I want to select all columns that start in one of the four following ways: CB, LB, LW, CW but not any columns that have the string "con."
My current approach is:
tester <- df_ans[,names(df_ans) %in% colnames(df_ans)[grepl("^(LW|LB|CW|CB)[A-Z_0-9]*",colnames(df_ans))]]
tester <- tester[,names(tester) %in% colnames(tester)[!grepl("con",colnames(tester))]]
Is there a better / more efficient way to do this in a library like dplyr?
We can use matches
library(dplyr)
df %>%
select(matches("^(CB|LB|LW|CW)"), -matches("con"))
# CB1 LB2 CW3 LW20
#1 3 9 6 1
#2 3 3 4 5
#3 7 7 7 7
#4 5 8 7 2
#5 6 3 3 3
data
set.seed(24)
df <- as.data.frame(matrix(sample(1:9, 10 * 5, replace = TRUE),
ncol = 10, dimnames = list(NULL, c("CB1", "LB2", "CW3", "WC1",
"LW20", "conifer", "hercon", "other", "other2", "other3"))))
Try this:
nms <- names(df_ans)
df_ans[ grepl("^(LW|LB|CW|CB)", nms) & !grepl("con", nms) ]
Related
I was examining below code
library(dplyr)
DF = data.frame('A' = 1:3, 'B' =2:4)
Condition = 'A'
fn1 = function(x) x + 3
fn2 = function(x) x + 5
DF %>% mutate('aa' = 3:5) %>%
{if (Condition == 'A') {
bb = . %>% mutate('A1' = fn1(A), 'B1' = fn1(B))
bb
} else {
bb = . %>% mutate('A1' = fn2(A), 'B1' = fn2(B))
bb
}
}
Basically, I have 2 similar functions fn1 and fn2. Now based on some condition, I want to use one of these functions.
Above implementation is throwing below error -
Functional sequence with the following components:
1. mutate(., A1 = fn1(A), B1 = fn1(B))
Use 'functions' to extract the individual functions.
Can you please help be how to properly write the pipe sequence to execute above code?
We could use across within mutate
library(dplyr)
DF %>%
mutate(aa = 3:5, across(c(A, B), ~ if(Condition == 'A') fn1(.)
else fn2(.), .names = "{.col}1"))
-output
A B aa A1 B1
1 1 2 3 4 5
2 2 3 4 5 6
3 3 4 5 6 7
Also, an option is to get the functions in a list and convert the logical vector to numeric index for subsetting
DF %>%
mutate(aa = 3:5,
across(c(A, B), ~ list(fn2, fn1)[[1 + (Condition == 'A')]](.),
.names = "{.col}1"))
-output
A B aa A1 B1
1 1 2 3 4 5
2 2 3 4 5 6
3 3 4 5 6 7
Based on the comments, if we need a custom name for the new columns, create a named vector and replace with str_replace_all
library(stringr)
nm1 <- setNames(c("XXX", "YYY"), names(DF)[1:2])
DF %>%
mutate(aa = 3:5,
across(c(A, B), ~ list(fn2, fn1)[[1 + (Condition == 'A')]](.),
.names = "{str_replace_all(.col, nm1)}"))
A B aa XXX YYY
1 1 2 3 4 5
2 2 3 4 5 6
3 3 4 5 6 7
I have a df where one variable is an integer. I'd like to split this column into it's individual digits. See my example below
Group Number
A 456
B 3
C 18
To
Group Number Digit1 Digit2 Digit3
A 456 4 5 6
B 3 3 NA NA
C 18 1 8 NA
We can use read.fwf from base R. Find the max number of character (nchar) in 'Number' column (mx). Read the 'Number' column after converting to character (as.character), specify the 'widths' as 1 by replicating 1 with mx and assign the output to new 'Digit' columns in the data
mx <- max(nchar(df1$Number))
df1[paste0("Digit", seq_len(mx))] <- read.fwf(textConnection(
as.character(df1$Number)), widths = rep(1, mx))
-output
df1
# Group Number Digit1 Digit2 Digit3
#1 A 456 4 5 6
#2 B 3 3 NA NA
#3 C 18 1 8 NA
data
df1 <- structure(list(Group = c("A", "B", "C"), Number = c(456L, 3L,
18L)), class = "data.frame", row.names = c(NA, -3L))
Another base R option (I think #akrun's approach using read.fwf is much simpler)
cbind(
df,
with(
df,
type.convert(
`colnames<-`(do.call(
rbind,
lapply(
strsplit(as.character(Number), ""),
`length<-`, max(nchar(Number))
)
), paste0("Digit", seq(max(nchar(Number))))),
as.is = TRUE
)
)
)
which gives
Group Number Digit1 Digit2 Digit3
1 A 456 4 5 6
2 B 3 3 NA NA
3 C 18 1 8 NA
Using splitstackshape::cSplit
splitstackshape::cSplit(df, 'Number', sep = '', stripWhite = FALSE, drop = FALSE)
# Group Number Number_1 Number_2 Number_3
#1: A 456 4 5 6
#2: B 3 3 NA NA
#3: C 18 1 8 NA
Updated
I realized I could use max function for counting characters limit in each row so that I could include it in my map2 function and save some lines of codes thanks to an accident that led to an inspiration by dear #ThomasIsCoding.
