My data are as follow:
Row x y
1 1 2
2 2 3
3 3 4
4 4 3
5 5 NA
6 1 NA
7 2 NA
8 3 NA
9 4 NA
10 5 7
11 1 NA
12 2 NA
13 3 NA
14 4 NA
15 5 NA
I wish to delete Row 11 to 15 since y are NA for ALL cycles of x (y euqal to NA whatever value x takes for Row 11 to 15). I am not going to delete other rows since there is at lease one number of y not NA when x moves from 1 to 5 (Like from Row 6 to 10, y is 7 when x is 5, thus I keep Row 6 to 10). I wish to know how should I write a R code to accompolish this.
using base R, Taking into assumption that x is arranged and that all start from 1.
subset(df,!ave(is.na(y),cumsum(c(1,diff(x)<0)),FUN=all))
Row x y
1 1 1 2
2 2 2 3
3 3 3 4
4 4 4 3
5 5 5 NA
6 6 1 NA
7 7 2 NA
8 8 3 NA
9 9 4 NA
10 10 5 7
using tidyverse:
df%>%
group_by(m = cumsum(c(1,diff(x)<0)))%>%
filter(!all(is.na(y)))
# A tibble: 10 x 4
# Groups: m [2]
Row x y m
<int> <int> <int> <dbl>
1 1 1 2 1
2 2 2 3 1
3 3 3 4 1
4 4 4 3 1
5 5 5 NA 1
6 6 1 NA 2
7 7 2 NA 2
8 8 3 NA 2
9 9 4 NA 2
10 10 5 7 2
of course you can unselect then remove m
Related
I have a data frame grouped by 'id' and a variable 'age' which contains missing values, NA.
Within each 'id', I want to replace missing values of 'age', but only "fill up" before the first non-NA value.
data <- data.frame(id=c(1,1,1,1,1,1,2,2,2,2,2,3,3,3,3,3),
age=c(NA,6,NA,8,NA,NA,NA,NA,3,8,NA,NA,NA,7,NA,9))
id age
1 1 NA
2 1 6 # first non-NA in id = 1. Fill up from here
3 1 NA
4 1 8
5 1 NA
6 1 NA
7 2 NA
8 2 NA
9 2 3 # first non-NA in id = 2. Fill up from here
10 2 8
11 2 NA
12 3 NA
13 3 NA
14 3 7 # first non-NA in id = 3. Fill up from here
15 3 NA
16 3 9
Expected output:
1 1 6
2 1 6
3 1 NA
4 1 8
5 1 NA
6 1 NA
7 2 3
8 2 3
9 2 3
10 2 8
11 2 NA
12 3 7
13 3 7
14 3 7
15 3 NA
16 3 9
I tried using fill with .direction = "up" like this:
library(dplyr)
library(tidyr)
data1 <- data %>% group_by(id) %>%
fill(!is.na(age[1]), .direction = "up")
You could use cumall(is.na(age)) to find the positions before the first non-NA value.
library(dplyr)
data %>%
group_by(id) %>%
mutate(age2 = replace(age, cumall(is.na(age)), age[!is.na(age)][1])) %>%
ungroup()
# A tibble: 16 × 3
id age age2
<dbl> <dbl> <dbl>
1 1 NA 6
2 1 6 6
3 1 NA NA
4 1 8 8
5 1 NA NA
6 1 NA NA
7 2 NA 3
8 2 NA 3
9 2 3 3
10 2 8 8
11 2 NA NA
12 3 NA 7
13 3 NA 7
14 3 7 7
15 3 NA NA
16 3 9 9
Another option (agnostic about where the missing and non-missing values start) could be:
data %>%
group_by(id) %>%
mutate(rleid = with(rle(is.na(age)), rep(seq_along(lengths), lengths)),
age2 = ifelse(rleid == min(rleid[is.na(age)]),
age[rleid == (min(rleid[is.na(age)]) + 1)][1],
age))
id age rleid age2
<dbl> <dbl> <int> <dbl>
1 1 NA 1 6
2 1 6 2 6
3 1 NA 3 NA
4 1 8 4 8
5 1 NA 5 NA
6 1 NA 5 NA
7 2 NA 1 3
8 2 NA 1 3
9 2 3 2 3
10 2 8 2 8
11 2 NA 3 NA
12 3 NA 1 7
13 3 NA 1 7
14 3 7 2 7
15 3 NA 3 NA
16 3 9 4 9
This question already has answers here:
R: Assign variable labels of data frame columns
(4 answers)
Closed 2 years ago.
