说我有一个数据框，像这样：
# Set RNG seed
set.seed(33550336)
# Create dummy data frame
df <- data.frame(PC1 = runif(20),
PC2 = runif(20),
PC3 = runif(20),
A = runif(20),
B = runif(20),
loc = sample(LETTERS[1:2], 20, replace = TRUE),
seas = sample(c("W", "S"), 20, replace = TRUE))
# > head(df)
# PC1 PC2 PC3 A B loc seas
# 1 0.8636470 0.02220823 0.7553348 0.4679607 0.0787467 A S
# 2 0.3522257 0.42733152 0.2412971 0.6691419 0.1194121 A W
# 3 0.5257408 0.44293320 0.3225228 0.0934192 0.2966507 B S
# 4 0.0667227 0.90273594 0.6297959 0.1962124 0.4894373 A W
# 5 0.3751383 0.50477920 0.6567203 0.4510632 0.4742191 B S
# 6 0.9197086 0.32024904 0.8382138 0.9907894 0.9335657 A S
我对计算PC1，PC2和PC3与按{{1分组的每个变量A和B之间的相关性感兴趣}}和loc。因此，例如，基于this answer，我可以执行以下操作：
seas这给了我我想要的：# Correlation of variable A and PC1 per loc & seas combination
df %>%
group_by(loc, seas) %>%
summarise(cor = cor(PC1, A)) %>%
ungroup
# # A tibble: 4 x 3
# loc seas cor
# <fct> <fct> <dbl>
# 1 A S 0.458
# 2 A W 0.748
# 3 B S -0.0178
# 4 B W -0.450
和PC1的每种组合在A和loc之间的相关性。 很棒。
我正在努力进行的工作是对seas变量和其他变量（在示例中为PC*和A）的每种组合进行推断以执行计算。我的预期结果是上面的小标题，但是B和其他变量的每个组合都有一列。我可以做很多事情... PC*，cor(PC2, A)，cor(PC3, A)等，但是大概有一种简洁的编码计算方法。我怀疑它牵涉到cor(PC1, B)，但我不能完全理解……有人可以启发我吗？
我选择了格罗腾迪克（G. Grothendieck）的solution below，但这需要进行一些重组才能将其转换为所需的格式。我已经发布了我在这里使用的代码，以防其他人使用它。
do给出，
# Perform calculation
res <- by(df[1:5], df[-(1:5)], cor)
# Combinations of loc & seas
comb <- expand.grid(dimnames(res))
# loc seas
# 1 A S
# 2 B S
# 3 A W
# 4 B W
# A matrix corresponding to a loc & seas
# Plus the loc & seas themselves
restructure <- function(m, n){
# Convert to data frame
# Add rownames as column
# Retains PCs as rows, but not columns
# Gather variables to long format
# Unite PC & variable names
# Spread to a single row
# Add combination of loc & seas
m %>%
data.frame %>%
rownames_to_column() %>%
filter(grepl("PC", rownames(m))) %>%
select(-contains("PC")) %>%
gather(variable, value, -rowname) %>%
unite(comb, rowname, variable) %>%
spread(comb, value) %>%
bind_cols(n)
}
# Restructure each list element & combine into data frame
do.call(rbind, lapply(1:length(res), function(x)restructure(res[[x]], comb[x, ])))
答案 0 :(得分：7)
像这样使用// Copy assets to out dir. Add your own globs as needed.
new CopyWebpackPlugin([
{ from: "fonts/**" },
{ from: "**/*.jpg" },
{ from: "**/*.png" },
{ from: "assets/**" }, // Add your asset folder path like this
], { ignore: [`${relative(appPath, appResourcesFullPath)}/**`] }),
：
by给予：
By <- by(df[1:5], df[-(1:5)], cor)
根据发布者对所需内容的进一步讨论，定义> By
loc: A
seas: S
PC1 PC2 PC3 A B
PC1 1.00000000 -0.3941583 0.1872622 0.4576316 -0.00925106
PC2 -0.39415826 1.0000000 -0.6797708 0.3522161 0.20916667
PC3 0.18726218 -0.6797708 1.0000000 -0.2003091 0.37414025
A 0.45763159 0.3522161 -0.2003091 1.0000000 0.57292305
B -0.00925106 0.2091667 0.3741403 0.5729230 1.00000000
-----------------------------------------------------------------------------------------------------------------------------
loc: B
seas: S
PC1 PC2 PC3 A B
PC1 1.00000000 -0.52651449 0.07120701 -0.01779813 -0.7432814
PC2 -0.52651449 1.00000000 -0.05448583 -0.35011878 0.4632416
PC3 0.07120701 -0.05448583 1.00000000 0.80342399 0.4580262
A -0.01779813 -0.35011878 0.80342399 1.00000000 0.5558740
B -0.74328144 0.46324158 0.45802622 0.55587404 1.0000000
-----------------------------------------------------------------------------------------------------------------------------
loc: A
seas: W
PC1 PC2 PC3 A B
PC1 1.0000000 -0.79784422 0.0932317 0.7483545 0.49639477
