Question

我想为每个人获得因子分数。因素存储在数据框factors中我需要获取与data对应的值factors的另一个数据框中的值的平均值，并将其存储在{{{}}的新列中1}}。我为可怕的解释道歉。我希望我的榜样有所帮助，我很乐意回答问题！

data

我会尝试将其分解为步骤（据我理解的过程）：

在数据框

factors<-data.frame(c(NA,2,NA),c(NA,3,1))
colnames(factors)<-c("v1","v2")
row.names(factors)<-c("col1data","col2data","col3data")
factors

data<-data.frame(c(2,4,2),c(1,1,2),c(3,3,3))
colnames(data)<-c("col1data","col2data","col3data")
row.names(data)<-c("person1","person2","person3")
data
#in dataframe factors row col2data is present (i.e. not NA) under factor V1
#go into dataframe data for each person and make a new column called v1 that holds the value of col2data
#do this for factor v2 and average the values to come up with one number for each person. Final result
data<-data.frame(c(2,4,2),c(1,1,2),c(3,3,3),c(1,1,2),c(2,3,2.5))
colnames(data)<-c("col1data","col2data","col3data","v1","v2(avg col2 and col3)")
row.names(data)<-c("person1","person2","person3")
data

的列中查找非NA的行名称

将行名称与数据框factors列匹配。

在data中对匹配的行名称进行求和并存储在名为data中列的列名称的新变量中（例如data），每个人

< / p>

Answer 1

您可以将data的行方式限制为相应的列：

cbind(data, apply(factors, 2, function(x) rowMeans(data[,!is.na(x),drop=FALSE])))
#         col1data col2data col3data v1  v2
# person1        2        1        3  1 2.0
# person2        4        1        3  1 2.0
# person3        2        2        3  2 2.5

Answer 2

我按照您的方式说明了您如何理解流程作为代码中的注释，以查看流程中每个步骤的执行位置。

factors<-data.frame(c(NA,2,NA),c(NA,3,1))
colnames(factors)<-c("v1","v2")
row.names(factors)<-c("col1data","col2data","col3data")
factors

data<-data.frame(c(2,4,2),c(1,1,2),c(3,3,3))
colnames(data)<-c("col1data","col2data","col3data")
row.names(data)<-c("person1","person2","person3")
data

#find row names in a column of dataframe factors that are not NA
not_na_rows_v1 <- rownames(factors)[!is.na(factors$v1)]
not_na_rows_v2 <- rownames(factors)[!is.na(factors$v2)]
not_na_rows_v1
not_na_rows_v2
#match row names to dataframe data columns.
#Sum matching row names in data and store in new variable called the column name of the column in data (eg v1) for each person
###*note*### apply(...,1 ,mean) takes the mean for each row (the "1" means by row, "2" would mean by column)
data[, 'v1'] <- data[, not_na_rows_v1]
data[, 'v2'] <- apply(data[, not_na_rows_v2], 1, mean)
data

查找与另一个矩阵中的相同位置对应的值

2 个答案: