R 字符串变量中的伪变量
我想从该数据集中创建虚拟变量：R 字符串变量中的伪变量,r,R,我想从该数据集中创建虚拟变量： DF<-structure(list(A = c(1, 2, 3, 4, 5), B = c("1,3,2", "2,1,3,6", "3,2,5,1,7", "3,7,4,2,6,5", "4,10,7,3,5,6")), .Names = c("A", "B"), row.names = c(NA, 5L), class = "data.frame") > DF A B 1
DF<-structure(list(A = c(1, 2, 3, 4, 5), B = c("1,3,2", "2,1,3,6",
"3,2,5,1,7", "3,7,4,2,6,5", "4,10,7,3,5,6")), .Names = c("A", "B"),
row.names = c(NA, 5L), class = "data.frame")
> DF
A B
1 1 1,3,2
2 2 2,1,3,6
3 3 3,2,5,1,7
4 4 3,7,4,2,6,5
5 5 4,10,7,3,5,6
strsplitifelsedummie提前感谢您的帮助。使用
ifelsestrsplitcols <- 1:max( as.numeric( unlist(strsplit(DF$B,","))))
df <- t(apply(DF["B"] , 1 , function(x) ifelse( cols %in% as.numeric( unlist( strsplit(x , ",") ) ) , 1 , 0 ) ) )
colnames(df) <- cols
df
# 1 2 3 4 5 6 7 8 9 10
#1 1 1 1 0 0 0 0 0 0 0
#2 1 1 1 0 0 1 0 0 0 0
#3 1 1 1 0 1 0 1 0 0 0
#4 0 1 1 1 1 1 1 0 0 0
#5 0 0 1 1 1 1 1 0 0 1
cols我知道已经有了一个很好且相当有效的答案，但我们也可以使用另一种方法来获得相同的结果
tmp <- strsplit(DF$B, ",")
label <- 1:max(as.numeric(unlist(tmp)))
tmp <- lapply(tmp, function(x)
sapply(label, function(y) (x == y)))
t(sapply(tmp, colSums))
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
## [1,] 1 1 1 0 0 0 0 0 0 0
## [2,] 1 1 1 0 0 1 0 0 0 0
## [3,] 1 1 1 0 1 0 1 0 0 0
## [4,] 0 1 1 1 1 1 1 0 0 0
## [5,] 0 0 1 1 1 1 1 0 0 1
tmp更新：
在下面添加了基准
Update2：为@Anada的解决方案添加了bechmarks。哇，太快了
增加了evern更大数据集的基准测试，@Anada的解决方案以更大的优势领先
DT <- data.table(DF)
DT.long <- DT[, list(vals=as.numeric(unlist(strsplit(B, ",")))), by=A]
cols <- DT.long[, max(vals)]
rows <- DT.long[, max(A)]
matrix(as.numeric(DT.long[, (1:cols) %in% vals, by=A]$V1), ncol=cols,
byrow=TRUE, dimnames=list(seq(rows), seq(cols)))
# 1 2 3 4 5 6 7 8 9 10
# 1 1 1 1 0 0 0 0 0 0 0
# 2 1 1 1 0 0 1 0 0 0 0
# 3 1 1 1 0 1 0 1 0 0 0
# 4 0 1 1 1 1 1 1 0 0 0
# 5 0 0 1 1 1 1 1 0 0 1
原始答复：
如下所示，KnownMax
和UnknownMax
甚至优于data.table
解决方案。尽管如此，我怀疑如果有10e6+行，那么data.table
解决方案将是最快的。（只需修改本文底部的参数，就可以对其进行基准测试）
解决方案1:KnownMax
如果你知道B中的最大值，那么你就有了一个很好的两行：
maximum <- 10
results <- t(sapply(strsplit(DF$B, ","), `%in%`, x=1:maximum)) + 0
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,] 1 1 1 0 0 0 0 0 0 0
# [2,] 1 1 1 0 0 1 0 0 0 0
# [3,] 1 1 1 0 1 0 1 0 0 0
# [4,] 0 1 1 1 1 1 1 0 0 0
# [5,] 0 0 1 1 1 1 1 0 0 1
解决方案3:DT
根据@dickoa的请求，这里有一个带有data.table
的选项
DT <- data.table(DF)
DT.long <- DT[, list(vals=as.numeric(unlist(strsplit(B, ",")))), by=A]
cols <- DT.long[, max(vals)]
rows <- DT.long[, max(A)]
matrix(as.numeric(DT.long[, (1:cols) %in% vals, by=A]$V1), ncol=cols,
byrow=TRUE, dimnames=list(seq(rows), seq(cols)))
