I want to extract a data frame using a formula, which specifies which columns to select and some crossing overs among columns.
I know model.frame function. However it does not give me the crossing overs:
For example:
df <- data.frame(x = c(1,2,3,4), y = c(2,3,4,7), z = c(5,6, 9, 1))
f <- formula('z~x*y')
model.frame(f, df)
output:
> df
x y z
1 1 2 5
2 2 3 6
3 3 4 9
4 4 7 1
> f <- formula('z~x*y')
> model.frame(f, df)
z x y
1 5 1 2
2 6 2 3
3 9 3 4
4 1 4 7
I hope to get:
z x y x*y
1 5 1 2 2
2 6 2 3 6
3 9 3 4 12
4 1 4 7 28
Is there a package that could achieve this functionality? (It would be perfect if I can get the resulting matrix as a sparse matrix because the crossed columns will be highly sparse)
You can use model.matrix:
> model.matrix(f, df)
(Intercept) x y x:y
1 1 1 2 2
2 1 2 3 6
3 1 3 4 12
4 1 4 7 28
attr(,"assign")
[1] 0 1 2 3
If you want to save the result as a sparse matrix, you can use the Matrix package:
> mat <- model.matrix(f, df)
> library(Matrix)
> Matrix(mat, sparse = TRUE)
4 x 4 sparse Matrix of class "dgCMatrix"
(Intercept) x y x:y
1 1 1 2 2
2 1 2 3 6
3 1 3 4 12
4 1 4 7 28
Related
SO. The following might serve as a small example of the real list.
a <- data.frame(
x = c("A","A","A","A","A"),
y = c(1,2,3,4,5),
z = c(1,2,3,4,5))
b <- data.frame(
x = c("A","A","A","A","A"),
y = c(1,2,3,4,5),
z = c(1,2,3,4,5))
c <- data.frame(
x = c("A","A","A","A","A"),
y = c(1,2,3,4,5),
z = c(1,2,3,4,5))
l <- list(a,b,c)
From the second column to last column - on every data frame - i want to add the sums as a new column to each data frame.
I tried:
lapply(l, function(x) rowSums(x[2:ncol(x)]))
which returns the correct sums, but doesn't add them to the data frames.
I also tried:
lapply(l, transform, sum = y + z)
which gives me the correct results but is not flexible enough, because i don't always know how many columns there are for each data frame and what names they have. The only thing i know, is, that i have to start from second column to end. I tried to combine these two approaches but i can't figure out, how to do it exactly.
Thanks
Try this. You can play around index in columns and exclude the first variable so that there is not issues about how many additional variables you have in order to obtain the rowsums. Here the code:
#Compute rowsums
l1 <- lapply(l,function(x) {x$RowSum<-rowSums(x[,-1],na.rm=T);return(x)})
Output:
l1
[[1]]
x y z RowSum
1 A 1 1 2
2 A 2 2 4
3 A 3 3 6
4 A 4 4 8
5 A 5 5 10
[[2]]
x y z RowSum
1 A 1 1 2
2 A 2 2 4
3 A 3 3 6
4 A 4 4 8
5 A 5 5 10
[[3]]
x y z RowSum
1 A 1 1 2
2 A 2 2 4
3 A 3 3 6
4 A 4 4 8
5 A 5 5 10
Here's how to combine your attempts. I used data[-1] instead of data[2:ncol(data)] because it seems simpler, but either should work.
lapply(l, function(data) transform(data, sum = rowSums(data[-1])))
Unfortunately, transform will be confused if the name of the argument to your anonymous function is the same as a column name - data[-1] needs to look at the data frame, not a particular column. (I originally use function(x) instead of function(data), and this caused an error because there is a column named x. From this perspective, Duck's answer is a little safer.)
