I have few columns which I need to convert to factors
for cols in ['col1','col2']:
df$cols<-as.factor(as.character(df$cols))
Error
for cols in ['col1','col2']:
Error: unexpected symbol in "for cols"
> df$cols<-as.factor(as.character(df$cols))
Error in `$<-.data.frame`(`*tmp*`, cols, value = integer(0)) :
replacement has 0 rows, data has 942
The syntax showed also use the python for loop and python list. Instead it would be a vector of strings in `R
for (col in c('col1','col2')) {
df[[col]] <- factor(df[[col]])
}
NOTE: here we use [[ instead of $ and the braces {}. The factor can be directly applied instead of as.character wrapping
Or with lapply where it can be done easily (without using any packages)
df[c('col1', 'col2')] <- lapply(df[c('col1', 'col2')], factor)
Or in dplyr, where it can be done more easily
library(dplyr)
df <- df %>%
mutate_at(vars(col1, col2), factor)
In complement to #akrun solution, with data.table, this can be done easily:
library(data.table)
setDT(df)
df[,c("col1","col2") := lapply(.SD, function(c) as.factor(as.character(c))), .SDcols = c("col1","col2")]
Note that df is updated by reference (:=) so no need for reassignment
Related
I am trying to write a function to dynamically group_by every combination of a character vector.
This is how I set it up my list:
stuff <- c("type", "country", "color")
stuff_ListStr <- do.call("c", lapply(seq_along(stuff), function(i) combn(stuff, i, FUN = list)))
stuff_ListChar <- sapply(stuff_ListStr, paste, collapse = ", ")
stuff_ListSym <- lapply(stuff_ListChar, as.symbol)
Then I threw it into a loop.
b <- list()
for (each in stuff_ListSym) {
a <- answers_wfh %>%
group_by(!!each) %>%
summarize(n=n())
b <- append(b, a)
}
So essentially I want to replicate this
... group_by(type),
... group_by(country),
... group_by(type, country),
... and the rest of the combinations. Then I want put all the summaries into one list (a list of tibbles/lists)
It's totally failing. This is my error message:
Error: Column `type, country` is unknown.
Not only that, b is not giving me what I want. It's a list with length 12 already when I only expected 2 before it failed. One tibble grouped by 'type' and the second by 'country'.
I'm new to R in general but thought tidy eval was really cool and wanted to try. Any tips here?
I think you have a problem of standard evaluation. !! is sometimes not enough to unquote variables and get dplyr to work. Use !!! and rlang::syms for multiple unquotes
b <- list()
for (each in stuff_ListSym) {
a <- answers_wfh %>%
group_by(!!!rlang::syms(each)) %>%
summarize(n=n())
b <- append(b, a)
}
I think lapply would be better in your situation than for since you want to end-up with a list
Since you use variable names as arguments of functions, you might be more comfortable with data.table than dplyr. If you want the equivalent data.table implementation:
library(data.table)
setDT(answers_wfh)
lapply(stuff_ListSym, function(g) answers_wfh[,.(n = .N), by = g])
You can have a look at this blog post I wrote on the subject of SE vs NSE in dplyr and data.table
I think stuff_ListStr is enough to get what you want. You cold use group_by_at which accepts character vector.
library(dplyr)
library(rlang)
purrr::map(stuff_ListStr, ~answers_wfh %>% group_by_at(.x) %>% summarize(n=n()))
A better option is to use count but count does not accept character vectors so using some non-standard evaluation.
purrr::map(stuff_ListStr, ~answers_wfh %>% count(!!!syms(.x)))
I'm trying to do a string search and replace across multiple columns in R. My code:
# Get columns of interest
selected_columns <- c(368,370,372,374,376,378,380,382,384,386,388,390,392,394)
#Perform grepl across multiple columns
df[,selected_columns][grepl('apples',df[,selected_columns],ignore.case = TRUE)] <- 'category1'
However, I'm getting the error:
Error: undefined columns selected
Thanks in advance.
grep/grepl works on vectors/matrix and not on data.frame/list. According to the?grep`
x - a character vector where matches are sought, or an object which can be coerced by as.character to a character vector.
