我无法在网上找到这个问题的解决方案,就像看起来那么简单。这是: 我有一个像df1这样的数据框:
PIM WDR MYC OBX
ILMN_1651282 0.555764 0.675233 0.5908629 0.4897703
ILMN_1651354 0.6963458 0.8588675 0.9216328 0.88705
ILMN_1651358 0.7501548 0.6766059 0.8157319 0.9373666
ILMN_1652716 0.5505098 0.5802357 0.7342341 0.5953167
ILMN_1654324 0.9294231 0.9311051 0.8424824 0.888825
ILMN_1654639 0.9197155 0.4687101 0.678938 0.4309232
ILMN_1655418 0.690068 0.6345875 0.9595042 0.6132203
和像这样的数据框文件命名为df2:
PIM WDR MYC OBX
ILMN_1651282 -1 -1 -1 -1
ILMN_1651354 -1 1 1 1
ILMN_1651358 1 1 1 1
ILMN_1652716 -1 -1 -1 -1
ILMN_1654324 -1 -1 -1 -1
ILMN_1654639 -1 -1 -1 -1
ILMN_1655418 1 1 -1 1
我的截止值为0.8。在df1中,每个高于0.8的值都在0中变化。所有低于0.8的值必须用df2(1& -1)中的值替换
创建了df2:
PIMvsEV<-list()
for (x in 1:nrow(df1)) {
t<-(if (mean(PIM[,x]) > mean(EV[,x])) {print(1)} else
if (mean(PIM[,x]) < mean(EV[,x])) {print(-1)} )
PIMvsEV[[x]]<-matrix(t)
}
WDRvsEV<-list()
for (x in 1:nrow(df1)) {
t<-(if (mean(WDR[,x]) > mean(EV[,x])) {print(1)} else
if (mean(WDR[,x]) < mean(EV[,x])) {print(-1)} )
WDRvsEV[[x]]<-matrix(t)
}
OBXvsEV<-list()
for (x in 1:nrow(cdf1)) {
t<-(if (mean(OBX[,x]) > mean(EV[,x])) {print(1)} else
if (mean(OBX[,x]) < mean(EV[,x])) {print(-1)} )
OBXvsEV[[x]]<-matrix(t)
}
MYCvsEV<-list()
for (x in 1:nrow(df1)) {
t<-(if (mean(MYC[,x]) > mean(EV[,x])) {print(1)} else
if (mean(MYC[,x]) < mean(EV[,x])) {print(-1)} )
MYCvsEV[[x]]<-matrix(t)
}
dataframe<-as.data.frame(cbind(as.matrix(PIMvsEV), as.matrix(WDRvsEV)))
dataframe<-as.data.frame(cbind(as.matrix(dataframe), as.matrix(MYCvsEV)))
dataframe<-as.data.frame(cbind(as.matrix(dataframe), as.matrix(OBXvsEV)))
row.names(dataframe)<-colnames(ttest)
colnames(dataframe)<-c("PIM","WDR","MYC","OBX")
有什么想法?非常感谢,
Lisanne
答案 0 :(得分:1)
这应该做你想要的吗?
一些数据:
df1 <- as.data.frame(matrix(rnorm(100),10,10))
df2 <- matrix(sample(c(-1,1),100,T),10,10)
ifelse
的矢量化使用:
ifelse( as.matrix(df1) > 0.8, 0, df2)
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
[1,] 0 0 1 -1 1 1 -1 -1 1 -1
[2,] -1 1 1 -1 -1 -1 -1 -1 1 -1
[3,] -1 -1 -1 0 0 -1 1 0 1 -1
[4,] 0 -1 -1 1 -1 1 -1 1 1 1
[5,] -1 -1 -1 -1 0 1 -1 1 0 0
[6,] -1 1 1 1 -1 1 0 0 1 1
[7,] -1 1 1 0 -1 -1 -1 -1 -1 -1
[8,] 1 1 1 0 -1 -1 1 -1 0 -1
[9,] 1 0 0 1 1 0 -1 -1 0 -1
[10,] -1 -1 -1 -1 1 -1 -1 -1 -1 1
或者我们可以做到
(df1<= 0.8)*df2
答案 1 :(得分:1)
示例数据:
df1 <- as.data.frame(matrix(rnorm(100),10,10))
df2 <- as.data.frame(matrix(sample(c(-1,1),100,T),10,10))
data.frames代码
res <- df1
res[df1>0.8] <- 0
res[df1<=0.8] <- df2[df1<=0.8]