I know I can do this in other ways, but I am just curious.
dfDice = sample(1:6, 10000, replace = TRUE) %>%
data.frame()
The above creates a data.frame, where the column header is called '.'.
So my first question is can I pipe the column header into my code? I have tried putting it in my data.frame() function but it just creates a new column.
And my second question is, can I pipe multiple columns into a data.frame, or would I have to do something like this?:
dfDice = (num = sample(1:6, 10000, replace = TRUE) %>%
data.frame(letters = sample(LETTERS, 10000, replace = TRUE))
Again, I know this is not the best way to create a data.frame, I am just curious from a learning perspective and trying to fully understand piping.
So my first question is can I pipe the column header into my code? I
have tried putting it in my data.frame() function but it just creates
a new column.
For single columns you have two options:
dfDice <- sample(1:6, 10000, replace = TRUE) %>%
data.frame() %>%
setNnames("num")
dfDice <- sample(1:6, 10000, replace = TRUE) %>%
data.frame(num = .)
And my second question is, can I pipe multiple columns into a
data.frame?
sample(1:6, 5, replace = TRUE) %>%
cbind(sample(LETTERS, 5, replace = TRUE)) %>%
as.data.frame() %>%
setNames(c("num", "letters"))
To assign names, you can use a predefined vector and use setNames
library(dplyr)
cols <- "a"
sample(1:6, 10, replace = TRUE) %>%
data.frame() %>%
setNames(cols)
Or can also name dynamically without knowing number of columns beforehand.
sample(1:6, 10, replace = TRUE) %>%
data.frame() %>%
setNames(letters[seq_along(.)])
For 2nd question simplest option would be
data.frame(a = sample(1:6, 10, replace = TRUE),
b = sample(LETTERS, 10, replace = TRUE))
OR if you want to use piping maybe ?
sample(1:6, 10, replace = TRUE) %>%
data.frame() %>%
setNames(cols) %>%
bind_cols(b = sample(LETTERS, 10, replace = TRUE))
Related
In R I've got a dataset like this one:
df <- data.frame(
ID = c(1:30),
x1 = seq(0, 1, length.out = 30),
x2 = seq(100, 3000, length.out = 30),
category = gl(3, 10, labels = c("NEGATIVE", "NEUTRAL", "POSITIVE"))
)
Now I want to add a new column with randomized boolean values, but inside each category the proportion of TRUE and FALSE values should be the same (i.e. the randomizing process should generate the same count of true and false values, in the above data frame 5 TRUEs and 5 FALSEs in each of the 3 categories). How to do this?
You can sample a vector of "TRUE" and "FALSE" values without replacement so you have a randomized and balanced column in your data-frame.
sample(rep(c("TRUE","FALSE"),each=5),10,replace=FALSE)
Based on Yacine Hajji answer:
addRandomBool <- function(df, p){
n <- ceiling(nrow(df) * p)
df$bool <- sample(rep(c("TRUE","FALSE"), times = c(n, nrow(df) - n)))
df
}
Reduce(rbind, lapply(split(df, df$category), addRandomBool, p = 0.5))
where parametar p determines the proportion of TRUE.
This will sample within each group from a vector of 5 TRUE and 5 FALSE without replacement. It will assume that there are always 10 records per group.
library(dplyr)
library(tidyr)
df <- data.frame(
ID = c(1:30),
x1 = seq(0, 1, length.out = 30),
x2 = seq(100, 3000, length.out = 30),
category = gl(3, 10, labels = c("NEGATIVE", "NEUTRAL", "POSITIVE"))
)
set.seed(pi)
df %>%
group_by(category) %>%
nest() %>%
mutate(data = lapply(data,
function(df){ # Function to saple and assign the new_col
df$new_col <- sample(rep(c(FALSE, TRUE),
each = 5),
size = 10,
replace = FALSE)
df
})) %>%
unnest(cols = "data")
This next example is a little more generalized, but still assumes (approximately) even distribution of TRUE and FALSE within a group. But it can accomodate variable group sizes, and even groups with odd numbers of records (but will favor FALSE for odd numbers of records)
library(dplyr)
library(tidyr)
df <- data.frame(
ID = c(1:30),
x1 = seq(0, 1, length.out = 30),
x2 = seq(100, 3000, length.out = 30),
category = gl(3, 10, labels = c("NEGATIVE", "NEUTRAL", "POSITIVE"))
)
set.seed(pi)
df %>%
group_by(category) %>%
nest() %>%
mutate(data = lapply(data,
function(df){
df$new_col <- sample(rep(c(FALSE, TRUE),
length.out = nrow(df)),
size = nrow(df),
replace = FALSE)
df
})) %>%
unnest(cols = "data")
Maintaining Column Order
A couple of options to maintain the column order:
First, you can save the column order before you do your group_by - nest, and then use select to set the order when you're done.