library(dplyr)
library(tidyr)
library(purrr)
library(stringr)
df %>%
rowwise() %>%
mutate(map2_dfc(Number, 1:max(nchar(Number)), ~ str_sub(.x, .y, .y))) %>%
unnest(cols = !c(Group, Number)) %>%
rename_with(~ str_replace(., "\\.\\.\\.", "Digit"), .cols = !c(Group, Number)) %>%
mutate(across(!c(Group, Number), as.numeric, na.rm = TRUE))
# A tibble: 3 x 5
Group Number Digit1 Digit2 Digit3
<chr> <dbl> <dbl> <dbl> <dbl>
1 A 456 4 5 6
2 B 3 3 NA NA
3 C 18 1 8 NA
Data
df <- tribble(
~Group, ~Number,
"A", 456,
"B", 3,
"C", 18
)
Two base r methods:
no_cols <- max(nchar(as.character(df1$Number)))
# Using `strsplit()`:
cbind(df1, setNames(data.frame(do.call(rbind,
lapply(strsplit(as.character(df1$Number), ""),
function(x) {
length(x) <- no_cols
x
}
)
)
), paste0("Digit", seq_len(no_cols))))
# Using `regmatches()` and `gregexpr()`:
cbind(df1, setNames(data.frame(do.call(rbind,
lapply(regmatches(df1$Number, gregexpr("\\d", df1$Number)),
function(x) {
length(x) <- no_cols
x
}
)
)
), paste0("Digit", seq_len(no_cols))))
I have data like this:
Name Rating
Tom 3
Tom 4
Tom 2
Johnson 5
Johnson 7
But I'd like it so each unique name is instead a column, with the ratings below, in each row. How can I approach this?
Here is a good way of doing it
x <- data.frame(c("Tom", "Tom", "Tom", "Johnson", "Johnson"), c(3,4,2,5,7))
colnames(x) <- c("Name", "Rating")
n <- unique(x[,1])
m <- max(table(x[,1]))
c <- data.frame(matrix(, ncol = length(n), nrow = m))
for (i in 1:length(n)) {
l <- x[which(x[,1] == n[i]), 2]
l2 <- rep("", m - length(l))
c[,i] <- c(l, l2)
}
colnames(c) <- n
Results:
Tom Johnson
1 3 5
2 4 7
3 2
Here is a way using CRAN package reshape.
library(reshape2)
d <- dcast(mydata, Rating ~ Name, value.var = "Rating")[-1]
d
# Johnson Tom
#1 NA 2
#2 NA 3
#3 NA 4
#4 5 NA
#5 7 NA
As you can see, there are too many NA values in this result. One way of getting rid of them could be:
d <- lapply(d, function(x) x[!is.na(x)])
n <- max(sapply(d, length))
d <- do.call(cbind.data.frame, lapply(d, function(x) c(x, rep(NA, n - length(x)))))
d
# Johnson Tom
#1 5 2
#2 7 3
#3 NA 4
Well, this does the job but introduces some NAs.
Edit: Replace the NAs with some other Rating.
mydata<-data.frame(Name=c("Tom","Tom","Tom","Johnson","Johnson"),Rating=c(3,4,2,5,7))
library(reshape2)
library(tidyverse)
mydata1<-mydata %>%
mutate(Name=as.factor(Name)) %>%
melt(id.var="Name") %>%
dcast(variable+value~Name) %>%
select(-value) %>%
rename(Name=variable) %>%
select_if(is.numeric)
mydata1 %>%
mutate(Johnson=as.factor(Johnson),Tom=as.factor(Tom)) %>%
mutate(Johnson=fct_explicit_na(Johnson,na_level = "No Rating"),
Tom=fct_explicit_na(Tom,na_level = "No Rating"))
Johnson Tom
1 No Rating 2
2 No Rating 3
3 No Rating 4
4 5 No Rating
5 7 No Rating
I have a data frame where column "A" has 6 distinct values. Column "B" has float values. By using dplyr, I can group by column "A" and find mean of column "B" of each group as follows:
mydf %>% group_by(A) %>% summarize(Mean = mean(B, na.rm=TRUE))
My utter aim is to find rows in each group whose "B" values are higher than the group average. How can I achieve this (using base R or dplyr)?
