R, how to overlay two geom_bars in ggplot2? - r

I would like to plot one bar on the top of the other one in R.
First the count of all the elements having 0, then the count of all the elements having 1, on top of it.
I tried this piece of code in R:
library(ggplot2)
var <- c(0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0)
ggplot(data.frame(var), aes(factor(var), fill=factor(var))) + geom_bar(stat="count", position="stack")
but it generated this plot:
Which is not what I want.
I would like to get something like this (I made it with KolourPaint):
Any suggestion on how to do that? Thanks!


The problem is that you've supplied a variable to the x aesthetic, factor(var), but then from what you say, you don't actually want it there. You can use some dummy variable as x in your aes: a single number or letter, or even just a blank.
Also note that count is the default stat for geom_bar, so you don't have to explicitly supply stat = "count".
library(tidyverse)
var <- c(0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0)
ggplot(data.frame(var), aes(x = "", fill = factor(var))) +
geom_bar(position = "stack")
Created on 2018-05-08 by the reprex package (v0.2.0).


A quick and dirty solution is to add an additional variable to use on your x axis.
var <- c(0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0)
var=as.data.frame(var)
var$var1=1
ggplot(data.frame(var), aes(factor(var1), fill=factor(var))) + geom_bar(stat="count", position="stack")

Related

R Margins: Incorrect number of dimensions

I want to calculate the margins for an independent variable at the values of another independent variable. All variables (including the dependent variable) are binary.
model1 <- glm(data = TrialDF, formula = dep ~ indep1*indep2, family=binomial)
margins::margins(model1, data = TrialDF, variables = "indep1",
at = list("indep2" =c(0,1)))
However, I get the following error:
Error in dat[, not_numeric, drop = FALSE] :
incorrect number of dimensions
I also tried variations of this command by using factor variables or list("indep2" = 0:1), but I always get the same error messages. What does that mean?
The data is:
TrialDF <-structure(list(dep = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1,
1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0,
1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1,
1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1,
0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1,
0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1,
1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), indep1 = c(1,
0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1,
0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1,
1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1,
0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0,
0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0,
1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1,
0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1,
1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1,
1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0,
0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0,
1, 0, 0, 1, 1, 1, 1, 1), indep2 = c(1, 0, 1, 1, 1, 0, 1, 0, 0,
0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0,
0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1,
0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1,
0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0,
0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1,
0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0,
1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1,
0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0,
1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1,
1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1,
0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1
)), row.names = c(NA, -240L), class = c("data.table", "data.frame"
))

This error is related to the problem described in https://github.com/leeper/prediction/pull/34. You can get past it by coercing the data to a data frame with data=data.frame(TrialDF):
> margins(model1, data=data.frame(TrialDF), variables="indep1", at=list("indep2"=c(0,1)))
Average marginal effects at specified values
glm(formula = dep ~ indep1 * indep2, family = binomial, data = TrialDF)
at(indep2) indep1
0 0.6776
1 0.1496

Summing in a list in dplyr

I have a tibble which has the date, day of the quarter, and a list of whether the days in quarter are sales days or not (1/0). I would like to sum the list of sales days (to get # of sales days left in the quarter) by summing from day_num to the end of sales_day_list So for the first day of Q2, 1993, I'd like to sum from 1:end of sales_day_list to get 64. On day 2, I'd like to sum from 2:end of sales_day_list etc.
Here is an example of my data (the output from dput)
my_data= structure(list(DW_DATE_ID = structure(c(733622400, 733708800,
733795200, 733881600, 733968000, 734054400), class = c("POSIXct",
"POSIXt"), tzone = "UTC"), day_num = 1:6, sales_day_list = list(
c(1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1,
1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0,
1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1,
0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1,
1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0), c(1, 0, 0,
1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1,
0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1,
1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1,
1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0,
0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0), c(1, 0, 0, 1, 1, 1,
1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1,
1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0,
0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1,
1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1,
1, 1, 1, 0, 0, 1, 1, 1, 0), c(1, 0, 0, 1, 1, 1, 1, 1, 0,
0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1,
1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1,
1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0,
1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1,
0, 0, 1, 1, 1, 0), c(1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1,
1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0,
1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1,
0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1,
1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1,
1, 1, 0), c(1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1,
0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1,
1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1,
1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0,
0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0))), row.names = c(NA,
-6L), class = c("tbl_df", "tbl", "data.frame"))