set.seed(pi)
orig_col <- names(df) # original column order
df %>%
group_by(category) %>%
nest() %>%
mutate(data = lapply(data,
function(df){
df$new_col <- sample(rep(c(FALSE, TRUE),
length.out = nrow(df)),
size = nrow(df),
replace = FALSE)
df
})) %>%
unnest(cols = "data") %>%
select_at(c(orig_col, "new_col")) # Restore the column order
Or you can use a base R solution that doesn't change the column order in the first place
df <- split(df, df["category"])
df <- lapply(df,
function(df){
df$new_col <- sample(rep(c(FALSE, TRUE),
length.out = nrow(df)),
size = nrow(df),
replace = FALSE)
df
})
do.call("rbind", c(df, list(make.row.names = FALSE)))
There are likely a dozen other ways to do this, and probably more efficient ways that I'm not thinking of.
I need to label a values in a lot of variables with sjlabelled::set_labels. Here is a reproducable example and what already works:
library(data.table)
library(sjlabelled)
lookup <- data.table(id = paste0("q", 1:5),
answers = paste(paste0("atext", 1:5), paste0("btext", 1:5)
, paste0("ctext", 1:5), sep = ";"))
data <- data.table(q1 = sample(1:3, 10, replace = TRUE),
q2 = sample(1:3, 10, replace = TRUE),
q3 = sample(1:3, 10, replace = TRUE),
q4 = sample(1:3, 10, replace = TRUE),
q5 = sample(1:3, 10, replace = TRUE))
data$q1 <- set_labels(data$q1, labels = unlist(strsplit(lookup[id == "q1", answers], split = ";")))
get_labels(data$q1)
So the labels for the different answers (=values) are seperated by a semicolon. I am able to make it work if I call the variables by id but as you can see in the example code but I am struggling with the task if I want to "loop" through all variables.
The goal is to be able to export the datatable (or dataframe) as an SPSS file. If it works with other packages I would also be happy.
Match the column names of data with id, split the answers on ; and pass the labels as a list.
library(sjlabelled)
data <- set_labels(data, labels = strsplit(lookup$answers[match(names(data), lookup$id)], ';'))
get_labels(data)
#$q1
#[1] "atext1" "btext1" "ctext1"
#$q2
#[1] "atext2" "btext2" "ctext2"
#$q3
#[1] "atext3" "btext3" "ctext3"
#$q4
#[1] "atext4" "btext4" "ctext4"
#$q5
#[1] "atext5" "btext5" "ctext5"
I'd like to compare element by element from two data.frame called df1 and df2. From they, I'd like to build a new data.frame called out. If the elements are equals, then the element in out is 1, otherwise is 0.
MWE
set.seed(1)
df1 <- data.frame(Q1 = sample(letters[1:5], 2, replace = TRUE),
Q2 = sample(letters[1:5], 2, replace = TRUE))
set.seed(2)
df2 <- data.frame(Q1 = sample(letters[1:5], 2, replace = TRUE),
Q2 = sample(letters[1:5], 2, replace = TRUE))
Expected output
out <- data.frame(Q1 = c(0, 0), Q2 = c(1, 0))
If the datasets are created with stringsAsFactors = FALSE while creating the data.frame - factor makes it difficult as the attributes would create difficulty in doing the comparison)
+(df1 == df2)
Or if it is factor convert to character columns with type.convert
+(type.convert(df1, as.is = TRUE) == type.convert(df2, as.is = TRUE))
Or make use of matrix hack way of changing to character
+(as.matrix(df1) == as.matrix(df2))
I have repeated measures data with two ratings (reliable and fast) repeated on two different objects, (each survey respondent rates each object using the same two ratings measures). I would like to have two columns, one for object 1 and one for object 2, with the ratings displayed in two separate rows.
In the reference manual there is reference to using a | separator to compare two variables, but the example given is for mrsets not means, I'm not sure how to do the same with means and keep them in separate data frame columns.
