我想生成相关性，它可能是基本的，但我无法得到它。需要你的帮助!!
我正在尝试为用户指定的变量生成相关性（即需要生成相关性的变量不固定。在不同的场景中可能会有所不同，因此需要在向量中输入和存储 str_char ）
对于这些变量中的每一个，我需要生成与值变量的相关性，并且相关性应基于类型变量中的组。
以下是样本虚拟数据。我的实际数据有更多的列和行。
library("plyr")
library("data.table")
set.seed(1200)
id <- 1:100
bills <- sample(1:20,100,replace = T)
nos <- sample(1:80,100,replace = T)
stru <- sample(c("A","B","C","D"),100,replace = T)
var1 <- sample(1:80,100,replace = T)
var2 <- sample(1:80,100,replace = T)
v1 <- sample(1:80,100,replace = T)
v2 <- sample(1:80,100,replace = T)
a1 <- sample(1:80,100,replace = T)
b1 <- sample(1:80,100,replace = T)
type <- sample(1:7,100,replace = T)
value <- sample(100:1000,100,replace = T)
df1 <- as.data.table(data.frame(id,bills,nos,stru,var1,var2,v1,v2,a1,b1,type,value))
#storing the variables for which need to generate correlation. This would change in different scenarios and one would need to update this variable.
str_char <- c("bills","nos","stru","var2","v1","b1")
len <- length(str_char)
#Since the variables are not fixed using for loop. To tackle the requirement of generating correlation by group using ddply
corr<-data.frame()
for (i in 1:len){
df1$var1 <- df1[,which(colnames(df1) == str_char[i])]
var1 <- str_char[i]
temp1 <- ddply(
df1
, .(type)
, summarize
, var1=cor(value,var1,method="spearman")
)
corr <- as.data.frame(cbind(corr,temp1))
}
这会为 corr 生成一个空数据框。不知道我哪里错了。我希望将行中的类型和每个变量放在具有相关值的单元格的列中。
一旦我有了具有相关值的数据框，我想识别相关性为> 1的变量。 0.2并将它们存储在矢量中。
请你帮忙建议我出错的地方或建议一些更好的方法来满足这个要求。
谢谢！
答案 0 :(得分：2)
data.table没有＆＃34;复杂的技巧＆＃34;是必须的。可以使用by参数（而不是split()）和.SDcols参数来指定要在cor()调用中使用的列。所以，它非常直接data.table语法：
# without stru because it is factor not numeric
str_char <- c("bills", "nos", "var2", "v1", "b1")
df1[, lapply(.SD, function(x) cor(value, x, method = "spearman")),
keyby = type, .SDcol = str_char]
type bills nos var2 v1 b1 1: 1 -0.58026951 0.16493506 -0.07664827 0.11627152 -0.05595326 2: 2 0.02646100 0.22246750 0.40308468 0.38943918 -0.10121018 3: 3 -0.11389551 0.36446564 -0.16438528 0.00000000 -0.04100238 4: 4 -0.45645233 -0.21585955 -0.19560440 0.28351648 -0.08580863 5: 5 -0.18596606 -0.23776224 -0.06304738 -0.03508794 0.39860140 6: 6 -0.72346726 -0.04175824 0.24862501 -0.30583077 -0.31718139 7: 7 -0.02649032 -0.08810594 0.48398529 0.30143033 0.50165047
# with stru after coersion of factor to numeric
str_char <- c("bills", "nos", "stru", "var2", "v1", "b1")
result <- df1[, lapply(.SD, function(x) cor(value, as.numeric(x), method = "spearman")),
keyby = type, .SDcol = str_char]
result
type bills nos stru var2 v1 b1 1: 1 -0.58026951 0.16493506 0.08202645 -0.07664827 0.11627152 -0.05595326 2: 2 0.02646100 0.22246750 0.21968328 0.40308468 0.38943918 -0.10121018 3: 3 -0.11389551 0.36446564 -0.11769798 -0.16438528 0.00000000 -0.04100238 4: 4 -0.45645233 -0.21585955 -0.37551547 -0.19560440 0.28351648 -0.08580863 5: 5 -0.18596606 -0.23776224 0.39444627 -0.06304738 -0.03508794 0.39860140 6: 6 -0.72346726 -0.04175824 0.28585837 0.24862501 -0.30583077 -0.31718139 7: 7 -0.02649032 -0.08810594 -0.05718863 0.48398529 0.30143033 0.50165047
请注意，keyby代替by使结果与LAP's answer中的结果相同，以便进行比较。
此外，OP已要求在结果中附加一个新列，其中包含具有最高cor()值＆gt;的3个顶部变量的名称。每个type为0.2。
从宽格式到长格式重塑result之后，可以最方便地查找前3个值：
# reshape from wide to long
melt(result, id.vars = "type")[
# select by value
value > 0.2][
# order by descending value and pick the first 3 (if any)
order(-value), toString(head(variable, 3L)), keyby = type]
type V1 1: 2 var2, v1, nos 2: 3 nos 3: 4 v1 4: 5 b1, stru 5: 6 stru, var2 6: 7 b1, var2, v1
通过更新加入来追加result：
result[
melt(result, id.vars = "type")[value > 0.2][
order(-value), toString(head(variable, 3L)), keyby = type],
on = "type", selected := V1][
# beautify result
is.na(selected), selected := ""][]
