问题描述
此问题是答案.
在中使用
函数? c(...%*%...)
和 sum(... * ...)
有什么区别> dplyr
中的> group_by()
What is the difference between using c(... %*% ...)
and sum(... * ...)
in a group_by()
function from dplyr
?
这两个代码都给出相同的结果:
Both of these code give the same result:
#1
library(dplyr) # 1.0.0
library(tidyr)
df1 %>%
group_by(Date, Market) %>%
group_by(Revenue = c(Quantity %*% Price),
TotalCost = c(Quantity %*% Cost),
Product, .add = TRUE) %>%
summarise(Sold = sum(Quantity)) %>%
pivot_wider(names_from = Product, values_from = Sold)
#2
library(dplyr) # 1.0.0
library(tidyr)
df1 %>%
group_by(Date, Market) %>%
group_by(Revenue = sum(Quantity * Price),
TotalCost = sum(Quantity * Cost),
Product, .add = TRUE) %>%
summarise(Sold = sum(Quantity)) %>%
pivot_wider(names_from = Product, values_from = Sold)
# A tibble: 2 x 7
# Groups: Date, Market, Revenue, TotalCost [2]
# Date Market Revenue TotalCost Apple Banana Orange
# <chr> <chr> <dbl> <dbl> <int> <int> <int>
#1 6/24/2020 A 135 37.5 35 20 20
#2 6/25/2020 A 25 15 10 15 NA
是 c(...%*%...)
和 sum(... * ...)
之一?
原始答案中的数据:
df1 <- structure(list(Date = c("6/24/2020", "6/24/2020", "6/24/2020",
"6/24/2020", "6/25/2020", "6/25/2020"), Market = c("A", "A",
"A", "A", "A", "A"), Salesman = c("MF", "RP", "RP", "FR", "MF",
"MF"), Product = c("Apple", "Apple", "Banana", "Orange", "Apple",
"Banana"), Quantity = c(20L, 15L, 20L, 20L, 10L, 15L), Price = c(1L,
1L, 2L, 3L, 1L, 1L), Cost = c(0.5, 0.5, 0.5, 0.5, 0.6, 0.6)),
class = "data.frame", row.names = c("1",
"2", "3", "4", "5", "6"))
推荐答案
我将这些注释编译为答案,如果我错过任何内容,其他人都可以加入.
I'll compile the comments into an answer, others can jump in if I miss anything.
-
%*%
和*
是完全不同的运算符:*
执行逐元素乘法,而%*%
执行线性代数矩阵乘法.这些是非常不同的操作,演示如下:
%*%
and*
are drastically different operators:*
does element-wise multiplication, and%*%
does linear algebra matrix multiplication. Those are very different operations, demonstrated with:
1:4 * 2:5
# [1] 2 6 12 20
1:4 %*% 2:5
# [,1]
# [1,] 40
sum(1:4 * 2:5)
# [1] 40
如果您要从两个向量相乘中查找单个汇总统计信息,并且从线性代数中求矩阵乘法就很有意义,那么%*%
是适合您的工具.
If you are looking for a single summary statistic from multiply two vectors, and the matrix-multiply from linear algebra makes sense, then %*%
is the right tool for you.
关于声明性代码应该说些什么;虽然您可以执行第三项操作( sum(.*.)
),但对我来说,最好使用%*%
,这有两个原因:
there should be something said about declarative code; while you can do the third operation (sum(.*.)
), to me it may be better to use %*%
, for two reasons:
-
声明性意图.我是说我有两个矩阵,我打算做线性代数";上.
Declarative intent. I am saying that I have two matrices that I intend to do "linear algebra" on.
保护措施.如果存在任何尺寸不匹配的情况(例如, sum(1:4 * 2:3)
仍在语法上起作用,但 1:4%*%2:3
则不能),我想马上知道.使用 sum(.*.)
,这种不匹配将被全世界默默忽略(我认为回收可能是一个大问题的一个原因).
Safeguards. If there is any dimensional mismatch (e.g., sum(1:4 * 2:3)
still works syntactically but 1:4 %*% 2:3
does not), I want to know it right away. With sum(.*.)
, the mismatch is silently ignored to the world (one reason I think recycling can be a big problem).
原因不是 性能:而对于较小的向量/矩阵,%*%
的性能与 sum(.*.)
,随着数据大小变大,%*%
相对更昂贵.
The reason is not performance: while with smaller vectors/matrices %*%
's performance is on par with sum(.*.)
, as the size of the data gets larger, %*%
is relatively more expensive.
m1 <- 1:100 ; m2 <- m1+1 ; m3 <- 1:100000; m4 <- m3+1
microbenchmark::microbenchmark(sm1 = sum(m1*m2), sm2 = m1%*%m2, lg1 = sum(m3*m4), lg2 = m3%*%m4)
# Unit: nanoseconds
# expr min lq mean median uq max neval
# sm1 800 1100 112900 1600 2100 11083600 100
# sm2 1100 1400 2143 1900 2450 10200 100
# lg1 239700 249550 411235 270800 355300 11102800 100
# lg2 547900 575550 634763 637850 678250 780500 100
到目前为止所有的讨论都在 vectors 上,它们实际上是一维矩阵(就 %*%
似乎认为......甚至还不够准确).一旦开始使用真正的矩阵,交换它们就会变得更加困难...实际上,我不知道一种模拟%*%
的简便方法(缺少 for
循环等):
All of the discussion so far has been on vectors, which are effectively 1d matrices (as far as %*%
seems to think ... though even that is not fully accurate). Once you start getting into true matrices, it becomes more difficult to interchange them ... in fact, I don't know of an easier way to emulate %*%
(short of for
loops, etc):
m1 %*% m2
# [,1] [,2] [,3] [,4]
# [1,] 22 49 76 103
# [2,] 28 64 100 136
t(sapply(seq_len(nrow(m1)), function(i) sapply(seq_len(ncol(m2)), function(j) sum(m1[i,] * m2[,j]))))
# [,1] [,2] [,3] [,4]
# [1,] 22 49 76 103
# [2,] 28 64 100 136
(虽然嵌套的- sapply
可能不是最快的非%*%
方式来处理矩阵Y的东西,但%*%
是更快 1-2个数量级,因为它是 .Internal
,并针对"数学进行编译和 meant ." 这样.)
(And while that nested-sapply
may not be the fastest non-%*%
way to do the matrix-y stuff, %*%
is 1-2 orders of magnitude faster, since it is .Internal
and compiled and meant for "Math!" like this.)
最底线,而%*%
确实在内部使用 *
运算符(对于其中几个步骤之一),而其他两个则不然不同的.哎呀,也许还可以用相同的方式比较 *
和 ^
...具有相似的结果.
Bottom line, while %*%
does use the *
operator internally (for one of a couple steps), the two are otherwise different. Heck, one might also compare *
and ^
in the same vein ... with a similar outcome.
干杯!
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