Count combinations of two variables excluding rows that repeat ID - r

Question

I have a data on countries and want to summarize it and create a table.
> head(data)
country year score members
A 1989 0 7
A 1990 0 7
A 1991 0 7
A 1992 0 7
A 1993 0 7
A 1994 0 7
The table should show the relationship between country "score" and the number of "members" – put differently, I want to see how many states with score 0,1 or 2 have "members"(ranging from 1 to 7).
I want to set it like this:
score members==1 members==2 members==3 members==4 members==5 members==6 members==7
0 1 0
1 2 0
2 0 1 and so on..
To do this I run the following:
library(dplyr)
table <- data %>%
group_by(score) %>%
summarise(
m1 = sum(members==1, na.rm=TRUE),
m2 = sum(members==2, na.rm=TRUE),
m3 = sum(members==3, na.rm=TRUE),
m4 = sum(members==4, na.rm=TRUE),
m5 = sum(members==5, na.rm=TRUE),
m6 = sum(members==6, na.rm=TRUE),
m7 = sum(members==7, na.rm=TRUE)
)
This gives:
score m1 m2 m3 m4 m5 m6 m7
0 0 2 0 0 0 3 30
1 15 3 11 11 3 18 3
2 3 0 2 2 0 6 9
.
.
I need a little help here. As you see it has calculated the total number of observations, whereas I want to count each country only once.
How do I summarize this data to have the total number of countries for each members-level?
Here's a sample of my data for reproducibility:
data <-
structure(list(country = structure(c(1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L,
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L,
5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L), .Label = c("A", "B", "C", "D", "E", "F"), class = "factor"),
year = c(1989L, 1990L, 1991L, 1992L, 1993L, 1994L, 1995L,
1996L, 1997L, 1998L, 1999L, 2000L, 2001L, 2002L, 2003L, 2004L,
2005L, 2006L, 2007L, 2008L, 2010L, 1989L, 1990L, 1991L, 1992L,
1993L, 1994L, 1995L, 1996L, 1997L, 1998L, 1999L, 2000L, 2001L,
2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L, 2009L, 2010L,
2011L, 1989L, 1991L, 1993L, 1994L, 1995L, 1996L, 1997L, 1999L,
2000L, 2001L, 2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L,
2010L, 1989L, 1990L, 1991L, 1992L, 1993L, 1994L, 1995L, 1996L,
1997L, 1998L, 1999L, 2000L, 2001L, 2002L, 2003L, 2004L, 2005L,
2006L, 2007L, 2008L, 2009L, 2010L, 2011L, 1991L, 1992L, 1993L,
1994L, 1995L, 1997L, 1998L, 1999L, 2000L, 2001L, 2002L, 2003L,
2004L, 2005L, 2006L, 2007L, 2008L, 2010L, 1991L, 1992L, 1993L,
1994L, 1995L, 1997L, 1998L, 1999L, 2000L, 2001L, 2002L, 2003L,
2004L, 2005L, 2006L, 2007L, 2008L, 2010L), score = c(0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 1L, 0L, 1L, 1L, 0L, 1L, 1L, 0L, 1L, 1L,
1L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 2L, 2L,
2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L,
2L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L,
1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 2L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L
), members = c(7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L,
7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L,
7L, 7L, 7L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 7L,
7L, 7L, 7L, 7L, 7L, 7L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L,
4L, 4L, 4L, 4L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L)), .Names = c("country", "year", "score",
"members"), class = "data.frame", row.names = c(NA, -121L))

Answer 1

I believe you need this:
library(reshape2)
dcast(aggregate(country~score+members, data=data, FUN=function(x) length(unique(x))),
score~members, value.var="country", fill=0L)
# score 1 2 3 4 5 6 7
#1 0 0 1 0 0 0 1 2
#2 1 1 1 2 2 1 3 2
#3 2 1 0 1 2 0 1 1
Or, to put it the dplyr/tidyr way:
data %>%
group_by(members, score) %>%
summarise(n=n_distinct(country)) %>%
spread(members, n, fill=0L)
## A tibble: 3 x 8
# score 1 2 3 4 5 6 7
#* <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 0 0 1 0 0 0 1 2
#2 1 1 1 2 2 1 3 2
#3 2 1 0 1 2 0 1 1

Answer 2

As the OP is using dplyr methods, we can do this by grouping with 'score', 'members' to get the number of elements (n()), and then spread (from tidyr) to reshape it to 'wide' format.
library(dplyr)
library(tidyr)
data %>%
group_by(score, members) %>%
summarise(n = n()) %>%
mutate(members = paste0("m", members)) %>%
spread(members, n, fill = 0)
# score m1 m2 m3 m4 m5 m6 m7
# <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 0 0 2 0 0 0 3 30
#2 1 15 3 11 11 3 18 3
#3 2 3 0 2 2 0 6 9
If we need to also get the counts by 'country', just add 'country' in the group_by
data %>%
group_by(country, score, members) %>%
summarise(n = n()) %>%
mutate(members = paste0("m", members)) %>%
spread(members, n, fill = 0)
If the expected output is the one showed in the other posts, an option using data.table would be to convert the 'data.frame' to 'data.table' (setDT(data), and dcast from 'long' to 'wide' specifying the fun.aggregate as uniqueN of the 'value.var' variable i.e. 'country' where uniqueN returns the length of unique elements in the 'country' column. The fill=0 specifies to occupy 0 for those combinations that are not available. By default, it returns as NA.
library(data.table)
dcast(setDT(data), score~members, value.var= 'country', fun.aggregate = uniqueN, fill = 0)
# score 1 2 3 4 5 6 7
#1: 0 0 1 0 0 0 1 2
#2: 1 1 1 2 2 1 3 2
#3: 2 1 0 1 2 0 1 1

Answer 3

It seems the crux of the issue is having the duplicated rows for each year? In which case you can remove them with distinct, then it's a simple crosstab. You could use the %$% exposition pipe from magrittr:
library(dplyr)
library(magrittr)
data %>%
distinct(country, score, members) %$%
table(score, members)
members
score 1 2 3 4 5 6 7
0 0 1 0 0 0 1 2
1 1 1 2 2 1 3 2
2 1 0 1 2 0 1 1
Or a regular pipe and tabyl from the janitor package:
library(dplyr)
library(janitor)
data %>%
distinct(country, score, members) %>%
tabyl(score, members)
score 1 2 3 4 5 6 7
0 0 1 0 0 0 1 2
1 1 1 2 2 1 3 2
2 1 0 1 2 0 1 1