We can loop over the columns (lapply) and replace the values based on the match
df[, selected_columns] <- lapply(df[, selected_columns],
function(x) replace(x, grepl('apples', x, ignore.case = TRUE), 'category1'))
Or with dplyr
library(dplyr)
library(stringr)
df %>%
mutate_at(selected_columns, ~ replace(., str_detect(., 'apples'), 'category1'))
Assuming you want to partially match a cell and replace it, you could use rapply() and replace cell contents that have "apples" with "category1" using gsub():
df[selected_columns] <- rapply(df[selected_columns], function(x) gsub("apples", "category1", x), how = "replace")
Just keep in mind the difference between grepl()/gsub() (with and without boundaries in your regex), and %in%/match() when searching for strings.
I am performing Data Analysis and cleaning in R using tidyverse.
I have a Data Frame with 23 columns containing values 'NO','STEADY','UP' and 'down'.
I want to change all the values in these 23 columns to 0 in case of 'NO','STEADY' and 1 in other case.
What i did is, i created a list by name keys in which i have kept all my columns, After that i am using for loop, ifelse statements and mutate.
Please have a look at the code below
# Column names are kept in the list by name keys
keys = c('metformin', 'repaglinide', 'nateglinide', 'chlorpropamide', 'glimepiride',
'glipizide', 'glyburide', 'pioglitazone', 'rosiglitazone', 'acarbose', 'miglitol',
'insulin', 'glyburide-metformin', 'tolazamide', 'metformin-pioglitazone',
'metformin-rosiglitazone', 'glimepiride-pioglitazone', 'glipizide-metformin',
'troglitazone', 'tolbutamide', 'acetohexamide')
After that, i used following code to get the desired result :
for (col in keys){
Dataset = Dataset %>%
mutate(col = ifelse(col %in% c('No','Steady'),0,1)) }
I was expecting that, it will do the changes that i require, but nothing happens after this. (NO ERROR MESSAGE AND NO DESIRED RESULT)
After that, i researched further and executed following code
for (col in keys){
print(col)}
It gives me elements of list as characters like - "metformin"
So, i thought - may be this is the issue. Hence, i used the below code to caste the keys as symbols :
keys_new = sym(keys)
After that i again ran the same code:
for (col in keys_new){
Dataset = Dataset %>%
mutate(col = ifelse(col %in% c('No','Steady'),0,1))}
It gives me following Error -
Error in match(x, table, nomatch = 0L) :
'match' requires vector arguments
After all this. I also tried to create a function to get the desired results, but that too didn't worked:
change = function(name){
Dataset = Dataset %>%
mutate(name = ifelse(name %in% c('No','Steady'),0,1),
name = as.factor(name))
return(Dataset)}
for (col in keys){
change(col)}
This didn't perform any action. (NO ERROR MESSAGE AND NO DESIRED RESULT)
When keys_new is placed in this code:
for (col in keys_new){
change(col)}
I got the same Error :
Error in match(x, table, nomatch = 0L) :
'match' requires vector arguments
PLEASE GUIDE
There's no need to loop or keep track of column names. You can use mutate_all -
Dataset %>%
mutate_all(~ifelse(. %in% c('No','Steady'), 0, 1))
Another way, thanks to Rui Barradas -
Dataset %>%
mutate_all(~as.integer(!. %in% c('No','Steady')))
There's a simpler way using mutate_at and case_when.
Dataset %>% mutate_at(keys, ~case_when(. %in% c("NO", "STEADY") ~ 0, TRUE ~ 1))
mutate_at will only mutate the columns specified in the keys variable. case_when then lets you replace one value by another by some condition.
This answer for using mutate through forloop.
I don't have your data, so i tried to make my own data, i changed the keys into a tibble using enframe then spread it into columns and used the row number as a value for each column, then check if the value is higher than 10 or not.
