 0
8 1 0
9 1 1 # should be 0
10 0 0
11 1 1
12 1 1
13 1 1
14 0 0
15 1 0
16 1 1 # should be 0
17 0 0
18 1 1
19 1 1
20 1 1
21 1 1
We can use rle from base R to change the elements that have less than 3 repeating 1s to 0
inverse.rle(within.list(rle(df$a), values[values == 1 & lengths <3] <- 0))
#[1] 1 1 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 1 1 1 1
Or use rleid from data.table
library(data.table)
library(dplyr)
df %>%
group_by(grp = rleid(a)) %>%
mutate(b1 = if(n() <3 & all(a == 1)) 0 else a) %>%
ungroup %>%
select(-grp)
Does anyone has an idea how to achieve the following: start from here
df <- data.frame(var = c(0,0,1,1,0,0,0,1,1,0,0,0,0,1,1))
and achieve this:
df <- data.frame(var = c(0,0,1,1,0,0,0,1,1,0,0,0,0,1,1),
newvar = c(0,0,1,1,0,0,0,2,2,0,0,0,0,3,3))
Here is an option with rle by replacing the 'values' that are not 0 with the sequence of those values and then call inverse_rle to get the full vector
df$newvar <- inverse.rle(within.list(rle(df$var),
values[values!=0] <- seq_along(values[values!=0])))
df
# var newvar
#1 0 0
#2 0 0
#3 1 1
#4 1 1
#5 0 0
#6 0 0
#7 0 0
#8 1 2
#9 1 2
#10 0 0
#11 0 0
#12 0 0
#13 0 0
#14 1 3
#15 1 3
you can try following:
df %>%
mutate(n=ifelse(var==lead(var,default = 0),1,0)) %>%
mutate(n2=ifelse(var==0,0,n)) %>%
mutate(res=ifelse(var==1, cumsum(n2),0))
var n n2 res
1 0 1 0 0
2 0 0 0 0
3 1 1 1 1
4 1 0 0 1
5 0 1 0 0
6 0 1 0 0
7 0 0 0 0
8 1 1 1 2
9 1 0 0 2
10 0 1 0 0
11 0 1 0 0
12 0 1 0 0
13 0 0 0 0
14 1 1 1 3
15 1 0 0 3
Then select(var, res) only the columns you need.
An efficient solution:
df %>%
mutate(temp= var - lag(var,default=df$var[1])) %>%
mutate(newvar= var * cumsum(temp>0))
or without the additional column:
df %>%
mutate(newvar= var - lag(var,default=df$var[1])) %>%
mutate(newvar= var * cumsum(newvar>0))
var temp newvar
1 0 0 0
2 0 0 0
3 1 1 1
4 1 0 1
5 0 -1 0
6 0 0 0
7 0 0 0
8 1 1 2
9 1 0 2
10 0 -1 0
11 0 0 0
12 0 0 0
13 0 0 0
14 1 1 3
15 1 0 3
Another one for fun:
library(data.table)
setDT(df)
tmp = 0
df[, newvar := if(var[1] != 0) tmp <- tmp + 1 else 0, by = rleid(var)][]
And another one:
df[, newvar := var * cumsum(diff(c(0, var)) == 1)]
# or if still a data.frame
within(df, newvar <- var * cumsum(diff(c(0, var)) == 1))
Let's say I have 3 vectors (strings of 10):
X <- c(1,1,0,1,0, 1,1, 0, NA,NA)
H <- c(0,0,1,0,NA,1,NA,1, 1, 1 )
I <- c(0,0,0,0,0, 1,NA,NA,NA,1 )
Data.frame Y contains 10 columns and 6 rows:
1 2 3 4 5 6 7 8 9 10
0 1 0 0 1 1 1 0 1 0
1 1 1 0 1 0 1 0 0 0
0 0 0 0 1 0 0 1 0 1
1 0 1 1 0 1 1 1 0 0
0 0 0 0 0 0 1 0 0 0
1 1 0 1 0 0 0 0 1 1
I'd like to use vector X, H en I to make column selections in data.frame Y, using "1's" and "0's" in the vector as selection criterium .
So the results for vector X using the '1' as selection criterium should be:
X <- c(1,1,0,1,0, 1,1, 0, NA,NA)
1 2 4 6 7
0 1 0 1 1
1 1 0 0 1
0 0 0 0 0
1 0 1 1 1
0 0 0 0 1
1 1 1 0 0
For vector H using the '1' as selection criterium:
H <- c(0,0,1,0,NA,1,NA,1, 1, 1 )
3 6 8 9 10
0 1 0 1 0
1 0 0 0 0
0 0 1 0 1
1 1 1 0 0
0 0 0 0 0
0 0 0 1 1
For vector I using the '1' as selection criterium:
I <- c(0,0,0,0,0, 1,NA,NA,NA,1 )
6 10
1 0
0 0
0 1
1 0
0 0
0 1
For convenience and speed I'd like to use a loop. It might be something like this:
all.ones <- lapply[,function(x) x %in% 1]
In the outcome (all.ones), the result for each vector should stay separate. For example:
X 1,2,4,6,7
H 3,6,8,9,10
I 6,10
The standard way of doing this is using the %in% operator:
Y[, X %in% 1]
To do this for multiple vectors (assuming you want an AND operation):
mylist = list(X, H, I, D, E, K)
Y[, Reduce(`&`, lapply(mylist, function(x) x %in% 1))]
The problem is the NA, use which to get round it. Consider the following:
x <- c(1,0,1,NA)
x[x==1]
[1] 1 1 NA
x[which(x==1)]
[1] 1 1
How about this?
idx <- which(X==1)
Y[,idx]
EDIT: For six vectors, do
idx <- which(X==1 & H==1 & I==1 & D==1 & E==1 & K==1)
Y[,idx]
Replace & with | if you want all columns of Y where at least one of the lists has a 1.