R 二进制矩阵的字符串列表
我想根据字符串列表创建一个二进制矩阵R 二进制矩阵的字符串列表,r,dplyr,data.table,R,Dplyr,Data.table,我想根据字符串列表创建一个二进制矩阵 dt = data.table(id = c('id1','id2','id3','id4','id5','id6'), sample = c("MER-1,MER-3,MER-4","MER-5","MER-2","MER-2,MER-3,MER-4,MER-5","MER_3","MER-5" )) dt id sample 1: id1 MER-1,MER-3,MER-4 2: id2
dt = data.table(id = c('id1','id2','id3','id4','id5','id6'), sample = c("MER-1,MER-3,MER-4","MER-5","MER-2","MER-2,MER-3,MER-4,MER-5","MER_3","MER-5" ))
dt
id sample
1: id1 MER-1,MER-3,MER-4
2: id2 MER-5
3: id3 MER-2
4: id4 MER-2,MER-3,MER-4,MER-5
5: id5 MER_3
6: id6 MER-5
结果应该是:
m_count = matrix(c(1,0,1,1,0, 0,0,0,0,1, 0,1,0,0,0, 0,1,1,1,1, 0,0,1,0,0, 0,0,0,0,1), nrow = 6, ncol = 5)
m_count
MER-1 MER-2 MER-3 MER-4 MER-5
id1 1 0 0 1 0
id2 0 0 0 1 0
id3 1 0 0 0 0
id4 1 1 0 0 0
id5 0 0 1 1 0
id6 0 1 1 0 1
我可以循环遍历列表中的每个元素,并填充矩阵,但考虑到表的大小,这将非常缓慢。有没有更快/更优雅的方式?也许用dplyr/tidyverse?
谢谢 使用
dt
从最后的注释中修复问题中的打字错误,使用分隔行
逐行展开数据,然后使用表
计算计数
library(data.table)
library(dplyr)
library(tidyr)
dt %>%
separate_rows(sample, sep = ",") %>%
table
给予:
sample
id MER-1 MER-2 MER-3 MER-4 MER-5
id1 1 0 1 1 0
id2 0 0 0 0 1
id3 0 1 0 0 0
id4 0 1 1 1 1
id5 0 0 1 0 0
id6 0 0 0 0 1
注
库(data.table)
dt您也可以使用库splitstackshape
:
table(cSplit(dt, "sample", sep = ",", direction = "long"))
sample
id MER-1 MER-2 MER-3 MER-4 MER-5
id1 1 0 1 1 0
id2 0 0 0 0 1
id3 0 1 0 0 0
id4 0 1 1 1 1
id5 0 0 1 0 0
id6 0 0 0 0 1
或者使用专门为此场景创建的cSplit_e
(由@A5C1D2H2I1M1N2O1R2T1提供):
您可以使用strsplit
:
table(dt[,unlist(strsplit(sample,",")),by=id])
V1
id MER-1 MER-2 MER-3 MER-4 MER-5 MER_3
id1 1 0 1 1 0 0
id2 0 0 0 0 1 0
id3 0 1 0 0 0 0
id4 0 1 1 1 1 0
id5 0 0 0 0 0 1
id6 0 0 0 0 1 0
为此,您应该使用cSplit_e
。@A5C1D2H2I1M1N2O1R2T1看起来是一个不错的选项,但是,我无法让它处理此数据cSplit_e(dt,“sample”,sep=“,”,fill=0)
抛出一个错误。在其中添加一个type=“character”
。@A5C1D2H2I1M1N2O1R2T1谢谢,这真是一个简洁的函数:)
cSplit_e(dt, "sample", sep = ",", type = "character", fill = 0, drop = TRUE)
table(dt[,unlist(strsplit(sample,",")),by=id])
V1
id MER-1 MER-2 MER-3 MER-4 MER-5 MER_3
id1 1 0 1 1 0 0
id2 0 0 0 0 1 0
id3 0 1 0 0 0 0
id4 0 1 1 1 1 0
id5 0 0 0 0 0 1
id6 0 0 0 0 1 0