I have a data frame (df) and I was wondering how to return the row number(s) for a particular value (2585) in the 4th column (height_chad1) of the same data frame?
I've tried:
row(mydata_2$height_chad1, 2585)
and I get the following error:
Error in factor(.Internal(row(dim(x))), labels = labs) :
a matrix-like object is required as argument to 'row'
Is there an equivalent line of code that works for data frames instead of matrix-like objects?
Any help would be appreciated.
Use which(mydata_2$height_chad1 == 2585)
Short example
df <- data.frame(x = c(1,1,2,3,4,5,6,3),
y = c(5,4,6,7,8,3,2,4))
df
x y
1 1 5
2 1 4
3 2 6
4 3 7
5 4 8
6 5 3
7 6 2
8 3 4
which(df$x == 3)
[1] 4 8
length(which(df$x == 3))
[1] 2
count(df, vars = "x")
x freq
1 1 2
2 2 1
3 3 2
4 4 1
5 5 1
6 6 1
df[which(df$x == 3),]
x y
4 3 7
8 3 4
As Matt Weller pointed out, you can use the length function.
The count function in plyr can be used to return the count of each unique column value.
which(df==my.val, arr.ind=TRUE)
Related
Suppose I have a dataframe that looks like this:
> data <- data.frame(x = c(1,1,2,2,3,4,5,6), y = c(1,2,3,4,5,6,7,8))
> data
x y
1 1 1
2 1 2
3 2 3
4 2 4
5 3 5
6 4 6
7 5 7
8 6 8
I want to use mutate and case_when to create a new id variable that will identify rows using the variable x, and give rows missing x a unique id. In other words, I should have the same id for rows one and two, rows three and four, while rows 5-8 should have their own unique ids. Suppose I want to generate these id values with a function:
id_function <- function(x, n){
set.seed(x)
res <- character(n)
for(i in seq(n)){
res[i] <- paste0(sample(c(letters, LETTERS, 0:9), 32), collapse="")
}
res
}
id_function(1, 1)
[1] "4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf"
I am trying to use this function on the RHS of a case_when expression like this:
data %>%
mutate(my_id = id_function(1234, nrow(.)),
my_id = dplyr::case_when(!is.na(x) ~ id_function(x, 1),
TRUE ~ my_id))
But the RHS does not seem to be vectorized and I get the same value for all non-missing values of x:
x y my_id
1 1 1 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
2 1 2 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
3 2 3 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
4 2 4 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
5 NA 5 0vnws5giVNIzp86BHKuOZ9ch4dtL3Fqy
6 NA 6 IbKU6DjvW9ypitl7qc25Lr4sOwEfghdk
7 NA 7 8oqQMPx6IrkGhXv4KlUtYfcJ5Z1RCaDy
8 NA 8 BRsjumlCEGS6v4ANrw1bxLynOKkF90ao
I'm sure there's a way to vectorize the RHS, what am I doing wrong? Is there an easier approach to solving this problem?
I guess rowwise() would do the trick:
data %>%
rowwise() %>%
mutate(my_id = id_function(x, 1))
x y my_id
1 1 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
1 2 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
2 3 uof7FhqC3lOXkacp54MGZJLUR6siSKDb
2 4 uof7FhqC3lOXkacp54MGZJLUR6siSKDb
3 5 e5lMJNQEhtj4VY1KbCR9WUiPrpy7vfXo
4 6 3kYcgR7109DLbxatQIAKXFeovN8pnuUV
5 7 bQ4ok7OuDgscLUlpzKAivBj2T3m6wrWy
6 8 0jSn3Jcb2HDA5uhvG8g1ytsmRpl6CQWN
purrr map functions can be used for non-vectorized functions. The following will give you a similar result. map2 will take the two arguments expected by your id_function.
library(tidyverse)
data %>%
mutate(my_id = map2(x, 1, id_function))
Output
x y my_id
1 1 1 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
2 1 2 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
3 2 3 uof7FhqC3lOXkacp54MGZJLUR6siSKDb
4 2 4 uof7FhqC3lOXkacp54MGZJLUR6siSKDb
5 3 5 e5lMJNQEhtj4VY1KbCR9WUiPrpy7vfXo
6 4 6 3kYcgR7109DLbxatQIAKXFeovN8pnuUV
7 5 7 bQ4ok7OuDgscLUlpzKAivBj2T3m6wrWy
8 6 8 0jSn3Jcb2HDA5uhvG8g1ytsmRpl6CQWN
My dataframe looks like this
data = data.frame(ID=c(1,2,3,4,5,6,7,8,9,10),
Gender=c('Male','Female','Female','Female','Male','Female','Male','Male','Female','Female'))
And I have a reference list that looks like this -
ref=list(Male=1,Female=2)
I'd like to replace values in the Gender column using this reference list, without adding a new column to my dataframe.