I am trying to assign variable labels into my columns in R. I was able to create values list using the following code:
var.labels= dataframe$var name
now I want to use this list as variable labels for my columns in the data frame. I tried this code but it did not work:
label(dataframe) = as.list(var.labels[match(names(dataframe), names(var.labels))]
Thank you for your help.
Maybe this?
var.labels <- dataframe$var name
colnames(dataframe) <- var.labels
Example:
df <- as.data.frame(replicate(n = 13, expr = sample(c(1:8, NA), 13, replace = TRUE)))
df$names <- LETTERS[1:13]
df
#> V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 names
#>1 1 NA 5 3 2 5 7 NA 3 NA NA NA 1 A
#>2 7 5 1 6 4 3 1 6 4 3 NA 3 NA B
#>3 7 3 1 2 7 2 6 4 6 1 7 3 2 C
#>4 3 1 5 1 5 6 1 2 2 NA 8 5 1 D
#>5 3 6 3 7 4 2 6 7 7 NA 1 2 8 E
#>6 NA 4 4 1 8 3 8 NA 6 3 8 4 NA F
#>7 1 8 3 8 1 3 2 4 7 4 2 1 2 G
#>8 NA NA 2 3 5 4 5 1 4 7 8 5 3 H
#>9 7 NA 3 2 7 NA 2 8 7 NA 6 8 6 I
#>10 6 8 5 3 6 5 5 3 4 8 NA 5 1 J
#>11 1 7 8 5 1 2 3 NA NA 3 2 6 7 K
#>12 8 4 1 8 7 NA 6 6 5 6 7 NA 2 L
#>13 5 8 5 1 2 1 6 3 NA 1 7 3 5 M
colnames(df) <- df$names
#> A B C D E F G H I J K L M NA
#>1 1 NA 5 3 2 5 7 NA 3 NA NA NA 1 A
#>2 7 5 1 6 4 3 1 6 4 3 NA 3 NA B
#>3 7 3 1 2 7 2 6 4 6 1 7 3 2 C
#>4 3 1 5 1 5 6 1 2 2 NA 8 5 1 D
#>5 3 6 3 7 4 2 6 7 7 NA 1 2 8 E
#>6 NA 4 4 1 8 3 8 NA 6 3 8 4 NA F
#>7 1 8 3 8 1 3 2 4 7 4 2 1 2 G
#>8 NA NA 2 3 5 4 5 1 4 7 8 5 3 H
#>9 7 NA 3 2 7 NA 2 8 7 NA 6 8 6 I
#>10 6 8 5 3 6 5 5 3 4 8 NA 5 1 J
#>11 1 7 8 5 1 2 3 NA NA 3 2 6 7 K
#>12 8 4 1 8 7 NA 6 6 5 6 7 NA 2 L
#>13 5 8 5 1 2 1 6 3 NA 1 7 3 5 M
# Finally, remove the names column
df[14] <- NULL
i got this df:
df <- data.frame(month = c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4),
day = c(1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5),
flow = c(2,5,7,8,5,4,6,7,9,2,NA,1,6,10,2,NA,NA,NA,NA,NA))
and i want to reach this result:
month day flow dayofminflow
1 1 1 2 1
2 1 2 5 1
3 1 3 7 1
4 1 4 8 1
5 1 5 5 1
6 2 1 4 5
7 2 2 6 5
8 2 3 7 5
9 2 4 9 5
10 2 5 2 5
11 3 1 NA 2
12 3 2 1 2
13 3 3 6 2
14 3 4 10 2
15 3 5 2 2
16 4 1 NA NA
17 4 2 NA NA
18 4 3 NA NA
19 4 4 NA NA
20 4 5 NA NA
I was using this solution, but it returns NA when the first value is NA:
newdf <- df %>% group_by(month) %>% mutate(Val=day[flow==min(flow)][1])
And this solution returns an error when all data is NA:
library(dplyr)
df <- df %>%
group_by(month) %>%
mutate(dayminflowofthemonth = day[which.min(flow)]) %>%
ungroup
You would just change the default na.rm = TRUE in min() from the first solution to ignore NAs?
