在dplyr中分组的众多变量之间的相关性

Question

假设我有一个数据框,如下所示:

# 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

我感兴趣的是计算之间的相关性PC1,PC2以及PC3与每个变量A,并B通过分组loc和seas.所以,例如,基于这个答案,我可以做到以下几点:

# 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 

这给了我我想要的东西:之间的相关性PC1,并A为每个组合的loc和seas.真棒.

我正在与被推断此进行计算的每个组合挣扎PC*变量和其他变量(即A和B,在本例中).我的预期结果是紧接在上面的tibble,但每个组合的列PC*和其他变量.我能做到这一点长手... cor(PC2, A),cor(PC3, A),cor(PC1, B)等,但想必是有编码的计算的简洁的方式.我怀疑它涉及到do,但我不能完全理解它...有人可以开导我吗？

---

解

我在下面使用了G. Grothendieck的解决方案,但这需要进行一些重组才能使其达到所需的格式.我已经发布了我在这里使用的代码,以防它对其他人有用.

# 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, ])))

这使,

#         PC1_A       PC1_B      PC2_A       PC2_B      PC3_A     PC3_B loc seas
# 1  0.45763159 -0.00925106  0.3522161  0.20916667 -0.2003091 0.3741403   A    S
# 2 -0.01779813 -0.74328144 -0.3501188  0.46324158  0.8034240 0.4580262   B    S
# 3  0.74835455  0.49639477 -0.3994917 -0.05233889 -0.5902400 0.3606690   A    W
# 4 -0.45025181 -0.66721038 -0.9899521 -0.80989058  0.7606430 0.3738706   B    W

Answer 1

使用by这样:

By <- by(df[1:5], df[-(1:5)], cor)

赠送:

> 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

添加

基于海报对所需内容的进一步讨论,定义onerow接受相关矩阵或数据帧的函数(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.该if报表onerow是不需要的,但不会伤害,对adply行代码,但我们已经包括它,以便onerow在随后的下面一个例子简单的方式也可以作为很好.

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)

赠送:

  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

或者by完全摆脱并使用它给出相同的输出:

library(plyr)
ddply(df, -(1:5), onerow)

或使用dplyr:

library(dplyr)
df %>%
  group_by_at(-(1:5)) %>%
  do( onerow(.) %>% t %>% as.data.frame ) %>%
  ungroup