我有一个示例数据框，如下所示：
CNC[1:9,]
ID CNDIST1 CNDIST2
C1 0,0,136,1 0,0,4,2
C2 0,0,141,1 0,0,4,1
C3 6,8,126 0,0,4
C4 6,11,125 0,0,6
C5 0,0,141,0,0,0,0 0,0,3,0,1,0,1
C6 0,0,139,0,0 0,0,3,1,1
C7 0,0,141,0 0,0,4,2
C8 0,0,141,0 0,0,4,2
C9 31,44,61 2,2,2
使用dput（）的相同数据框：
dput(CNC[1:9,])
structure(list(ID = structure(1:9, .Label = c("C1",
"C2", "C3", "C4",
"C5", "C6", "C7",
"C8", "C9"), class = "factor"),
CNDIST1 = structure(c(1L, 5L, 8L, 7L, 4L, 2L, 3L, 3L,
6L), .Label = c("0,0,136,1", "0,0,139,0,0", "0,0,141,0",
"0,0,141,0,0,0,0", "0,0,141,1", "31,44,61", "6,11,125", "6,8,126"
), class = "factor"), CNDIST2 = structure(c(5L, 4L, 3L,
6L, 1L, 2L, 5L, 5L, 7L), .Label = c("0,0,3,0,1,0,1", "0,0,3,1,1",
"0,0,4", "0,0,4,1", "0,0,4,2", "0,0,6", "2,2,2"), class = "factor")), .Names = c("ID",
"CNDIST1", "CNDIST2"), row.names = c(NA, 9L), class = "data.frame")
我正在使用下面的Rcode来做chisq.test。第3列中的值形成概率向量＆＃39; p＆＃39;相同长度的数字向量＆＃39; x＆＃39;来自第2栏
read.table("report.dat2",header=T,sep="\t")->CNC
chi.pval=vector()
for(i in 1:nrow(CNC)){
as.numeric(unlist(strsplit(as.character(CNC$CNDIST1[i]),",")))->x
as.numeric(unlist(strsplit(as.character(CNC$CNDIST2[i]),",")))->p
chi.pval[i]<-chisq.test(x,p+0.001,rescale.p=T)$p.value ###add 0.001 to 'p' vector to remove '0'
}
CNC1<-cbind(CNC,chi.pval)
write.table(CNC1,'chi.test.txt',sep='\t',quote=F,row.names=F)
代码返回错误：
Error in chisq.test(x, p + 0.001, rescale.p = T) :
'x' and 'y' must have at least 2 levels
In addition: Warning messages:
1: In chisq.test(x, p + 0.001, rescale.p = T) :
Chi-squared approximation may be incorrect
代码显示在某些行和退出时执行chisq.test时出错。但是它会对数据帧的某些行进行测试。有没有人提供线索来找出解决这个问题的方法？
dput（CNC [1：9，]））缺少DIST1和DIST2中的某些值。看起来好像发生了重复值的事情。