Here's my attempt
do.call(dplyr::recode, c(list(data), ref))
Which gives me the following error -
no applicable method for 'recode' applied to an object of class
"data.frame"
Any inputs would be greatly appreciated
An option would be do a left_join after stacking the 'ref' list to a two column data.frame
library(dplyr)
left_join(data, stack(ref), by = c('Gender' = 'ind')) %>%
select(ID, Gender = values)
A base R approach would be
unname(unlist(ref)[as.character(data$Gender)])
#[1] 1 2 2 2 1 2 1 1 2 2
In base R:
data$Gender = sapply(data$Gender, function(x) ref[[x]])
You can use factor, i.e.
factor(data$Gender, levels = names(ref), labels = ref)
#[1] 1 2 2 2 1 2 1 1 2 2
You can unlist ref to give you a named vector of codes, and then index this with your data:
transform(data,Gender=unlist(ref)[as.character(Gender)])
ID Gender
1 1 1
2 2 2
3 3 2
4 4 2
5 5 1
6 6 2
7 7 1
8 8 1
9 9 2
10 10 2
Surprisingly, that one works as well:
data$Gender <- ref[as.character(data$Gender)]
#> data
# ID Gender
# 1 1 1
# 2 2 2
# 3 3 2
# 4 4 2
# 5 5 1
# 6 6 2
# 7 7 1
# 8 8 1
# 9 9 2
# 10 10 2
I have a list of events and sequences. I would like to print the sequences in a separate table if event = x is included somewhere in the sequence. See table below:
Event Sequence
1 a 1
2 a 1
3 x 1
4 a 2
5 a 2
6 a 3
7 a 3
8 x 3
9 a 4
10 a 4
In this case I would like a new table that includes only the sequences where Event=x was included:
Event Sequence
1 a 1
2 a 1
3 x 1
4 a 3
5 a 3
6 x 3
Base R solution:
d[d$Sequence %in% d$Sequence[d$Event == "x"], ]
Event Sequence
1: a 1
2: a 1
3: x 1
4: a 3
5: a 3
6: x 3
data.table solution:
library(data.table)
setDT(d)[Sequence %in% Sequence[Event == "x"]]
As you can see syntax/logic is quite similar between these two solutions:
Find event's that are equal to x
Extract their Sequence
Subset table according to specified Sequence
We can use dplyr to group the data and filter the sequence with any "x" in it.
library(dplyr)
df2 <- df %>%
group_by(Sequence) %>%
filter(any(Event %in% "x")) %>%
ungroup()
df2
# A tibble: 6 x 2
Event Sequence
<chr> <int>
1 a 1
2 a 1
3 x 1
4 a 3
5 a 3
6 x 3
DATA
df <- read.table(text = " Event Sequence
1 a 1
2 a 1
3 x 1
4 a 2
5 a 2
6 a 3
7 a 3
8 x 3
9 a 4
10 a 4",
header = TRUE, stringsAsFactors = FALSE)
I would like to get a data frame that contains only data that is within 2 SD per each numeric column.
I know how to do it for a single column but how can I do it for a bunch of columns at once?
Here is the toy data frame:
df <- read.table(text = "target birds wolfs Country
3 21 7 a
3 8 4 b
1 2 8 c
1 2 3 a
1 8 3 a
6 1 2 a
6 7 1 b
6 1 5 c",header = TRUE)
Here is the code line for getting only the data that is under 2 SD for a single column(birds).How can I do it for all numeric columns at once?
df[!(abs(df$birds - mean(df$birds))/sd(df$birds)) > 2,]
target birds wolfs Country
2 3 8 4 b
3 1 2 8 c
4 1 2 3 a
5 1 8 3 a
6 6 1 2 a
7 6 7 1 b
8 6 1 5 c
We can use lapply to loop over the dataset columns and subset the numeric vectors (by using a if/else condition) based on the mean and sd.
lapply(df, function(x) if(is.numeric(x)) x[!(abs((x-mean(x))/sd(x))>2)] else x)
EDIT:
I was under the impression that we need to remove the outliers for each column separately. But, if we need to keep only the rows that have no outliers for the numeric columns, we can loop through the columns with lapply as before, instead of returning 'x', we return the sequence of 'x' and then get the intersect of the list element with Reduce. The numeric index can be used for subsetting the rows.
lst <- lapply(df, function(x) if(is.numeric(x))
seq_along(x)[!(abs((x-mean(x))/sd(x))>2)] else seq_along(x))
df[Reduce(intersect,lst),]
I'm guessing that you are trying to filter your data set by checking that all of the numeric columns are within 2 SD (?)
In that case I would suggest to create two filters. 1 one that will indicate numeric columns, the second one that will check that all of them within 2 SD. For the second condition, we can use the built in scale function
indx <- sapply(df, is.numeric)
indx2 <- rowSums(abs(scale(df[indx])) <= 2) == sum(indx)
df[indx2,]
# target birds wolfs Country
# 2 3 8 4 b
# 3 1 2 8 c
# 4 1 2 3 a
# 5 1 8 3 a
# 6 6 1 2 a
# 7 6 7 1 b
# 8 6 1 5 c
I'm trying to select the column with the highest value for each row in a data.frame. So for instance, the data is set up as such.
> df <- data.frame(one = c(0:6), two = c(6:0))
> df
one two
1 0 6
2 1 5
3 2 4
4 3 3
5 4 2
6 5 1
7 6 0
Then I'd like to set another column based on those rows. The data frame would look like this.
> df
one two rank
1 0 6 2
2 1 5 2
3 2 4 2
4 3 3 3
5 4 2 1
6 5 1 1
7 6 0 1
I imagine there is some sort of way that I can use plyr or sapply here but it's eluding me at the moment.
There might be a more efficient solution, but
ranks <- apply(df, 1, which.max)
ranks[which(df[, 1] == df[, 2])] <- 3
edit: properly spaced!