# 1 2 3 4 5 6 7 8 9 10
# 1 1 1 1 0 0 0 0 0 0 0
# 2 1 1 1 0 0 1 0 0 0 0
# 3 1 1 1 0 1 0 1 0 0 0
# 4 0 1 1 1 1 1 1 0 0 0
# 5 0 0 1 1 1 1 1 0 0 1
使用OP data.frame，其中结果为5 x 10
Unit: microseconds
expr min lq median uq max neval
KnownMax 106.556 114.692 122.4915 129.406 6427.521 50
UnknownMax 114.470 122.561 128.9780 136.384 158.346 50
DT.withAssign 3000.777 3099.729 3198.8175 3291.284 10415.315 50
DT.withOutAssign 2637.023 2739.930 2814.0585 2903.904 9376.747 50
lapply.Dickoa 7031.791 7315.781 7438.6835 7634.647 14314.687 50
apply.SimonO101 430.350 465.074 487.9505 522.938 7568.442 50
forLoop.Ananda 81.415 91.027 99.7530 104.588 265.394 50
使用稍大的data.frame（下图），其中结果为1000 x 100
删除lappy.Dickoa
，因为我的编辑可能会减慢它的速度，并且它会崩溃
更大的集合，结果为10000 x 600
Unit: milliseconds
expr min lq median uq max neval
KnownMax 1583.5902 1631.6214 1658.6168 1724.9557 1902.3923 50
UnknownMax 1597.1215 1655.9634 1690.7550 1735.5913 1804.2156 50
DT.withAssign 586.4675 641.7206 660.7330 716.0100 1193.4806 50
DT.withOutAssign 587.0492 628.3731 666.3148 717.5575 776.2671 50
apply.SimonO101 1916.6589 1995.2851 2044.9553 2079.6754 2385.1028 50
forLoop.Ananda 163.4549 172.5627 182.6207 211.9153 315.0706 50
使用以下命令：
library(microbmenchmark)
library(data.table)
KnownMax <- quote(t(sapply(strsplit(DF$B, ","), `%in%`, x=1:maximum)) + 0)
UnknownMax <- quote({ splat <- strsplit(DF$B, ","); maximum <- max(as.numeric(unlist(splat))); t(sapply(splat, `%in%`, x=1:maximum)) + 0})
DT.withAssign <- quote({DT <- data.table(DF); DT.long <- DT[, list(vals=as.numeric(unlist(strsplit(B, ",")))), by=A]; cols <- DT.long[, max(vals)]; rows <- DT.long[, max(A)] ; matrix(as.numeric(DT.long[, (1:cols) %in% vals, by=A]$V1), ncol=cols, byrow=TRUE, dimnames=list(seq(rows), seq(cols)))})
DT.withOutAssign <- quote({DT.long <- DT[, list(vals=as.numeric(unlist(strsplit(B, ",")))), by=A]; cols <- DT.long[, max(vals)]; rows <- DT.long[, max(A)] ; matrix(as.numeric(DT.long[, (1:cols) %in% vals, by=A]$V1), ncol=cols, byrow=TRUE, dimnames=list(seq(rows), seq(cols)))})
lapply.Dickoa <- quote({ tmp <- strsplit(DF$B, ","); label <- 1:max(as.numeric(unlist(tmp))); tmp <- lapply(tmp, function(x) as.data.frame(lapply(label, function(y) (x == y)))); unname(t(sapply(tmp, colSums))) })
apply.SimonO101 <- quote({cols <- 1:max( as.numeric( unlist(strsplit(DF$B,",")))); t(apply(DF["B"] , 1 , function(x) ifelse( cols %in% as.numeric( unlist( strsplit(x , ",") ) ) , 1 , 0 ) ) ) })
forLoop.Ananda <- quote({b = strsplit(DF$B, ","); ncol = max(as.numeric(unlist(b))); temp = lapply(b, as.numeric); m = matrix(0, nrow = nrow(DF), ncol = ncol) ; for (i in 1:nrow(DF)) { m[i, temp[[i]]] = 1 }; m })