df %>%
group_by(month) %>%
mutate(dayofminflow = day[which(min(flow, na.rm = TRUE) == flow)][1])
# A tibble: 20 x 4
# Groups: month [4]
month day flow dayofminflow
<dbl> <dbl> <dbl> <dbl>
1 1 1 2 1
2 1 2 5 1
3 1 3 7 1
4 1 4 8 1
5 1 5 5 1
6 2 1 4 5
7 2 2 6 5
8 2 3 7 5
9 2 4 9 5
10 2 5 2 5
11 3 1 NA 2
12 3 2 1 2
13 3 3 6 2
14 3 4 10 2
15 3 5 2 2
16 4 1 NA NA
17 4 2 NA NA
18 4 3 NA NA
19 4 4 NA NA
20 4 5 NA NA
Though you get a warning no non-missing arguments to min; returning Inf from month 4 since all flow values are NA.
I have a dataset:
a day day.1.time day.2.time day.3.time day.4.time day.5.time
1 NA 2 4 5 7 10 4
2 NA 5 4 1 1 6 NA
3 NA 3 7 9 6 7 4
4 NA 3 6 8 8 4 5
5 NA 3 5 2 4 5 6
6 NA 3 87 3 2 1 78
7 NA 1 NA 7 5 9 54
8 NA 5 6 6 3 2 3
9 NA 2 5 10 9 8 3
10 NA 3 9 4 10 3 3
I am trying to use the day column value to match with the day.x.time column to replace the missing value in column a. For instance, in the first row, the first value in the day column is 2, then we should use day.2.time value 5 to replace the first value in column a.
If the day.x.time value is missing, we should use -1 day or +1 day to replace the missing in column a. For instance, in the second row, the day column shows 5, so we should use the value in day.5.time column, but it's also a missing value. In this case, we should use the value in day.4.time column to replace the missing value in column a.
You can use dat = data.frame(a = rep(NA,10), day = c(2,5,3,3,3,3,1,5,2,3), day.1.time = c(4,4,7,6,5,87,NA,6,5,9), day.2.time = sample(10), day.3.time = sample(10), day.4.time = sample(10), day.5.time = c(4,NA,4,5,6,78,54,3,3,3)) to generate the sample data.
I have tried grep(paste0("^day."dat$day,".time$", names(dat)) to match with the column but my code isn't matching in every row, so any help would be appreciated!
Here is one way to do this.
The first part is easy to match day column with the corresponding day.x.time column. We can do this using matrix subsetting.
cols <- grep('day\\.\\d+\\.time', names(dat))
dat$a <- dat[cols][cbind(1:nrow(dat), dat$day)]
dat
# a day day.1.time day.2.time day.3.time day.4.time day.5.time
#1 3 2 4 3 3 3 4
#2 NA 5 4 4 10 2 NA
#3 1 3 7 8 1 8 4
#4 4 3 6 6 4 5 5
#5 6 3 5 10 6 7 6
#6 8 3 87 5 8 9 78
#7 NA 1 NA 1 7 10 54
#8 3 5 6 7 9 1 3
#9 2 2 5 2 5 6 3
#10 2 3 9 9 2 4 3
To fill values where day.x.time column is NA we can select the closest non-NA value in that row.
inds <- which(is.na(dat$a))
dat$a[inds] <- mapply(function(x, y)
na.omit(unlist(dat[x, cols[order(abs(y- seq_along(cols)))]])[1:4])[1],
inds, dat$day[inds])
dat
# a day day.1.time day.2.time day.3.time day.4.time day.5.time
#1 3 2 4 3 3 3 4
#2 2 5 4 4 10 2 NA
#3 1 3 7 8 1 8 4
#4 4 3 6 6 4 5 5
#5 6 3 5 10 6 7 6
#6 8 3 87 5 8 9 78
#7 1 1 NA 1 7 10 54
#8 3 5 6 7 9 1 3
#9 2 2 5 2 5 6 3
#10 2 3 9 9 2 4 3
Using sapply to loop over the rows and subset by day[i] + 2 column.