PC2 -0.7978442 1.00000000 -0.3526315 -0.3994917 -0.05233889
PC3 0.0932317 -0.35263151 1.0000000 -0.5902400 0.36066898
A 0.7483545 -0.39949171 -0.5902400 1.0000000 0.18081316
B 0.4963948 -0.05233889 0.3606690 0.1808132 1.00000000
-----------------------------------------------------------------------------------------------------------------------------
loc: B
seas: W
PC1 PC2 PC3 A B
PC1 1.0000000 0.3441459 0.1135686 -0.4502518 -0.6672104
PC2 0.3441459 1.0000000 -0.8447551 -0.9899521 -0.8098906
PC3 0.1135686 -0.8447551 1.0000000 0.7606430 0.3738706
A -0.4502518 -0.9899521 0.7606430 1.0000000 0.8832408
B -0.6672104 -0.8098906 0.3738706 0.8832408 1.0000000
函数，该函数接受相关矩阵或数据帧（在后一种情况下，它将前5列转换为相关矩阵），产生一行输出。对于onerow代码行，if中的onerow语句不是必需的，但不会受到伤害，但是我们已经包含了它，因此adply也可以在一个简单的环境中工作在下面的后续示例中也是如此。
onerow给予：
library(plyr)
onerow <- function(x) {
if (is.data.frame(x)) x <- cor(x[1:5])
dtab <- as.data.frame.table(x[4:5, 1:3])
with(dtab, setNames(Freq, paste(Var2, Var1, sep = "_")))
}
adply(By, 1:2, onerow)
或者完全摆脱 loc seas PC1_A PC1_B PC2_A PC2_B PC3_A PC3_B
1 A S 0.45763159 -0.00925106 0.3522161 0.20916667 -0.2003091 0.3741403
2 B S -0.01779813 -0.74328144 -0.3501188 0.46324158 0.8034240 0.4580262
3 A W 0.74835455 0.49639477 -0.3994917 -0.05233889 -0.5902400 0.3606690
4 B W -0.45025181 -0.66721038 -0.9899521 -0.80989058 0.7606430 0.3738706
，并使用它给出相同的输出：
by或使用dplyr：
library(plyr)
ddply(df, -(1:5), onerow)
答案 1 :(得分：3)
这是通过[AttributeUsage(AttributeTargets.Class | AttributeTargets.Method)]
public class MyActionFilterAttribute : ActionFilterAttribute
{
public override void OnActionExecuting(HttpActionContext actionContext)
{
if(actionContext.ActionDescriptor.GetCustomAttributes<SkipMyActionFilterAttribute>().Any())
{
return;
}
//Do something
}
}
的解决方案，其中我们使用tidyverse指定所有summarise_at并将它们与PC[0-9]关联。与A相同的过程，然后简单地合并，即
B给出，
library(tidyverse) df %>% group_by(loc, seas) %>% summarise_at(vars(starts_with('PC')), funs(cor(., A))) %>% left_join(., df %>% group_by(loc, seas) %>% summarise_at(vars(starts_with('PC')), funs(cor(., B))), by = c('loc', 'seas'), suffix = c('.A', '.B'))
答案 2 :(得分：3)
我们可以在split中进行cor和base R
lapply(split(df[1:5], df[-(1:5)]), cor)
#$A.S
# PC1 PC2 PC3 A B
#PC1 1.00000000 -0.3941583 0.1872622 0.4576316 -0.00925106
#PC2 -0.39415826 1.0000000 -0.6797708 0.3522161 0.20916667
#PC3 0.18726218 -0.6797708 1.0000000 -0.2003091 0.37414025
#A 0.45763159 0.3522161 -0.2003091 1.0000000 0.57292305
#B -0.00925106 0.2091667 0.3741403 0.5729230 1.00000000
#$B.S
# PC1 PC2 PC3 A B
#PC1 1.00000000 -0.52651449 0.07120701 -0.01779813 -0.7432814
#PC2 -0.52651449 1.00000000 -0.05448583 -0.35011878 0.4632416
#PC3 0.07120701 -0.05448583 1.00000000 0.80342399 0.4580262
#A -0.01779813 -0.35011878 0.80342399 1.00000000 0.5558740
#B -0.74328144 0.46324158 0.45802622 0.55587404 1.0000000
#$A.W
# PC1 PC2 PC3 A B
#PC1 1.0000000 -0.79784422 0.0932317 0.7483545 0.49639477
#PC2 -0.7978442 1.00000000 -0.3526315 -0.3994917 -0.05233889
#PC3 0.0932317 -0.35263151 1.0000000 -0.5902400 0.36066898
#A 0.7483545 -0.39949171 -0.5902400 1.0000000 0.18081316
#B 0.4963948 -0.05233889 0.3606690 0.1808132 1.00000000
#$B.W
# PC1 PC2 PC3 A B
#PC1 1.0000000 0.3441459 0.1135686 -0.4502518 -0.6672104
#PC2 0.3441459 1.0000000 -0.8447551 -0.9899521 -0.8098906
#PC3 0.1135686 -0.8447551 1.0000000 0.7606430 0.3738706
#A -0.4502518 -0.9899521 0.7606430 1.0000000 0.8832408
#B -0.6672104 -0.8098906 0.3738706 0.8832408 1.0000000
或使用tidyverse
library(tidyverse)
df %>%
group_by_at(6:7) %>%
nest %>%
mutate(data = map(data, cor))