# slightly modified @Dickoa's alogrithm to allow for instances were B is only a single number.
# Instead of using `sapply(.)`, I used `as.data.frame(lapply(.))` which hopefully the simplification process in sapply is analogous in time to `as.data.frame`
identical(eval(lapply.Dickoa), eval(UnknownMax))
identical(eval(lapply.Dickoa), unname(eval(apply.SimonO101)))
identical(eval(lapply.Dickoa), eval(KnownMax))
identical(unname(as.matrix(eval(DT.withAssign))), eval(KnownMax))
# ALL TRUE
库（microbmenchmark）
库（数据表）
KnownMax更新
这里提到的函数现在已经移动到了CRAN上的一个名为“splitstackshape”的包中。CRAN上的版本比这个原始版本快得多。速度应该与您在回答结束时使用直接for
循环解决方案得到的速度相似。有关详细的基准测试，请参见@Ricardo的答案
安装它，并使用concat.split.expanded
获得所需的结果：
library(splitstackshape)
concat.split.expanded(DF, "B", fill = 0, drop = TRUE)
# A B_01 B_02 B_03 B_04 B_05 B_06 B_07 B_08 B_09 B_10
# 1 1 1 1 1 0 0 0 0 0 0 0
# 2 2 1 1 1 0 0 1 0 0 0 0
# 3 3 1 1 1 0 1 0 1 0 0 0
# 4 4 0 1 1 1 1 1 1 0 0 0
# 5 5 0 0 1 1 1 1 1 0 0 1
原创帖子
不久前，我编写了一个函数，不仅可以进行这种拆分，还可以进行其他拆分。可以找到名为concat.split（）
的函数
对于您的示例数据，用法如下：
## Keeping the original column
concat.split(DF, "B", structure="expanded")
# A B B_1 B_2 B_3 B_4 B_5 B_6 B_7 B_8 B_9 B_10
# 1 1 1,3,2 1 1 1 NA NA NA NA NA NA NA
# 2 2 2,1,3,6 1 1 1 NA NA 1 NA NA NA NA
# 3 3 3,2,5,1,7 1 1 1 NA 1 NA 1 NA NA NA
# 4 4 3,7,4,2,6,5 NA 1 1 1 1 1 1 NA NA NA
# 5 5 4,10,7,3,5,6 NA NA 1 1 1 1 1 NA NA 1
## Dropping the original column
concat.split(DF, "B", structure="expanded", drop.col=TRUE)
# A B_1 B_2 B_3 B_4 B_5 B_6 B_7 B_8 B_9 B_10
# 1 1 1 1 1 NA NA NA NA NA NA NA
# 2 2 1 1 1 NA NA 1 NA NA NA NA
# 3 3 1 1 1 NA 1 NA 1 NA NA NA
# 4 4 NA 1 1 1 1 1 1 NA NA NA
# 5 5 NA NA 1 1 1 1 1 NA NA 1
将NA重新编码为0必须手动完成——也许我会更新该函数以添加一个选项来执行此操作，同时实现以下更快的解决方案之一：）
游戏进行得有点晚了，但另一种策略使用了这样一个事实，即一个矩阵可以由另一个两列矩阵索引，指定要更新的行和列索引。所以
f2 <- function(DF) {
b <- strsplit(DF$B, ",", fixed=TRUE)
len <- vapply(b, length, integer(1)) # 'geometry'
b <- as.integer(unlist(b))
midx <- matrix(c(rep(seq_len(nrow(DF)), len), b), ncol=2)
m <- matrix(0L, nrow(DF), max(b))
m[midx] <- 1L
m
}
这可以通过使用fixed=TRUE
和避免b
的双重强制来提高效率，并通过强制为整数和使用seq_len（nrow（DF））
来避免0行DF的拐角情况，使其更加健壮
f1 <- function(DF) {
b = lapply(strsplit(DF$B, ",", fixed=TRUE), as.integer)
ncol = max(unlist(b))
m = matrix(0L, nrow = nrow(DF), ncol = ncol)
for (i in seq_len(nrow(DF)))
m[i, b[[i]]] = 1L
m
}
从f0到f1的2倍增长和for循环的相对效率对我来说都相对令人惊讶@AnandaMahto的解决方案具有更高的内存效率，在不增加太多性能成本的情况下提高了内存效率
