I have this data.frame:
a <- c(rep("1", 3), rep("2", 3), rep("3",3), rep("4",3), rep("5",3))
b <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15)
df <-data.frame(a,b)
a b
1 1 1
2 1 2
3 1 3
4 2 4
5 2 5
6 2 6
7 3 7
8 3 8
9 3 9
10 4 10
11 4 11
12 4 12
13 5 13
14 5 14
15 5 15
I want to have something like this:
a <- c(rep("2", 3), rep("3", 3))
b <- c(4,5,6,7,8,9)
dffinal<-data.frame(a,b)
a b
1 2 4
2 2 5
3 2 6
4 3 7
5 3 8
6 3 9
I could use the "subset" function, but its not working
sub <- subset(df,c(2,3) == a )
a b
5 2 5
8 3 8
This command only takes one row of "2" and "3" in column "a".
Any Help?
You're confusing == with %in%:
subset(df, a %in% c(2,3))
# a b
# 4 2 4
# 5 2 5
# 6 2 6
# 7 3 7
# 8 3 8
# 9 3 9
what about this?
library(dplyr)
df %>% filter(a == 2 | a==3)
a b
1 2 4
2 2 5
3 2 6
4 3 7
5 3 8
6 3 9
We can use data.table. We convert the 'data.frame' to 'data.table' (setDT(df)), and set the 'key' as column 'a', then we subset the rows.
library(data.table)
setDT(df, key= 'a')[c('2','3')]
# a b
#1: 2 4
#2: 2 5
#3: 2 6
#4: 3 7
#5: 3 8
#6: 3 9
Related
This question already has answers here:
Replace a value NA with the value from another column in R
(5 answers)
Closed 3 months ago.
I need to put the value of variable "A" in place of the NA of variable "B".
Example of my dataframe:
> df <- data.frame(A = seq(1, 10), B = c(1, NA, 3, 4, NA, NA, 7, 8, NA, NA))
> df
A B
1 1 1
2 2 NA
3 3 3
4 4 4
5 5 NA
6 6 NA
7 7 7
8 8 8
9 9 NA
10 10 NA
I want the above dataframe converted into this:
> df
A B
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
Using R base indexing
> df$B[is.na(df$B)] <- df$A[is.na(df$B)]
> df
A B
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
Use coalesce
library(dplyr)
df <- df %>%
mutate(B = coalesce(B, A))
-output
df
A B
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
I prefer coalesce. Here is one with an ifelse:
library(dplyr)
df %>%
mutate(B = ifelse(is.na(B), A, B))
A B
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
I have a dataframe.
dat <- data.frame(k=c("A","A","B","B","B","A","A","A"),
a=c(4,2,4,7,5,8,3,2),b=c(2,5,3,5,8,4,5,8),
stringsAsFactors = F)
k a b
1 A 4 2
2 A 2 5
3 B 4 3
4 B 7 5
5 B 5 8
6 A 8 4
7 A 3 5
8 A 2 8
I would like to subset contiguous blocks based on variable k. This would be a standard approach.
#using rle rather than levels
kval <- rle(dat$k)$values
for(i in 1:length(kval))
{
subdf <- subset(dat,dat$k==kval[i])
print(subdf)
#do something with subdf
}
k a b
1 A 4 2
2 A 2 5
6 A 8 4
7 A 3 5
8 A 2 8
k a b
3 B 4 3
4 B 7 5
5 B 5 8
k a b
1 A 4 2
2 A 2 5
6 A 8 4
7 A 3 5
8 A 2 8
So the subsetting above obviously does not work the way I intended. Any elegant way to get these results?
k a b
1 A 4 2
2 A 2 5
k a b
1 B 4 3
2 B 7 5
3 B 5 8
k a b
1 A 8 4
2 A 3 5
3 A 2 8
We can use rleid from data.table to create a grouping variable
library(data.table)
setDT(dat)[, grp := rleid(k)]
dat
# k a b grp
#1: A 4 2 1
#2: A 2 5 1
#3: B 4 3 2
#4: B 7 5 2
#5: B 5 8 2
#6: A 8 4 3
#7: A 3 5 3
#8: A 2 8 3
We can group by 'grp' and do all the operations within the 'grp' using standard data.table methods.
Here is a base R option to create 'grp'
dat$grp <- with(dat, cumsum(c(TRUE, k[-1]!= k[-length(k)])))
I have this data.frame:
a <- c(rep("1", 3), rep("2", 3), rep("3",3), rep("4",3), rep("5",3))
b <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15)
df <-data.frame(a,b)
a b
1 1 1
2 1 2
3 1 3
4 2 4
5 2 5
6 2 6
7 3 7
8 3 8
9 3 9
10 4 10
11 4 11
12 4 12
13 5 13
14 5 14
15 5 15
I want to have something like this:
a <- c(rep("2", 3), rep("3", 3))
b <- c(4,5,6,7,8,9)
dffinal<-data.frame(a,b)
a b
1 2 4
2 2 5
3 2 6
4 3 7
5 3 8
6 3 9
I could use the "subset" function, but its not working
sub <- subset(df,c(2,3) == a )
a b
5 2 5
8 3 8
This command only takes one row of "2" and "3" in column "a".