Upon inspection, the sales_day_list is the same for each row, which makes sense since the sales days shouldn't change throughout the quarter.
To solve this problem, get the day of the quarter and the sales days as vectors, then write a function to apply the sum across the sales days vector, subsetting it by the day of the quarter.
x <- my_data$day_num # days as vector
y <- my_data$sales_day_list[[1]] # can be any one of them
sapply(x, function(x){sum(y[x:length(y)])}) # desired output
You can assign the output of the last line back into your data frame.

First we can make a table of just the sales days, and count how many left in the qtr.
sales_days <- my_data[1,3] %>% # Grab row 1, column 3
tidyr::unnest() %>%
rename(sales_day = sales_day_list) %>%
mutate(day_num = row_number()) %>%
arrange(-day_num) %>%
mutate(remaining_sales = cumsum(sales_day)) %>%
arrange(day_num)
Then we can attach this to my_data:
my_data2 <- my_data[,1:2] %>%
left_join(sales_days)
head(my_data2)
# A tibble: 6 x 4
DW_DATE_ID day_num sales_day remaining_sales
<dttm> <int> <dbl> <dbl>
1 1993-04-01 00:00:00 1 1 64
2 1993-04-02 00:00:00 2 0 63
3 1993-04-03 00:00:00 3 0 63
4 1993-04-04 00:00:00 4 1 63
5 1993-04-05 00:00:00 5 1 62
6 1993-04-06 00:00:00 6 1 61

venn diagram with categorical data

I have three categorical vectors that represent symptoms. And I would like plot a venn diagram that show how many people have one two or three of them.
I tryed do
library(gplots)
venn(list(sym1, sym2, sym3))
but didn't work
Thank you
sym1=c(0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0,
0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1,
0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0,
0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1,
0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1,
0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0,
1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1,
0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0,
1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0,
1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0,
1, 0, 0, 1, 1, 0)
sym2=c(0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1,
1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1,
0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1,
0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1,
0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1,
1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1,
1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1,
1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1,
1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0,
0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1,
1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0,
1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0,
0, 0, 1, 0, 1, 0)
sym3=c(0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1,
0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0,
0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1,
0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1,
0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1,
0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1,
0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1,
1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1,
1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0,
0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0,
1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0,
0, 0, 0, 0, 1, 0)

Here is an example on how to achieve this with the library eulerr which provides much better looking (at least in my opinion) diagrams:
library(eulerr)
library(tidyverse)
set.seed(123) #for reproducible plot
data.frame(sym1, sym2, sym3)%>% #combine the vectors to a data frame
mutate_at(1:3, as.logical) %>% #convert to logical
euler(shape = "ellipse", input = "disjoint") %>% #calculate euler object, plot as ellipse
plot(quantities = T) plot it
with venn from gplots:
library(gplots)
data.frame(sym1, sym2, sym3)%>%
mutate_at(1:3, as.logical) %>%
venn()
From the help of venn:
Either a list list containing vectors of names or indices of group
intersections, or a data frame containing boolean indicators of group
intersectionship (see below)
In your case I trust the second options is desired.

R - Check different matrices with a possible lag

This issue is quite tricky to explain but I am sure some of you already faced it.
So I have two matrix.
Matrix 1 (mat 1) and
Matrix 2 (mat 2)
What I want to do is to record in a third matrix (mat3) the value of mat2, after checking for matrix 1, but with a LAG. Let me explain.
After the value 1 in matrix 1, I want to check if matrix 2 as a 1 too but within the range of a certain LAG, for example, 1 or 2 episodes after (column).
For example, row number 4 has a 1 in matrix 1 at the 6th column.
So I want to check if in matrix 2 for row number 4 it has a 1 directly after or after 2 or 3 more columns.
Do you understand the idea ?
mat1 = structure(c(0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0,
0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0,
0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0,
1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1,
0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0,
0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1), .Dim = c(10L, 21L), .Dimnames = list(NULL, c("wit5.020",
"wit5.021", "wit5.022", "wit5.023", "wit5.024", "wit5.025", "wit5.026",
"wit5.027", "wit5.028", "wit5.029", "wit5.030", "wit5.031", "wit5.032",
"wit5.033", "wit5.034", "wit5.035", "wit5.036", "wit5.037", "wit5.038",
"wit5.039", "wit5.040")))
mat2 = structure(c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0,
0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0,
0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0,
0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0,
0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0,
1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1,
0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0,
1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1,
0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0,
0, 1, 0, 1), .Dim = c(10L, 21L))
So mat3 - where I want to store the result of the check
mat3 = matrix(0, nrow = nrow(mat1), ncol = ncol(mat1))
So here is an example of a possible loop
in order to check the LAG - this loop doesn't work but it could give you an idea maybe of the solution.
I am not sure where to introduce the lag. I thought maybe in the i, but I am not sure.
for(j in 1:ncol(mat1)){
for(i in 1:nrow(mat1)){
if( mat1[i,j] == 1 & mat2[i,j] == 1 | mat2[i+1,j] == 1 | mat2[i+2,j] == 1) # lag here
{mat[i,j] <- 1}
else
{mat[i,j] <- 0}
}
}
Any ideas are very welcome.

Here's a simple way to do it:
lag <- 3 # or whatever lag you want
nr <- nrow(mat1)
nc <- ncol(mat1)
mat3 <- matrix(0, ncol=nc, nrow=nr)
for (r in 1:nr) {
for (c in 1:nc) {
if (mat1[r,c] == 1 && any(mat2[r,c:min(c+lag,nc)] == 1))
mat3[r,c] <- 1
}
}
Note the use of mat2[r,c:min(c+lag,nc)]. This selects all elements from current column c up through column c + lag, but it makes sure not to go past nc (the total number of columns). That is, this code is used to avoid an out-of-bounds error.
There's probably a faster, more vectory way of doing this, but the above code should work.