ncol = max(vapply(b, max, integer(1)))
好的，这困扰了我一段时间，但我认为这将是一个很好的使用Rcpp。所以我也写了一个小函数，看看是否能比@Ananda的
循环解决方案更快地得到一些东西。这个解决方案的运行速度似乎是原来的两倍（使用@RicardoSaporta发布的更大样本数据集）
注：我试图用更多的方法来教我如何使用Rcpp和C++，而不是提供一个有用的解决方案，但是所有的方法都是一样的……
我们的.cpp
文件
#include <Rcpp.h>
#include <string>
#include <sstream>
using namespace Rcpp;
//[[Rcpp::export]]
NumericMatrix expandR(CharacterVector x) {
int n = x.size();
std::vector< std::vector<int> > out; // list to hold numeric vectors
int tmax = 0;
for(int i = 0; i < n; ++i) {
std::vector<int> vect; // vector to hold split strings
std::string str = as<std::string>(x[i]);
std::stringstream ss(str);
int j = 0;
while (ss >> j) {
vect.push_back(j); // add integer to result vector
if (ss.peek() == ',') //split by ',' delim
ss.ignore();
}
int it = *std::max_element(vect.begin(), vect.end());
if( it > tmax )
tmax = it; //current max value
out.push_back(vect);
}
// Now we construct the matrix. tmax gives us number of columns, n is number of rows;
NumericMatrix mat(n,tmax);
for( int i = 0; i < n; ++i) {
NumericMatrix::Row zzrow = mat( i , _ );
std::vector<int> vec = out[i];
for( int j = 0; j < vec.size(); ++j ) {
zzrow[ (vec[j]-1) ] = 1; //don't forget R vs. C++ indexing
}
}
return mat;
}
使用@Ricardo提供的更大数据集，并与@Ananda的解决方案进行比较）
require（Rcpp）
要求（数据表）
种子（1）
最大不是一个特别快的解决方案，但是，对于那些喜欢tidyverse
可能性的人来说，它可能很有用：
DF %>%
mutate(B = str_split(B, fixed(","))) %>%
unnest() %>%
transmute(A,
var = as.numeric(B),
val = 1) %>%
complete(var = seq(min(var), max(var), 1), nesting(A)) %>%
spread(var, val, fill = 0)
A `1` `2` `3` `4` `5` `6` `7` `8` `9` `10`
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 1 1 0 0 0 0 0 0 0
2 2 1 1 1 0 0 1 0 0 0 0
3 3 1 1 1 0 1 0 1 0 0 0
4 4 0 1 1 1 1 1 1 0 0 0
5 5 0 0 1 1 1 1 1 0 0 1
DF%>%
变异（B=str_split（B，固定（“，”））%>%
unest（）%>%
蜕变，
var=作为数值（B），
val=1）%>%
完成（var=seq（最小（var），最大（var），1），嵌套（A））%>%
排列（var、val、fill=0）
A'1``2``3``4``5``6``7``8``9``10`
1 1 1 1 1 0 0 0 0 0 0 0
2 2 1 1 1 0 0 1 0 0 0 0
3 3 1 1 1 0 1 0 1 0 0 0
4 4 0 1 1 1 1 1 1 0 0 0
5 5 0 0 1 1 1 1 1 0 0 1
要使用更紧凑的列名，请执行以下操作：
DF %>%
mutate(B = str_split(B, fixed(","))) %>%
unnest() %>%
transmute(A,
var = as.numeric(B),
val = 1) %>%
complete(var = seq(min(var), max(var), 1), nesting(A)) %>%
spread(var, val, fill = 0) %>%
rename_at(2:length(.), ~ paste0("Col", 1:length(.)))
A Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Col9 Col10
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 1 1 0 0 0 0 0 0 0
2 2 1 1 1 0 0 1 0 0 0 0
3 3 1 1 1 0 1 0 1 0 0 0
4 4 0 1 1 1 1 1 1 0 0 0
5 5 0 0 1 1 1 1 1 0 0 1
DF%>%
变异（B=str_split（B，固定（“，”））%>%
unest（）%>%
蜕变，
var=作为数值（B），
val=1）%>%
完成（var=seq（最小（var），最大（var），1），嵌套（A））%>%
价差（var、val、fill=0）%>%
将_重命名为（2:length（.），~paste0（“Col”，1:length（.））
A列1列2列3列4列5列6列7列8列9列10
1 1 1 1 1 0 0 0 0 0 0 0
2 2 1 1 1 0 0 1 0 0 0 0
3 3 1 1 1 0 1 0 1 0 0 0
4 4 0 1 1 1 1 1 1 0 0 0
5 5 0 0 1 1 1 1 1 0 0 1
我正在寻找最有效的方法，所以if意味着我应该使用ifelse或strsplit
## Keeping the original column
concat.split(DF, "B", structure="expanded")
# A B B_1 B_2 B_3 B_4 B_5 B_6 B_7 B_8 B_9 B_10
# 1 1 1,3,2 1 1 1 NA NA NA NA NA NA NA
# 2 2 2,1,3,6 1 1 1 NA NA 1 NA NA NA NA
# 3 3 3,2,5,1,7 1 1 1 NA 1 NA 1 NA NA NA
# 4 4 3,7,4,2,6,5 NA 1 1 1 1 1 1 NA NA NA
# 5 5 4,10,7,3,5,6 NA NA 1 1 1 1 1 NA NA 1
## Dropping the original column
concat.split(DF, "B", structure="expanded", drop.col=TRUE)
# A B_1 B_2 B_3 B_4 B_5 B_6 B_7 B_8 B_9 B_10
# 1 1 1 1 1 NA NA NA NA NA NA NA
# 2 2 1 1 1 NA NA 1 NA NA NA NA
# 3 3 1 1 1 NA 1 NA 1 NA NA NA
# 4 4 NA 1 1 1 1 1 1 NA NA NA
# 5 5 NA NA 1 1 1 1 1 NA NA 1
temp <- concat.split(DF, "B", structure="expanded", drop.col=TRUE)
temp[is.na(temp)] <- 0
temp
# A B_1 B_2 B_3 B_4 B_5 B_6 B_7 B_8 B_9 B_10
# 1 1 1 1 1 0 0 0 0 0 0 0
# 2 2 1 1 1 0 0 1 0 0 0 0
# 3 3 1 1 1 0 1 0 1 0 0 0
# 4 4 0 1 1 1 1 1 1 0 0 0
# 5 5 0 0 1 1 1 1 1 0 0 1
b = strsplit(DF$B, ",")
ncol = max(as.numeric(unlist(b)))
temp = lapply(b, as.numeric)
## Set up an empty matrix
m = matrix(0, nrow = nrow(DF), ncol = ncol)
## Fill it in
for (i in 1:nrow(DF)) {
m[i, temp[[i]]] = 1
}
## View your result
m
f2 <- function(DF) {
b <- strsplit(DF$B, ",", fixed=TRUE)
len <- vapply(b, length, integer(1)) # 'geometry'
b <- as.integer(unlist(b))
midx <- matrix(c(rep(seq_len(nrow(DF)), len), b), ncol=2)
m <- matrix(0L, nrow(DF), max(b))
m[midx] <- 1L
m
}
f0 <- function(DF) {
b = strsplit(DF$B, ",")
ncol = max(as.numeric(unlist(b)))
temp = lapply(b, as.numeric)
m = matrix(0, nrow = nrow(DF), ncol = ncol)
for (i in 1:nrow(DF)) {
m[i, temp[[i]]] = 1
}
m
}
f1 <- function(DF) {
b = lapply(strsplit(DF$B, ",", fixed=TRUE), as.integer)
ncol = max(unlist(b))
m = matrix(0L, nrow = nrow(DF), ncol = ncol)
for (i in seq_len(nrow(DF)))
m[i, b[[i]]] = 1L
m
}
library(compiler)
f1c <- cmpfun(f1)
> library(microbenchmark)
> microbenchmark(f0(DF), f1(DF), f1c(DF), f2(DF))
Unit: milliseconds
expr min lq median uq max neval
f0(DF) 170.51388 180.25997 182.45772 188.23811 717.7511 100
f1(DF) 91.53578 97.14909 97.97195 100.24236 447.5900 100
f1c(DF) 79.39194 84.45712 85.71022 87.85763 411.8340 100
f2(DF) 76.45496 81.70307 82.50752 110.83620 398.6093 100
ncol = max(vapply(b, max, integer(1)))
#include <Rcpp.h>
#include <string>
#include <sstream>
using namespace Rcpp;
//[[Rcpp::export]]
NumericMatrix expandR(CharacterVector x) {