To use the column name in mutate you have to use !! and := in the mutate function
df <- enframe(c('metformin', 'repaglinide', 'nateglinide', 'chlorpropamide', 'glimepiride',
'glipizide', 'glyburide', 'pioglitazone', 'rosiglitazone', 'acarbose', 'miglitol',
'insulin', 'glyburide-metformin', 'tolazamide', 'metformin-pioglitazone',
'metformin-rosiglitazone', 'glimepiride-pioglitazone', 'glipizide-metformin',
'troglitazone', 'tolbutamide', 'acetohexamide')
) %>% spread(key = value,value = name)
keys = c('metformin', 'repaglinide', 'nateglinide', 'chlorpropamide', 'glimepiride',
'glipizide', 'glyburide', 'pioglitazone', 'rosiglitazone', 'acarbose', 'miglitol',
'insulin', 'glyburide-metformin', 'tolazamide', 'metformin-pioglitazone',
'metformin-rosiglitazone', 'glimepiride-pioglitazone', 'glipizide-metformin',
'troglitazone', 'tolbutamide', 'acetohexamide')
for (col in keys){
df = df %>%
mutate(!!as.character(col) := ifelse( df[col] > 10,0,100) )
}
I am kind of new to R and I want to apply the gsub function in columns 6 to 12 in my data frame called x. However, I am kind of stuck. I started using:
gsub("\\.", "",x[,c(6:12)])
But then it returned only a few rows.
Then, I tried to use this:
x1<-apply(x,c(6:12),function(x) gsub("\\.", "",x))
But I got the following error:
Error in if (d2 == 0L) { : missing value where TRUE/FALSE needed
Also tried this:
for (i in x[,c(6:12)])
{a<-data.frame(gsub("\\.", "",i))}
Does anybody have a tip or a solution?
It would also be great if someone showed me how to use an apply function and for.
Here is another solution. It returns all the columns of the original dataframe
library(dplyr)
mutate_at(x, 6:12, gsub("\\.", "", .))
Here is a solution in a data.table way. It worth considering when a table is big and time is a critical factor.
library(data.table) # load library
x = as.data.table(x) # convert data.frame into data.table
cols = names(x)[6:12] # define which columns to work with
x[ , (cols) := lapply(.SD, function(x) {gsub("\\.", "", x)}), .SDcols = cols] # replace
x # enjoy results
Adding the solution using apply from the comments, as this one really helped me:
x1 <- apply(x[,6:12], 2, function(x) gsub("\\.", "",x))
I have a dataframe with some 20 columns and some 10^7 rows. One of the columns is an id column that is a factor. I want to filter the rows by properties of the string representation of the levels of the factor. The code below achieves this, but seems to me to be really rather inelegant. In particular that I have to create a vector of the relevant ids seems to me should not be needed.
Any suggestions for streamlining this?
library(dplyr)
library(tidyr)
library(gdata)
dat <- data.frame(id=factor(c("xxx-nld", "xxx-jap", "yyy-aus", "zzz-ita")))
europ.id <- function(id) {
ctry.code <- substring(id, nchar(id)-2)
ctry.code %in% c("nld", "ita")
}
ids <- levels(dat$id)
europ.ids <- subset(ids, europ.campaign(ids))
datx <- dat %>% filter(id %in% europ.ids) %>% drop.levels
Docendo Discimus gave the right answer in comments. To explain it first see the error I kept getting in my different attempts
> dat %>% filter(europ.id(id))
Error in nchar(id) : 'nchar()' requires a character vector
Calls: %>% ... filter_impl -> .Call -> europ.id -> substring -> nchar
Then note that his solution works because grepl applies as.character to its argument if needed (from the man: a character vector where matches are sought, or an object which can be coerced by as.character to a character vector). This implicit application of as.character also happens if you use %in%. Since this solution is also perfectly performant, we can do the following
dat %>% filter(europ.id(as.character(id)) %>% droplevels
Or to make it read a bit nicer update the function to
europ.id <- function(id) {
ids <- as.character(id)
ctry.code <- substring(ids, nchar(ids)-2)
ctry.code %in% c("nld", "ita")
}
and use
dat %>% filter(europ.id(id)) %>% droplevels
which reads exactly like what I was looking for.