Any Help?
You're confusing == with %in%:
subset(df, a %in% c(2,3))
# a b
# 4 2 4
# 5 2 5
# 6 2 6
# 7 3 7
# 8 3 8
# 9 3 9
what about this?
library(dplyr)
df %>% filter(a == 2 | a==3)
a b
1 2 4
2 2 5
3 2 6
4 3 7
5 3 8
6 3 9
We can use data.table. We convert the 'data.frame' to 'data.table' (setDT(df)), and set the 'key' as column 'a', then we subset the rows.
library(data.table)
setDT(df, key= 'a')[c('2','3')]
# a b
#1: 2 4
#2: 2 5
#3: 2 6
#4: 3 7
#5: 3 8
#6: 3 9
This question already has answers here:
Replacing NAs with latest non-NA value
(21 answers)
Closed 7 years ago.
I have two data.frame as the following:
> a <- data.frame(x=c(1,2,3,4,5,6,7,8), y=c(1,3,5,7,9,11,13,15))
> a
x y
1 1 1
2 2 3
3 3 5
4 4 7
5 5 9
6 6 11
7 7 13
8 8 15
> b <- data.frame(x=c(1,5,7), z=c(2, 4, 6))
> b
x z
1 1 2
2 5 4
3 7 6
Then I use "join" for two data.frames:
> c <- join(a, b, by="x", type="left")
> c
x y z
1 1 1 2
2 2 3 NA
3 3 5 NA
4 4 7 NA
5 5 9 4
6 6 11 NA
7 7 13 6
8 8 15 NA
My requirement is to replace the NAs in the Z column by the last None-Na value before the current place. I want the result like this:
> c
x y z
1 1 1 2
2 2 3 2
3 3 5 2
4 4 7 2
5 5 9 4
6 6 11 4
7 7 13 6
8 8 15 6
This time (if your data is not too large) a loop is an elegant option:
for(i in which(is.na(c$z))){
c$z[i] = c$z[i-1]
}
gives:
> c
x y z
1 1 1 2
2 2 3 2
3 3 5 2
4 4 7 2
5 5 9 4
6 6 11 4
7 7 13 6
8 8 15 6
data:
library(plyr)
a <- data.frame(x=c(1,2,3,4,5,6,7,8), y=c(1,3,5,7,9,11,13,15))
b <- data.frame(x=c(1,5,7), z=c(2, 4, 6))
c <- join(a, b, by="x", type="left")
You might also want to check na.locf in the zoo package.
Say we have the following data
A <- c(1,2,2,2,3,4,8,6,6,1,2,3,4)
B <- c(1,2,3,4,5,1,2,3,4,5,1,2,3)
data <- data.frame(A,B)
How would one write a function so that for A, if we have the same value in the i+1th position, then the reoccuring row is removed.
Therefore the output should like like
data.frame(c(1,2,3,4,8,6,1,2,3,4), c(1,2,5,1,2,3,5,1,2,3))
My best guess would be using a for statement, however I have no experience in these
You can try
data[c(TRUE, data[-1,1]!= data[-nrow(data), 1]),]
Another option, dplyr-esque:
library(dplyr)
dat1 <- data.frame(A=c(1,2,2,2,3,4,8,6,6,1,2,3,4),
B=c(1,2,3,4,5,1,2,3,4,5,1,2,3))
dat1 %>% filter(A != lag(A, default=FALSE))
## A B
## 1 1 1
## 2 2 2
## 3 3 5
## 4 4 1
## 5 8 2
## 6 6 3
## 7 1 5
## 8 2 1
## 9 3 2
## 10 4 3
using diff, which calculates the pairwise differences with a lag of 1:
data[c( TRUE, diff(data[,1]) != 0), ]
output:
A B
1 1 1
2 2 2
5 3 5
6 4 1
7 8 2
8 6 3
10 1 5
11 2 1
12 3 2
13 4 3
Using rle
A <- c(1,2,2,2,3,4,8,6,6,1,2,3,4)
B <- c(1,2,3,4,5,1,2,3,4,5,1,2,3)
data <- data.frame(A,B)
X <- rle(data$A)
Y <- cumsum(c(1, X$lengths[-length(X$lengths)]))
View(data[Y, ])
row.names A B
1 1 1 1
2 2 2 2
3 5 3 5
4 6 4 1
5 7 8 2
6 8 6 3
7 10 1 5
8 11 2 1
9 12 3 2
10 13 4 3