Does this work:
> add_col <- function(df){
+ df[(ncol(df)+1)] = rowSums(df[2:ncol(df)])
+ df
+ }
> lapply(l, add_col)
[[1]]
x y z V4
1 A 1 1 2
2 A 2 2 4
3 A 3 3 6
4 A 4 4 8
5 A 5 5 10
[[2]]
x y z V4
1 A 1 1 2
2 A 2 2 4
3 A 3 3 6
4 A 4 4 8
5 A 5 5 10
[[3]]
x y z V4
1 A 1 1 2
2 A 2 2 4
3 A 3 3 6
4 A 4 4 8
5 A 5 5 10
>
With sum as column name:
> add_col <- function(df){
+ df['sum'] = rowSums(df[2:ncol(df)])
+ df
+ }
> lapply(l, add_col)
[[1]]
x y z sum
1 A 1 1 2
2 A 2 2 4
3 A 3 3 6
4 A 4 4 8
5 A 5 5 10
[[2]]
x y z sum
1 A 1 1 2
2 A 2 2 4
3 A 3 3 6
4 A 4 4 8
5 A 5 5 10
[[3]]
x y z sum
1 A 1 1 2
2 A 2 2 4
3 A 3 3 6
4 A 4 4 8
5 A 5 5 10
use tidyverse
library(tidyverse)
map(l, ~.x %>% mutate(Sum := apply(.x[-1], 1, sum)))
#> [[1]]
#> x y z Sum
#> 1 A 1 1 2
#> 2 A 2 2 4
#> 3 A 3 3 6
#> 4 A 4 4 8
#> 5 A 5 5 10
#>
#> [[2]]
#> x y z Sum
#> 1 A 1 1 2
#> 2 A 2 2 4
#> 3 A 3 3 6
#> 4 A 4 4 8
#> 5 A 5 5 10
#>
#> [[3]]
#> x y z Sum
#> 1 A 1 1 2
#> 2 A 2 2 4
#> 3 A 3 3 6
#> 4 A 4 4 8
#> 5 A 5 5 10
Created on 2020-09-30 by the reprex package (v0.3.0)
We can use map with mutate
library(purrr)
library(dplyr)
map(l, ~ .x %>%
mutate(sum = rowSums(select(., -1))))
Or with c_across
map(l, ~ .x %>%
rowwise() %>%
mutate(sum = sum(c_across(-1), na.rm = TRUE)) %>%
ungroup)
say I have a Data frame
g <- c("Smember_1", "Smember_1", "Smember_1", "Smember_2", "Smember_2", "Smember_2", "Smember_3", "Smember_3", "Smember_3")
m <- c(1,2,1,3,4,1,3,5,6)
df <- data.frame(g, m)
g m
1 Smember_1 1
2 Smember_1 2
3 Smember_1 1
4 Smember_2 3
5 Smember_2 4
6 Smember_2 1
7 Smember_3 3
8 Smember_3 5
9 Smember_3 6
I would like to remove Smember_ in from all the variables in the g column such that the data frame df looks like
> df
g m
1 1 1
2 1 2
3 1 1
4 2 3
5 2 4
6 2 1
7 3 3
8 3 5
9 3 6
I think you want
df$g <- gsub(".*(\\d+)$", "\\1", df$g)
df2$variable <- gsub("Smember_","", df2$variable)
worked!
What I want can be described as: give a data frame, contains all the case-control pairs. In the following example, y is the id for the case-control pair. There are 3 pairs in my data set. I'm doing a resampling with respect to the different values of y (the pair will be both selected or neither).
sample_df = data.frame(x=1:6, y=c(1,1,2,2,3,3))
> sample_df
x y
1 1 1
2 2 1
3 3 2
4 4 2
5 5 3
6 6 3
select_y = c(1,3,3)
select_y
> select_y
[1] 1 3 3
Now, I have computed a vector contains the pairs I want to resample, which is select_y above. It means the case-control pair number 1 will be in my new sample, and number 3 will also be in my new sample, but it will occur 2 times since there are two 3. The desired output will be:
x y
1 1
2 1
5 3
6 3
5 3
6 3
I can't find out an efficient way other than writing a for loop...
Solution:
Based on #HubertL , with some modifications, a 'vectorized' approach looks like:
sel_y <- as.data.frame(table(select_y))
> sel_y
select_y Freq
1 1 1
2 3 2
sub_sample_df = sample_df[sample_df$y%in%select_y,]
> sub_sample_df
x y
1 1 1
2 2 1
5 5 3
6 6 3
match_freq = sel_y[match(sub_sample_df$y, sel_y$select_y),]
> match_freq
select_y Freq
1 1 1
1.1 1 1
2 3 2
2.1 3 2
sub_sample_df$Freq = match_freq$Freq
rownames(sub_sample_df) = NULL
sub_sample_df
> sub_sample_df
x y Freq
1 1 1 1
2 2 1 1
3 5 3 2
4 6 3 2
selected_rows = rep(1:nrow(sub_sample_df), sub_sample_df$Freq)
> selected_rows
[1] 1 2 3 3 4 4
sub_sample_df[selected_rows,]
x y Freq
1 1 1 1
2 2 1 1
3 5 3 2
3.1 5 3 2
4 6 3 2
4.1 6 3 2
Another method of doing the same without a loop:
sample_df = data.frame(x=1:6, y=c(1,1,2,2,3,3))
row_names <- split(1:nrow(sample_df),sample_df$y)
select_y = c(1,3,3)
row_num <- unlist(row_names[as.character(select_y)])
ans <- sample_df[row_num,]
I can't find a way without a loop, but at least it's not a for loop, and there is only one iteration per frequency:
sample_df = data.frame(x=1:6, y=c(1,1,2,2,3,3))
select_y = c(1,3,3)
sel_y <- as.data.frame(table(select_y))
do.call(rbind,
lapply(1:max(sel_y$Freq),
function(freq) sample_df[sample_df$y %in%
sel_y[sel_y$Freq>=freq, "select_y"],]))
x y
1 1 1
2 2 1
5 5 3
6 6 3
51 5 3
61 6 3
Say we have the following data
A <- c(1,2,2,2,3,4,8,6,6,1,2,3,4)
B <- c(1,2,3,4,5,1,2,3,4,5,1,2,3)
data <- data.frame(A,B)
How would one write a function so that for A, if we have the same value in the i+1th position, then the reoccuring row is removed.
Therefore the output should like like
data.frame(c(1,2,3,4,8,6,1,2,3,4), c(1,2,5,1,2,3,5,1,2,3))
My best guess would be using a for statement, however I have no experience in these
You can try
data[c(TRUE, data[-1,1]!= data[-nrow(data), 1]),]
Another option, dplyr-esque:
library(dplyr)
dat1 <- data.frame(A=c(1,2,2,2,3,4,8,6,6,1,2,3,4),
B=c(1,2,3,4,5,1,2,3,4,5,1,2,3))
dat1 %>% filter(A != lag(A, default=FALSE))
## A B
## 1 1 1
## 2 2 2
## 3 3 5
## 4 4 1
## 5 8 2
## 6 6 3
## 7 1 5
## 8 2 1
## 9 3 2
## 10 4 3
using diff, which calculates the pairwise differences with a lag of 1:
data[c( TRUE, diff(data[,1]) != 0), ]
output:
A B
1 1 1
2 2 2
5 3 5
6 4 1
7 8 2
8 6 3
10 1 5
11 2 1
12 3 2
13 4 3
Using rle
A <- c(1,2,2,2,3,4,8,6,6,1,2,3,4)
B <- c(1,2,3,4,5,1,2,3,4,5,1,2,3)
data <- data.frame(A,B)
X <- rle(data$A)
Y <- cumsum(c(1, X$lengths[-length(X$lengths)]))
View(data[Y, ])
row.names A B
1 1 1 1
2 2 2 2
3 5 3 5
4 6 4 1
5 7 8 2
6 8 6 3
7 10 1 5
8 11 2 1
9 12 3 2
10 13 4 3
Lets say I have a data frame with the following structure:
DF <- data.frame(x = 0:4, y = 5:9)
> DF
x y
1 0 5
2 1 6
3 2 7
4 3 8
5 4 9
what is the most efficient way to turn 'DF' into a data frame with the following structure:
w x y
1 0 5
1 1 6
2 1 6
2 2 7
3 2 7
3 3 8
4 3 8
4 4 9
Where w is a length 2 window rolling through the dataframe 'DF.' The length of the window should be arbitrary, i.e a length of 3 yields
w x y
1 0 5
1 1 6
1 2 7
2 1 6
2 2 7
2 3 8
3 2 7
3 3 8
3 4 9
I am a bit stumped by this problem, because the data frame can also contain an arbitrary number of columns, i.e. w,x,y,z etc.
/edit 2: I've realized edit 1 is a bit unreasonable, as xts doesn't seem to deal with multiple observations per data point
My approach would be to use the embed function. The first thing to do is to create a rolling sequence of indices into a vector. Take a data-frame:
df <- data.frame(x = 0:4, y = 5:9)
nr <- nrow(df)
w <- 3 # window size
i <- 1:nr # indices of the rows
iw <- embed(i,w)[, w:1] # matrix of rolling-window indices of length w
> iw
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 2 3 4
[3,] 3 4 5
wnum <- rep(1:nrow(iw),each=w) # window number
inds <- i[c(t(iw))] # the indices flattened, to use below
dfw <- sapply(df, '[', inds)
dfw <- transform(data.frame(dfw), w = wnum)
> dfw
x y w
1 0 5 1
2 1 6 1
3 2 7 1
4 1 6 2
5 2 7 2
6 3 8 2
7 2 7 3
8 3 8 3
9 4 9 3