weight data with R Part II

Given is the following data frame:
structure(list(UH6401 = c(1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1,
1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0,
0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0,
1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0,
1, 0, 1, 1), UH6402 = c(1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1,
0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0,
1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0,
0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1,
1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1,
0, 1, 1), UH6403 = c(1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0,
1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0,
1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1,
1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1,
0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0,
1, 1), UH6404 = c(0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1,
0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1,
1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1,
1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0,
0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1,
1), UH6409 = c(1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0,
1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0,
1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0,
1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0
), UH6410 = c(1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0,
1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1,
1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1,
1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0,
0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0
), UH6411 = c(0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0,
1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1,
0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1,
1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1
), UH6412 = c(1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0,
1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1
), UH6503 = c(1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0,
1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0,
1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1,
1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1
), UH66 = c(1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1),
UH68 = c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0), UH6501a = c(1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), UH6405a = c(1,
0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0,
0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0,
0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1,
1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0,
1, 0, 1, 1), UH6407a = c(1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1,
1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0,
1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1), weight = c(405.002592353822,
479.360356183825, 526.548105855472, 810.005184707644, 312.321528531308,
930.961115757095, 567.383058387095, 475.323944260643, 1226.91439266118,
517.086839792615, 1200.2669656949, 810.005184707644, 656.723784884795,
605.370463928298, 668.467435759576, 558.112457492436, 793.751055244424,
479.360356183825, 1226.91439266118, 1606.54816212786, 1657.48609449633,
300.803580980276, 605.370463928298, 1140.55078447979, 669.102760422943,
810.005184707644, 1657.48609449633, 305.569853371963, 2994.30343152033,
762.922030382216, 479.360356183825, 1147.36030437824, 668.467435759576,
517.086839792615, 479.360356183825, 399.141865860217, 656.723784884795,
913.364738988386, 312.321528531308, 569.10576379231, 775.630259688922,
1207.22952429547, 1053.09621171094, 1140.55078447979, 314.857225320909,
668.467435759576, 2416.57081451012, 573.680152189121, 396.875527622212,
605.370463928298, 1036.3159447043, 3088.62283807823, 569.10576379231,
1140.55078447979, 2416.57081451012, 1147.36030437824, 762.922030382216,
702.064141140629, 351.032070570315, 629.714450641817, 517.086839792615,
1996.20228768022, 828.743047248167, 475.323944260643, 920.185794495882,
793.751055244424, 796.08788273764, 1197.42559758065, 405.002592353822,
418.584343119327, 300.803580980276, 654.76828203733, 2740.09421696516,
351.032070570315, 1069.6202614693, 2094.91447516374, 399.141865860217,
654.76828203733, 1003.65414063441, 573.680152189121, 851.074587580641,
913.364738988386, 762.922030382216, 1034.17367958523, 573.680152189121,
479.360356183825, 3208.8607844079, 654.76828203733, 908.055695892447,
328.361892442398, 1036.3159447043, 702.064141140629, 613.457196330588,
601.607161960551, 567.383058387095, 479.360356183825, 306.261087672466,
920.185794495882, 654.76828203733, 828.743047248167)), .Names = c("UH6401",
"UH6402", "UH6403", "UH6404", "UH6409", "UH6410", "UH6411", "UH6412",
"UH6503", "UH66", "UH68", "UH6501a", "UH6405a", "UH6407a", "weight"
), row.names = c(NA, 100L), class = "data.frame")
In social science we often have a weight variable to weight a case (row) by the factor of that variable to correct the sample to fit e.g. the population by age classes. If the weight variable of a row is "1.6" it means that this row need do be observed 1.6 times to fit the basis population.
In SPSS I would write
WEIGHT BY weight.
and all procedures after that command will weight the data accordingly.
In R I can do that with stabs with the command
xtabs(weight ~ UH6401, data=df)
But what if I want to do a SVD or PCA analysis? Here there is no function to weight data like it is in xtabs.
So the question is, is there a method to weight data in R like it is possible in SPSS?
The point with whole numbers would be easy, with the factor "2" we would just double the line, but what is with all the factors that are decimal?
UPDATE:
The SVD or PCA was just an example! Take any other statistical procedure.
In social science the samples are never perfect, but to do an statistical analysis with sample data, the sample needs to represent the basic population, but a sample mostly doesn't. So we try to fix that deficit with weights, so the sample represent the basic population!

First of all, doing PCA on this data doesn't make sense. Second, SPSS does not perform PCA but factor analysis, which is something else. I know they call it PCA, but it isn't.
The WEIGHT BY in SPSS is nothing more than a replication weight, and is exactly the same as doing your analysis by repeating your cases using rep(): complete madness. To link to your example: In SPSS, FACTOR (which is used for the socalled PCA) does not take fractional weights.
If you want to perform weighted procedures, the only sensible way of doing that is using the correct method/function/package for that. In statistics, there is no one-size-fits-all weight procedure, contrary to what SPSS likes to make you believe.
In your example : weighted PCA in R is contained in FactoMineR and aroma.light. But I strongly suggest you take also a look at the vegan package, as that contains a lot more useful ordination methods for the data you're describing.

You probably need to get acquainted with the search engines for R. Baron's RSiteSearch and Rseek:
This is one of the first hits on "weighted PCA" at Baron's site:
http://finzi.psych.upenn.edu/R/library/aroma.light/html/wpca.matrix.html
With the clarification in the comment to Joris Meys response, the answer is often that one needs to be clear that one is desires sample weights versus other types of weighting. Regression weighting is done with the survey package. Lumley's book on survey methods distinguishes among three types of weights. (The "weights" in the lm function are variance weights, NOT sample weights.)
Note: Both PCA and factor analysis (experimental) are included in the survey package. So maybe Dominick's question requestiong a unified approach to weighting in regression methods has a single "answer".

I am not sure if this would suite you. See the R package weights.

I have just found a Post in R-Bloggers which introduces a svydesign() function. As far as I know, this function from the 'survey' package is like SPSS function, allowing you to create a weighted data to use in further analysis. I find it more useful than using different functions from several packages in order to do multivariable analysis.
Note to #djhurio: The answer would have been better with code. It does seem a bit duplicative of my answer which pointed to the survey package that contains 'svydesign'. The cited webpage is still there 4 years later, but that might not always be the case.

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