Conditional calculating the numbers of values in c

2019-02-26 08:52发布

I have two vectors:

x <- c(1,1,1,1,1, 2,2,2,3,3,  3,3,3,4,4,  5,5,5,5,5 )
y <- c(2,2,1,3,2, 1,4,2,2,NA, 3,3,3,4,NA, 1,4,4,2,NA)

This question (Conditional calculating the numbers of values in column with R, part2) discussed how to find the number of values in w (don't count NA) for each x (from 1–5) and for each y (from 1–4).

Let's split X by groups: if x<=2, group I; if 2<x<=3, group II; and if 3<X<=5, group III. I need to find the number of different values in x by groups and by every value of y. I also need to find the mean of those values in x by the same groups. The output should be in this format:

y x    Result 1 (the number of distinct numbers in X); Result 2 (the mean)
1 I     ...
1 II    ...
1 III   ...     
...
4 I     ...
4 II    ...
4 III   ...

标签: r aggregation
2条回答
Viruses.
2楼-- · 2019-02-26 08:58
#Bring in data.table library
require(data.table)
data <- data.table(x,y)

#Summarize data
data[, list(x = mean(x, na.rm=TRUE)), by = 
       list(y, x.grp = cut(x, c(-Inf,2,3,5,Inf)))][order(y,x.grp)]

If you'd like the results to be NA when NAs are present, then just remove na.rm=TRUE from mean(.):

data[, list(x = mean(x)), by = 
       list(y, x.grp = cut(x, c(-Inf,2,3,5,Inf)))][order(y,x.grp)]
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淡お忘
3楼-- · 2019-02-26 09:11

My command of R code isn't great, so here's A Rather Ugly Function:

ARUF=function(x,y){df1=data.frame(x,y,group=NA);miny=min(y,na.rm=T)
maxy=max(y,na.rm=T);for(i in 1:length(df1$x))df1$group[i]=if(df1$x[i]<=2)'I'else
if(df1$x[i]>2&df1$x[i]<=3)'II'else if(df1$x[i]>3&df1$x[i]<=5)'III'else'NA'
Result1=c();Result2=c();for(i in miny:maxy){for(j in c('I','II','III')){
Result1=append(Result1,length(levels(factor(subset(df1,y==i&group==j)$x))))
Result2=append(Result2,mean(subset(df1,y==i&group==j)$x))}}
print(data.frame(y=rep(miny:maxy,rep(3,maxy+abs(miny-1))),
x=rep(c('I','II','III'),maxy+abs(miny-1)),Result1,Result2),row.names=F)}

With your x and y, ARUF(x,y) prints this data.frame:

y   x Result1  Result2
1   I       2 1.500000
1  II       0      NaN
1 III       1 5.000000
2   I       2 1.250000
2  II       1 3.000000
2 III       1 5.000000
3   I       1 1.000000
3  II       1 3.000000
3 III       0      NaN
4   I       1 2.000000
4  II       0      NaN
4 III       2 4.666667

I went a little out of my way to make ARUF robust with any integer values of y. I can't seem to break it by generating y randomly with rbinom, and I believe it should handle any real number values of x, so it should work for any other vectors of the same kind that you might have.

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