type bills nos stru var2 v1 b1 selected 1: 1 -0.58026951 0.16493506 0.08202645 -0.07664827 0.11627152 -0.05595326 2: 2 0.02646100 0.22246750 0.21968328 0.40308468 0.38943918 -0.10121018 var2, v1, nos 3: 3 -0.11389551 0.36446564 -0.11769798 -0.16438528 0.00000000 -0.04100238 nos 4: 4 -0.45645233 -0.21585955 -0.37551547 -0.19560440 0.28351648 -0.08580863 v1 5: 5 -0.18596606 -0.23776224 0.39444627 -0.06304738 -0.03508794 0.39860140 b1, stru 6: 6 -0.72346726 -0.04175824 0.28585837 0.24862501 -0.30583077 -0.31718139 stru, var2 7: 7 -0.02649032 -0.08810594 -0.05718863 0.48398529 0.30143033 0.50165047 b1, var2, v1
答案 1 :(得分：1)
我使用split获得基本rbind解决方案，以生成子集列表，按照您想要的方式计算相关性并data.table。我想使用data.frame会有更复杂的方法，但现在它可以做到这一点。
根据您提供的数据生成df1 <- data.frame(id,bills,nos,stru,var1,var2,v1,v2,a1,b1,type,value)
> head(df1)
id bills nos stru var1 var2 v1 v2 a1 b1 type value
1 1 4 74 A 36 1 54 75 9 31 2 139
2 2 8 36 D 75 73 10 72 43 55 6 743
3 3 10 12 B 64 60 39 22 62 40 4 574
4 4 11 33 B 11 73 69 33 29 38 1 409
5 5 10 32 B 73 66 37 34 29 58 6 620
6 6 12 39 D 38 39 40 56 68 29 6 539
：
split使用subsets <- split(df1, df1$type)
：
lapply使用嵌套的str_char解决方案循环遍历corlist <- lapply(subsets, function(x) lapply(str_char, function(y) cor(x[,"value"], as.numeric(x[,y]), method = "spearman")))
中的变量名称：
do.call使用嵌套的cormatrix <- do.call(rbind, lapply(corlist, function(x) do.call(c, x)))
创建相关系数矩阵：
colnames(cormatrix) <- str_char
为列分配名称：
> cormatrix
bills nos var2 v1 b1
1 -0.58026951 0.16493506 -0.07664827 0.11627152 -0.05595326
2 0.02646100 0.22246750 0.40308468 0.38943918 -0.10121018
3 -0.11389551 0.36446564 -0.16438528 0.00000000 -0.04100238
4 -0.45645233 -0.21585955 -0.19560440 0.28351648 -0.08580863
5 -0.18596606 -0.23776224 -0.06304738 -0.03508794 0.39860140
6 -0.72346726 -0.04175824 0.24862501 -0.30583077 -0.31718139
7 -0.02649032 -0.08810594 0.48398529 0.30143033 0.50165047
输出：
cormatrix使用相关系数＆gt;添加最多三个变量的类型和名称。 0.2（按值排序）到maxvector <- apply(cormatrix, 1, function(x) sort(x[which(x > .2)], decreasing = T))
maxvector <- lapply(maxvector, function(x) names(x)[1:3])
maxvector <- lapply(maxvector, function(x) x[!is.na(x)])
maxvector <- lapply(maxvector, function(x) paste(x, collapse = ","))
cormatrix <- cbind(type = 1:7, cormatrix, maxvector)
，请使用：
> cormatrix
type bills nos stru var2 v1 b1 maxvector
1 1 -0.5802695 0.1649351 0.08202645 -0.07664827 0.1162715 -0.05595326 ""
2 2 0.026461 0.2224675 0.2196833 0.4030847 0.3894392 -0.1012102 "var2,v1,nos"
3 3 -0.1138955 0.3644656 -0.117698 -0.1643853 0 -0.04100238 "nos"
4 4 -0.4564523 -0.2158596 -0.3755155 -0.1956044 0.2835165 -0.08580863 "v1"
5 5 -0.1859661 -0.2377622 0.3944463 -0.06304738 -0.03508794 0.3986014 "b1,stru"
6 6 -0.7234673 -0.04175824 0.2858584 0.248625 -0.3058308 -0.3171814 "stru,var2"
7 7 -0.02649032 -0.08810594 -0.05718863 0.4839853 0.3014303 0.5016505 "b1,var2,v1"
结果：
stru修改：我还通过转换为as.numeric（感谢@Uwe）重新加入new_col。
答案 2 :(得分：1)
这是一个整齐的尝试：
library(tidyverse)
df1 %>%
select(bills, nos, var2, v1, b1, type) %>% #select needed variables, one can also do: select(str_char, type), however `stru` is not numeric
group_by(type) %>% #group by type
do(correlation = as.data.frame(cor(.[1:5]))) %>% #correlation
unnest(correlation) %>% #convenient output
gather(key, value, bills:b1) %>% #for easier pairwise removal
filter(var != key) %>% #remove self correlation
arrange(type, var, key)
#output
# A tibble: 140 x 4
type var key value
<int> <fctr> <chr> <dbl>
1 1 b1 bills 0.01978168
2 1 b1 nos -0.40581082
3 1 b1 v1 -0.08507922
4 1 b1 var2 0.15430381
5 1 bills b1 0.01978168
6 1 bills nos 0.21208062
7 1 bills v1 -0.15127493
8 1 bills var2 -0.02983736
9 1 nos b1 -0.40581082
10 1 nos bills 0.21208062
# ... with 130 more rows