I want to define the first two columns of a data frame as rownames. Actually I want to do some calculations and the data frame has to be numeric for that.
data.frame <- data_frame(id=c("A1","B2"),name=c("julia","daniel"),BMI=c("20","49"))
The values for BMI are numerical (proved with is.numeric), but the over all data.frame not. How to define the first two columns (id and name) as rownames?
Thank you in advance for any suggestions
You can combine id and name column and then assign rownames
data.frame %>%
tidyr::unite(rowname, id, name) %>%
tibble::column_to_rownames()
# BMI
#A1_julia 20
#B2_daniel 49
In base R, you can do the same in steps as
data.frame <- as.data.frame(data.frame)
rownames(data.frame) <- paste(data.frame$id, data.frame$name, sep = "_")
data.frame[c('id', 'name')] <- NULL
Not sure if the code and result below is the thing you are after:
dfout <- `rownames<-`(data.frame(BMI = as.numeric(df$BMI)),paste(df$id,df$name))
such that
> dfout
BMI
A1 julia 20
B2 daniel 49
DATA
df <- structure(list(id = structure(1:2, .Label = c("A1", "B2"), class = "factor"),
name = structure(2:1, .Label = c("daniel", "julia"), class = "factor"),
BMI = structure(1:2, .Label = c("20", "49"), class = "factor")), class = "data.frame", row.names = c(NA,
-2L))
Related
I have a dataset called data1 that I need to split the first column into two columns. The issue I'm having is that there is no delimiter between what I need to split and the character lengths are different is many rows.
I would like to split it by the date and sex.
E.g
12/1/09male
1/9/20female
13/1/19female
4/12/12male
I've been trying this but because the values have a different amount of characters I'm stuck.
separate(data1, col = 1, into = c("date","sex"), sep = "")
Any help would be hugely appreciated!
An option is a positive look-behind and look-ahead to split on a digit followed by an "m" or "f".
df %>% separate(1, c("date", "sex"), sep = "(?<=\\d)(?=[mf])")
# date sex
#1 12/1/09 male
#2 1/9/20 female
#3 13/1/19 female
#4 4/12/12 male
For what it's worth, the same regexp pattern works in base R's strsplit
setNames(do.call(
rbind.data.frame,
strsplit(as.character(df[, 1]), "(?<=\\d)(?=[mf])", perl = T)),
c("date", "sex"))
Sample data
df <- read.table(text =
'12/1/09male
1/9/20female
13/1/19female
4/12/12male')
I am fairly new to R so I am sure this is not the most elegant solution. I first add a comma between the date and sex and then separate on the comma
a <- data.frame(row_1 = c("12/1/09male", "1/9/20female", "13/1/19female", "4/12/12male"))
a[, "row_1"] = str_replace(a$row_1, "(male|female)", ",\\1")
separate(a, row_1, ",", into = c("date", "sex"))
Using tidyr::extract, we can capture data into two parts. First capture the date (in the format d/m/y) and second capture all the remaining part of the string.
tidyr::extract(df, V1, c("date", "sex"), "(\\d+/\\d+/\\d+)(.*)")
# date sex
#1 12/1/09 male
#2 1/9/20 female
#3 13/1/19 female
#4 4/12/12 male
data
df <- structure(list(V1 = structure(c(2L, 1L, 3L, 4L), .Label = c("1/9/20female",
"12/1/09male", "13/1/19female", "4/12/12male"), class = "factor")),
class = "data.frame", row.names = c(NA,-4L))
Base R solution using gsub and some regex:
df_clean <- within(df, {
date <- as.Date(gsub("[A-Za-z]+", "", V1), format = "%d/%m/%y")
sex <- as.factor(gsub("\\d+|\\/", "", V1))
rm(V1)
}
)
Data:
df <- structure(list(V1 = structure(c(2L, 1L, 3L, 4L), .Label = c("1/9/20female",
"12/1/09male", "13/1/19female", "4/12/12male"), class = "factor")), class = "data.frame", row.names = c(NA,
-4L))
I am trying to clean some data. Below is an example of my data.
test1 test2 test3
jsb cjn kd N069W j N9DSW
I want to indicate what column has the pattern N0{num}{num}W in it. The {num} part can be any number between 0-9. This pattern can also appear anywhere in the string. Hence in this case my results would be as follows.
test1 test2 test3 col
jsb cjn kd N069W j N9DSW 2
Thanks in advance for any help.
We loop through the columns, use grepl to get a logical index and then with max.col get the column index of each row
max.col(data.frame(lapply(df1, grepl, pattern = "N0\\d{2}W")))
#[1] 2
data
df1 <- structure(list(test1 = "jsb cjn", test2 = "kd N069W j",
test3 = "N9DSW"), class = "data.frame", row.names = c(NA,
-1L))
you can also use the function str_detect() from the library stringr.
library(stringr)
str_detect('kd NO69W j', pattern = "NO\\d+W")
# [1] TRUE
Using apply:
df$col <- apply(df, 1, function(x) grep("N0\\d{2}W", x))
Data:
df <- structure(list(test1 = structure(1L, .Label = "jsb cjn", class = "factor"),
test2 = structure(1L, .Label = "kd N069W j", class = "factor"),
test3 = structure(1L, .Label = "N9DSW ", class = "factor")), class = "data.frame", row.names = c(NA,
-1L))
My problem statement is I have a list of dataframes as df1,df2,df3.Data is like
df1
a,b,c,d
1,2,3,4
1,2,3,4
df2
a,b,c,d
1,2,3,4
1,2,3,4
Now, for these two dataframe I should create a new dataframe taking aggregated column of those two dataframes ,for that I am using below code
for(i in 1:2){
assign(paste(final_val,i,sep=''),sum(assign(paste(df,i,sep='')))$d*100)}
I am getting the error:
Error in assign(paste(hvp_route_dsct_clust, i, sep = "")) :
argument "value" is missing, with no default
My output should look like
final_val1 <- 800
final_val2 <- 800
And for those values final_val1,final_val2 I should be creating dataframe dynamicaly
Can anybody please help me on this
If we need to use assign, get the object names from the global environment with ls by specifying the pattern 'df' followed by one or more numbers (\\d+), create another vector of 'final_val's ('nm1'), loop through the sequence of 'nm1', assign each of the element in 'nm2' to the value we got from extracting the column 'd' of each 'df's multiplied by 100 and taking its sum.
nm1 <- ls(pattern = "df\\d+")
nm2 <- paste0("final_val", seq_along(nm1))
for(i in seq_along(nm1)){
assign(nm2[i], sum(get(nm1[i])$d*100))
}
final_val1
#[1] 800
final_val2
#[1] 800
Otherwise, we place the datasets in a list, extract the 'd' column, multiply with 100 and do the column sums
unname(colSums(sapply(mget(nm1), `[[`, 'd') * 100))
#800 800
data
df1 <- structure(list(a = c(1L, 1L), b = c(2L, 2L), c = c(3L, 3L), d = c(4L,
4L)), .Names = c("a", "b", "c", "d"), class = "data.frame", row.names = c(NA,
-2L))
df2 <- structure(list(a = c(1L, 1L), b = c(2L, 2L), c = c(3L, 3L), d = c(4L,
4L)), .Names = c("a", "b", "c", "d"), class = "data.frame", row.names = c(NA,
-2L))
i have a R data set with >200 columns. I need to get what class each column is and get that into excel, with col name and its corresponding class as two columns
1. Using lapply/sapply with stack/melt
You could do this using lapply/sapply to get the class of each column and then using stack from base R or melt from reshape2 to get the 2 column data.frame.
res <- stack(lapply(df, class))
#or
library(reshape2)
res1<- melt(lapply(df, class))
Then use write.csv or using any of the specialized libraries for writing to excel data i.e. XLConnect, WriteXLS etc.
write.csv(res, file="file1.csv", row.names=FALSE, quote=FALSE)
.csv files can be opened in excel
2. From the output of str
Or you could use capture.output and regex to get the required info from the str and convert it to data.frame using read.table
v1 <- capture.output(str(df))
v2 <- grep("\\$", v1, value=TRUE)
res2 <- read.table(text=gsub(" +\\$ +(.*)\\: +([A-Za-z]+) +.*", "\\1 \\2", v2),
sep="",header=FALSE,stringsAsFactors=FALSE)
head(res2,2)
# V1 V2
#1 t02.clase Factor
#2 Std_A_CLI_monto_sucursal_1 chr
data
df <-structure(list(t02.clase = structure(c(1L, 1L, 1L), .Label = "AK",
class = "factor"),Std_A_CLI_monto_sucursal_1 = c("0", "0", "0"),
Std_A_CLI_monto_sucursal_2 = c(0, 0.01303586, 0), Std_A_CLI_monto_sucursal_3 =
c(0.051311597, 0.003442244, 0.017347593), Std_A_CLI_monto_sucursal_4 = c(0L,
0L, 0L), Std_A_CLI_promociones = c(0.4736842, 0.5, 0), Std_A_CLI_dias_cliente =
c(0.57061341, 0.55492154, 0.05991441), Std_A_CLI_sucursales = c(0.05555556,
0.05555556, 0.05555556)), .Names = c("t02.clase", "Std_A_CLI_monto_sucursal_1",
"Std_A_CLI_monto_sucursal_2", "Std_A_CLI_monto_sucursal_3",
"Std_A_CLI_monto_sucursal_4", "Std_A_CLI_promociones", "Std_A_CLI_dias_cliente",
"Std_A_CLI_sucursales"), row.names = c("1", "2", "3"), class = "data.frame")
I have many dataframes stored in a list, and I want to create weighted averages from these and store the results in a new dataframe. For example, with the list:
dfs <- structure(list(df1 = structure(list(A = 4:5, B = c(8L, 4L), Weight = c(TRUE, TRUE), Site = c("X", "X")),
.Names = c("A", "B", "Weight", "Site"), row.names = c(NA, -2L), class = "data.frame"),
df2 = structure(list(A = c(6L, 8L), B = c(9L, 4L), Weight = c(FALSE, TRUE), Site = c("Y", "Y")),
.Names = c("A", "B", "Weight", "Site"), row.names = c(NA, -2L), class = "data.frame")),
.Names = c("df1", "df2"))
In this example, I want to use columns A, B, and Weight for the weighted averages. I also want to move over related data such as Site, and want to sum the number of TRUE and FALSE. My desired result would look something like:
result <- structure(list(Site = structure(1:2, .Label = c("X", "Y"), class = "factor"),
A.Weight = c(4.5, 8), B.Weight = c(6L, 4L), Sum.Weight = c(2L,
1L)), .Names = c("Site", "A.Weight", "B.Weight", "Sum.Weight"
), class = "data.frame", row.names = c(NA, -2L))
Site A.Weight B.Weight Sum.Weight
1 X 4.5 6 2
2 Y 8.0 4 1
The above is just a very simple example, but my real data have many dataframes in the list, and many more columns than just A and B for which I want to calculate weighted averages. I also have several columns similar to Site that are constant in each dataframe and that I want to move to the result.
I'm able to manually calculate weighted averages using something like
weighted.mean(dfs$df1$A, dfs$df1$Weight)
weighted.mean(dfs$df1$B, dfs$df1$Weight)
weighted.mean(dfs$df2$A, dfs$df2$Weight)
weighted.mean(dfs$df2$B, dfs$df2$Weight)
but I'm not sure how I can do this in a shorter, less "manual" way. Does anyone have any recommendations? I've recently learned how to lapply across dataframes in a list, but my attempts have not been so great so far.
The trick is to create a function that works for a single data.frame, then use lapply to iterate across your list. Since lapply returns a list, we'll then use do.call to rbind the resulting objects together:
foo <- function(data, meanCols = LETTERS[1:2], weightCol = "Weight", otherCols = "Site") {
means <- t(sapply(data[, meanCols], weighted.mean, w = data[, weightCol]))
sumWeight <- sum(data[, weightCol])
others <- data[1, otherCols, drop = FALSE] #You said all the other data was constant, so we can just grab first row
out <- data.frame(others, means, sumWeight)
return(out)
}
In action:
do.call(rbind, lapply(dfs, foo))
---
Site A B sumWeight
df1 X 4.5 6 2
df2 Y 8.0 4 1
Since you said this was a minimal example, here's one approach to expanding this to other columns. We'll use grepl() and use regular expressions to identify the right columns. Alternatively, you could write them all out in a vector. Something like this:
do.call(rbind, lapply(dfs, foo,
meanCols = grepl("A|B", names(dfs[[1]])),
otherCols = grepl("Site", names(dfs[[1]]))
))
using dplyr
library(dplyr)
library('devtools')
install_github('hadley/tidyr')
library(tidyr)
unnest(dfs) %>%
group_by(Site) %>%
filter(Weight) %>%
mutate(Sum=n()) %>%
select(-Weight) %>%
summarise_each(funs(mean=mean(., na.rm=TRUE)))
gives the result
# Site A B Sum
#1 X 4.5 6 2
#2 Y 8.0 4 1
Or using data.table
library(data.table)
DT <- rbindlist(dfs)
DT[(Weight)][, c(lapply(.SD, mean, na.rm = TRUE),
Sum=.N), by = Site, .SDcols = c("A", "B")]
# Site A B Sum
#1: X 4.5 6 2
#2: Y 8.0 4 1
Update
In response to #jazzuro's comment, Using dplyr 0.3, I am getting
unnest(dfs) %>%
group_by(Site) %>%
summarise_each(funs(weighted.mean=stats::weighted.mean(., Weight),
Sum.Weight=sum(Weight)), -starts_with("Weight")) %>%
select(Site:B_weighted.mean, Sum.Weight=A_Sum.Weight)
# Site A_weighted.mean B_weighted.mean Sum.Weight
#1 X 4.5 6 2
#2 Y 8.0 4 1