A simple alternative with base R ave would be
df[df$b > ave(df$b, df$a) , ]
# a b
#4 1 4
#5 1 5
#9 2 9
#10 2 10
The default argument for ave is mean so no need to mention it explicitly, if there are NA values present in b modify it to
df[df$b > ave(df$b, df$a, FUN = function(x) mean(x,na.rm = TRUE)) , ]
Another solution with subset and ave as suggested by #Onyambu
subset(df,b>ave(b,a))
# a b
#4 1 4
#5 1 5
#9 2 9
#10 2 10
data
df <- data.frame(a = rep(c(1, 2), each = 5), b = 1:10)
df
# a b
#1 1 1
#2 1 2
#3 1 3
#4 1 4
#5 1 5
#6 2 6
#7 2 7
#8 2 8
#9 2 9
#10 2 10
You can just group and then filter:
mydf %>%
group_by(A) %>%
filter(B > mean(B, na.rm = TRUE)) %>%
ungroup()
Using Base R, I would go for this. It is not as elegant as dplyr.
mean.df <- aggregate(mydf$b, by =list(a = mydf$a), FUN = mean)
names(mean.df)[2] <- "mean"
mydf <- merge(mydf, mean.df, by = "a")
# Rows whose values are higher than mean
new.df <- subset(mydf, b > mean, select = -mean)
I like working with Data tables. So a data.table solution would be,
mydt <- data.table(mydf)
mydt[, mean := mean(b), by = a]
new.dt <- mydt[b > mean, -c("mean"), with = TRUE]
Another way to do it using base R and tapply:
mydf = cbind.data.frame(A=sample(6,20,rep=T),B=runif(20))
mydf.ave = tapply(mydf$B,mydf$A,mean)
newdf = mydf[mydf$B > mydf.ave[as.character(mydf$A)],]
(thus the one liner would be:mydf[mydf$B > tapply(mydf$B,mydf$A,mean)[as.character(mydf$A)],])
I would need to expand on this question: convert data frame of counts to proportions in R
I need to calculate proportion by one condition and retain the information of the dataset.
Reproducible example:
ID <- rep(c(1,2,3), each=3)
trial <- rep("a", 9)
variable1 <- sample(1:10, 9)
variable2 <- sample(1:10, 9)
variable3 <- sample(1:10, 9)
condition <- rep(c("i","j","k"), 3)
dat <- data.frame(cbind(ID, trial,variable1,variable2,variable3,condition))
For each variable I would like to have the proportion by the ID (i.e. 3 times)
Ideally the new variables would be stored in the same database as dat$variable1_p
I know how to do the trick by a series of for loops but I would like to learn how to use the apply function. Also to be able to expand it to more conditions if necessary.
We can use adply from the plyr package:
library(plyr)
adply(dat, 1, function(x)
c('variable1_p' = x$variable1 / sum(dat[x$ID == dat$ID,]$variable1)))
# ID trial variable1 variable2 variable3 condition variable1_p
# 1 1 a 3 5 4 i 0.20000000
# 2 1 a 8 9 9 j 0.53333333
# 3 1 a 4 4 8 k 0.26666667
# 4 2 a 7 10 5 i 0.50000000
# 5 2 a 6 8 10 j 0.42857143
# 6 2 a 1 1 7 k 0.07142857
# 7 3 a 10 6 3 i 0.47619048
# 8 3 a 9 7 6 j 0.42857143
# 9 3 a 2 3 2 k 0.09523810
Another option is to use dplyr, which would handle cases where there is more than one row per condition per ID:
library(dplyr)
dat %>%
group_by(ID, condition) %>%
mutate(sum_v1_cond = sum(variable1)) %>%
ungroup() %>%
group_by(ID) %>%
mutate(variable1_p = sum_v1_cond / sum(variable1)) %>%
select(-sum_v1_cond)
Edit - here's a full solution for variable1, variable2, and variable3:
adply(dat, 1, function(x)
c('variable1_p' = x$variable1 / sum(dat[x$ID == dat$ID,]$variable1),
'variable2_p' = x$variable2 / sum(dat[x$ID == dat$ID,]$variable2),
'variable3_p' = x$variable3 / sum(dat[x$ID == dat$ID,]$variable3)))
Data:
set.seed(123)
ID <- rep(c(1,2,3), each=3)
trial <- rep("a", 9)
variable1 <- sample(1:10, 9)
variable2 <- sample(1:10, 9)
variable3 <- sample(1:10, 9)
condition <- rep(c("i","j","k"), 3)
dat <- data.frame(ID, trial,variable1,variable2,variable3,condition,
stringsAsFactors = FALSE)