res <- transform(dat, a=sapply(1:nrow(dat), function(i) dat[i, dat$day[i] + 2]))
res
# a day day.1.time day.2.time day.3.time day.4.time day.5.time
# 1 5 2 4 5 7 10 4
# 2 NA 5 4 1 1 6 NA
# 3 6 3 7 9 6 7 4
# 4 8 3 6 8 8 4 5
# 5 4 3 5 2 4 5 6
# 6 2 3 87 3 2 1 78
# 7 NA 1 NA 7 5 9 54
# 8 3 5 6 6 3 2 3
# 9 10 2 5 10 9 8 3
# 10 10 3 9 4 10 3 3
Edit
The +/-2 days would require a decision rule, what to chose, if day is NA, but none of day - 1 and day + 1 is NA and both have the same values.
Here a solution that goes from day backwards and takes the first non-NA. If it is day one, as it's the case in row 7, we get NA.
res <- transform(dat, a=sapply(1:nrow(dat), function(i) {
days <- dat[i, -(1:2)]
day.value <- days[dat$day[i]]
if (is.na(day.value)) {
day.value <- tail(na.omit(unlist(days[1:dat$day[i]])), 1)
if (length(day.value) == 0) day.value <- NA
}
return(day.value)
}))
res
# a day day.1.time day.2.time day.3.time day.4.time day.5.time
# 1 10 2 4 10 1 2 4
# 2 10 5 4 1 3 10 NA
# 3 2 3 7 7 2 7 4
# 4 6 3 6 2 6 6 5
# 5 10 3 5 9 10 5 6
# 6 8 3 87 6 8 4 78
# 7 NA 1 NA 3 7 1 54
# 8 3 5 6 4 4 9 3
# 9 8 2 5 8 5 8 3
# 10 9 3 9 5 9 3 3
I have a simple data frame as follows
x = data.frame(id = seq(1,10),val = seq(1,10))
x
id val
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
I want to add 4 more columns. The first 2 are the previous two rows and the next two are the next two rows. For the first two rows and last two rows it needs to write out as NA.
How do I accomplish this using cast in the reshape package?
The final output would look like
1 1 NA NA 2 3
2 2 NA 1 3 4
3 3 1 2 4 5
4 4 2 3 5 6
... and so on...
Thanks much in advance
After your give the example , I change the solution
mat <- cbind(dat,
c(c(NA,NA),head(dat$id,-2)),
c(c(NA),head(dat$val,-1)),
c(tail(dat$id,-1),c(NA)),
c(tail(dat$val,-2),c(NA,NA)))
colnames(mat) <- c('id','val','idp','valp','idn','valn')
id val idp valp idn valn
1 1 1 NA NA 2 3
2 2 2 NA 1 3 4
3 3 3 1 2 4 5
4 4 4 2 3 5 6
5 5 5 3 4 6 7
6 6 6 4 5 7 8
7 7 7 5 6 8 9
8 8 8 6 7 9 10
9 9 9 7 8 10 NA
10 10 10 8 9 NA NA
Here is a soluting with sapply. First, choose the relative change for the new columns:
lags <- c(-2, -1, 1, 2)
Create the new columns:
newcols <- sapply(lags,
function(l) {
tmp <- seq.int(nrow(x)) + l;
x[replace(tmp, tmp < 1 | tmp > nrow(x), NA), "val"]})
Bind together:
cbind(x, newcols)
The result:
id val 1 2 3 4
1 1 1 NA NA 2 3
2 2 2 NA 1 3 4
3 3 3 1 2 4 5
4 4 4 2 3 5 6
5 5 5 3 4 6 7
6 6 6 4 5 7 8
7 7 7 5 6 8 9
8 8 8 6 7 9 10
9 9 9 7 8 10 NA
10 10 10 8 9 NA NA