int n = x.size();
std::vector< std::vector<int> > out; // list to hold numeric vectors
int tmax = 0;
for(int i = 0; i < n; ++i) {
std::vector<int> vect; // vector to hold split strings
std::string str = as<std::string>(x[i]);
std::stringstream ss(str);
int j = 0;
while (ss >> j) {
vect.push_back(j); // add integer to result vector
if (ss.peek() == ',') //split by ',' delim
ss.ignore();
}
int it = *std::max_element(vect.begin(), vect.end());
if( it > tmax )
tmax = it; //current max value
out.push_back(vect);
}
// Now we construct the matrix. tmax gives us number of columns, n is number of rows;
NumericMatrix mat(n,tmax);
for( int i = 0; i < n; ++i) {
NumericMatrix::Row zzrow = mat( i , _ );
std::vector<int> vec = out[i];
for( int j = 0; j < vec.size(); ++j ) {
zzrow[ (vec[j]-1) ] = 1; //don't forget R vs. C++ indexing
}
}
return mat;
}
require(Rcpp)
## source the function so it is available to use in R
sourceCpp("C:/path/to/file.cpp")
# Call it like any other R function
expandR(DF$B)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 1 1 1 0 0 0 0 0 0 0
[2,] 1 1 1 0 0 1 0 0 0 0
[3,] 1 1 1 0 1 0 1 0 0 0
[4,] 0 1 1 1 1 1 1 0 0 0
[5,] 0 0 1 1 1 1 1 0 0 1
require(Rcpp)
require(data.table)
set.seed(1)
maximum <- 600
rows <- 10000
DF <- data.frame(A=seq(rows), B=as.character(c(maximum, replicate(rows-1, paste(sample(maximum, sample(20), FALSE), collapse=",")))), stringsAsFactors=FALSE)
DT <- data.table(DF);
DT
## source in our c code
sourceCpp("C:/Users/sohanlon/Desktop/expandR2.cpp")
forLoop.Ananda <- quote({b = strsplit(DF$B, ","); ncol = max(as.numeric(unlist(b))); temp = lapply(b, as.numeric); m = matrix(0, nrow = nrow(DF), ncol = ncol) ; for (i in 1:nrow(DF)) { m[i, temp[[i]]] = 1 }; m })
rcpp.Simon <- quote({mm = expandR( DT$B )})
require(microbenchmark)
microbenchmark( eval(forLoop.Ananda) , eval(rcpp.Simon) , times = 5L )
Unit: milliseconds
expr min lq median uq max neval
eval(forLoop.Ananda) 173.3024 178.6445 181.5881 218.9619 227.9490 5
eval(rcpp.Simon) 115.8309 116.3876 116.8125 119.1971 125.6504 5
DF %>%
mutate(B = str_split(B, fixed(","))) %>%
unnest() %>%
transmute(A,
var = as.numeric(B),
val = 1) %>%
complete(var = seq(min(var), max(var), 1), nesting(A)) %>%
spread(var, val, fill = 0)
A `1` `2` `3` `4` `5` `6` `7` `8` `9` `10`
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 1 1 0 0 0 0 0 0 0
2 2 1 1 1 0 0 1 0 0 0 0
3 3 1 1 1 0 1 0 1 0 0 0
4 4 0 1 1 1 1 1 1 0 0 0
5 5 0 0 1 1 1 1 1 0 0 1
DF %>%
mutate(B = str_split(B, fixed(","))) %>%
unnest() %>%
transmute(A,
var = as.numeric(B),
val = 1) %>%
complete(var = seq(min(var), max(var), 1), nesting(A)) %>%
spread(var, val, fill = 0) %>%
rename_at(2:length(.), ~ paste0("Col", 1:length(.)))
A Col1 Col2 Col3 Col4 Col5 Col6 Col7 Col8 Col9 Col10
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1 1 1 1 0 0 0 0 0 0 0
2 2 1 1 1 0 0 1 0 0 0 0
3 3 1 1 1 0 1 0 1 0 0 0
4 4 0 1 1 1 1 1 1 0 0 0
5 5 0 0 1 1 1 1 1 0 0 1