如何计算R中不同唯一ID的值出现的次数？

Question

I would like to calculate the number of times a zero and one occurs for each ID. I have a single column with >500 unique IDs and each ID has a different number of times zeroes and ones occur. Thanks!

I am doing this in R using for loops

Answer 1

I guess something like this could help you:

#Example dataframe
dummy=data.frame(ID=c(10101,11110101,11111))

#Separate every character in ID column
Sepdummy=strsplit(as.character(dummy$ID), split="")

#Count how many times a value is repeated
dummy$Zeroes=unlist(lapply(Sepdummy, function(x) sum(as.numeric(x)==0)))
dummy$Ones=unlist(lapply(Sepdummy, function(x) sum(as.numeric(x)==1)))

Output looks like:

ID  Zeroes  Ones
10101   2   3
11110101    2   6
11111   0   5

The above will not work if your ID's aren't numeric. For that you can just use str_count() from stringr package (as pointed elsewhere in this post):

library(stringr)
#Example dataframe
dummy=data.frame(ID=c(10101,11110101,11111,"asd0110001df"))

#Count using str_count and add the results to the original dummy dataframe, so the results are all viewed in the same table.
dummy$Zeroes=str_count(dummy$ID, "0")
dummy$Ones=str_count(dummy$ID, "1")

Answer 2

Within the library stringr , you have the function str_count() which enable to count the number of occurences of a character in a string.

library(stringr)

str_count("abracadabra", "a")         # return 5
str_count("0010110", "0")             # return 4
str_count("001d021", "0|1")           # return 5
str_count(c("001", "123", "salut"))   # return (3, 1, 0)

Answer 3

Alternative guess, maybe your data frame looks like this?

library(dplyr)
set.seed(1)
data.df <- data.frame(id=c(rep(1,10),rep(2,10)), value=rbinom(20,1,.5))

count.df <- data.df%>%group_by(id)%>%summarize(ones=sum(value==1),zeros=sum(value==0))%>%ungroup()%>%as.data.frame

Answer 4

With the vector you gave above:

Transform the vect to a viable dataframe:

data=data.frame(matrix(vect,,2,byrow=T))
with(data,table(ID,Treatment))
         Treatment
ID         0  1
  100a002 16  8
  100a003 18  6

data:

data=read.table(text=" ID Treatment 
                100a002 1 
                100a002 0 
                100a002 0
                100a002 0 
                100a002 1 
                100a002 1 
                100a002 1 
                100a002 0
                100a002 0
                100a002 0 
                100a002 0 
                100a002 0
                100a002 0 
                100a002 0
                100a002 0
                100a002 0
                100a002 0
                100a002 0
                100a002 0
                100a002 0
                100a002 1 
                100a002 1 
                100a002 1
                100a002 1
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 0
                100a003 1
                100a003 1
                100a003 1
                100a003 1 
                100a003 1
                100a003 1",h=T,stringsAsFactors=F)

Answer 5

Most Efficient way to do it

dummy<- data.frame(id=c(rep(1,10),rep(2,10)), value=rbinom(20,1,.5))
library(data.table)
setDT(dummy)[, list(count_of_one = length(which(value==1)),count_of_zeroes = length(which(value==0))), by = id]

output

   id      count_of_one   count_of_zeroes
1:  1            5               5
2:  2            6               4