In the code below, the problem is that instead of placing the means side by side (for comparison) they are stacked on top of each other.
#library
library(expss)
library(magrittr)
#dummy data
set.seed(9)
df <- data.frame(
q1_reliable=sample(c(1,5), 100, replace = TRUE),
q1_fast=sample(c(1,5), 100, replace = TRUE),
q2_reliable=sample(c(1,5), 100, replace = TRUE),
q2_fast=sample(c(1,5), 100, replace = TRUE))
#table
df %>%
tab_cells(q1_reliable,q1_fast) %>%
tab_stat_mean(label = "") %>%
tab_cells(q2_reliable,q2_fast) %>%
tab_stat_mean(label = "") %>%
tab_pivot()
I discovered that if I add variable labels first and use 'tab_pivot(stat_position = "inside_columns")' it solved the problem.
#library
library(expss)
library(magrittr)
#dummy data
set.seed(9)
df <- data.frame(
q1_reliable=sample(c(1,5), 100, replace = TRUE),
q1_fast=sample(c(1,5), 100, replace = TRUE),
q2_reliable=sample(c(1,5), 100, replace = TRUE),
q2_fast=sample(c(1,5), 100, replace = TRUE)
)
#labels
df = apply_labels(df,
q1_reliable = "reliable",
q1_fast = "fast",
q2_reliable = "reliable",
q2_fast = "fast")
#table
df %>%
tab_cells(q1_reliable,q1_fast) %>%
tab_stat_mean(label = "") %>%
tab_cells(q2_reliable,q2_fast) %>%
tab_stat_mean(label = "") %>%
tab_pivot(stat_position = "inside_columns")
Like this data.table approach?
library(data.table)
#melt first
DT <- melt( setDT(df),
measure.vars = patterns( reliable = "reliable", fast = "fast"),
variable.name = "q")
#then summarise
DT[, lapply(.SD, mean), by = .(q), .SDcols = c("reliable", "fast")]
q reliable fast
1: 1 3.04 2.96
2: 2 2.92 2.96
I have a question similar to this one, but my dataset is a bit bigger: 50 columns with 1 column as UID and other columns carrying either TRUE or NA, I want to change all the NA to FALSE, but I don't want to use explicit loop.
Can plyr do the trick? Thanks.
UPDATE #1
Thanks for quick reply, but what if my dataset is like below:
df <- data.frame(
id = c(rep(1:19),NA),
x1 = sample(c(NA,TRUE), 20, replace = TRUE),
x2 = sample(c(NA,TRUE), 20, replace = TRUE)
)
I only want X1 and X2 to be processed, how can this be done?
If you want to do the replacement for a subset of variables, you can still use the is.na(*) <- trick, as follows:
df[c("x1", "x2")][is.na(df[c("x1", "x2")])] <- FALSE
IMO using temporary variables makes the logic easier to follow:
vars.to.replace <- c("x1", "x2")
df2 <- df[vars.to.replace]
df2[is.na(df2)] <- FALSE
df[vars.to.replace] <- df2
tidyr::replace_na excellent function.
df %>%
replace_na(list(x1 = FALSE, x2 = FALSE))
This is such a great quick fix. the only trick is you make a list of the columns you want to change.
Try this code:
df <- data.frame(
id = c(rep(1:19), NA),
x1 = sample(c(NA, TRUE), 20, replace = TRUE),
x2 = sample(c(NA, TRUE), 20, replace = TRUE)
)
replace(df, is.na(df), FALSE)
UPDATED for an another solution.
df2 <- df <- data.frame(
id = c(rep(1:19), NA),
x1 = sample(c(NA, TRUE), 20, replace = TRUE),
x2 = sample(c(NA, TRUE), 20, replace = TRUE)
)
df2[names(df) == "id"] <- FALSE
df2[names(df) != "id"] <- TRUE
replace(df, is.na(df) & df2, FALSE)
You can use the NAToUnknown function in the gdata package
df[,c('x1', 'x2')] = gdata::NAToUnknown(df[,c('x1', 'x2')], unknown = 'FALSE')
With dplyr you could also do
df %>% mutate_each(funs(replace(., is.na(.), F)), x1, x2)
It is a bit less readable compared to just using replace() but more generic as it allows to select the columns to be transformed. This solution especially applies if you want to keep NAs in some columns but want to get rid of NAs in others.