r语言 - 是否有一个有效的替代表()?

Question

我使用以下命令:

table(factor("list",levels=1:"n")

with "list":(示例)a = c(1,3,4,4,3)和levels = 1:5，也考虑到2和5。对于真正的大数据集，我的代码似乎非常无效。

有谁知道一个隐藏的库或代码片段，使它更快?

Answer 1

我们可以从collapse中使用fnobs，这将是有效的

library(collapse)
fnobs(df, g = df$X1)

---

在base R中，tabulate比table更高效

tabulate(df$X1)
[1]  9  6 15 13 11  9  7  9 11 10

Answer 2

TL;DR是base::tabulate。

总结，基本目标是性能，所以我准备了所有提供的解决方案的microbenchmark。我用小向量和大向量，两种不同的情况。对于我机器上的collapse包，我必须下载最新的Rcpp包1.0.7(以抑制崩溃)。即使我添加Rcpp解决方案也比base::tabulate慢。

suppressMessages(library(janitor))
suppressMessages(library(collapse))
suppressMessages(library(dplyr))
suppressMessages(library(cpp11))
# source https://stackoverflow.com/questions/31001392/rcpp-version-of-tabulate-is-slower-where-is-this-from-how-to-understand
Rcpp::cppFunction('IntegerVector tabulate_rcpp(const IntegerVector& x, const unsigned max) {
IntegerVector counts(max);
for (auto& now : x) {
if (now > 0 && now <= max)
counts[now - 1]++;
}
return counts;
}')
set.seed(1234)
a = c(1,3,4,4,3)
levels = 1:5
df <- data.frame(X1 = a)

microbenchmark::microbenchmark(tabulate_rcpp = {tabulate_rcpp(df$X1, max(df$X1))},
base_table = {base::table(factor(df$X1, 1:max(df$X1)))},
stats_aggregate = {stats::aggregate(. ~ X1, cbind(df, n = 1), sum)},
graphics_hist = {hist(df$X1, plot = FALSE, right = FALSE)[c("breaks", "counts")]},
janitor_tably = {adorn_totals(tabyl(df, X1))},
collapse_fnobs = {fnobs(df, df$X1)},
base_tabulate = {tabulate(df$X1)},
dplyr_count = {count(df, X1)})
#> Unit: microseconds
#>             expr      min        lq       mean    median        uq       max
#>    tabulate_rcpp    2.959    5.9800   17.42326    7.9465    9.5435   883.561
#>       base_table   48.524   59.5490   72.42985   66.3135   78.9320   153.216
#>  stats_aggregate  829.324  891.7340 1069.86510  937.4070 1140.0345  2883.025
#>    graphics_hist  148.561  170.5305  221.05290  188.9570  228.3160   958.619
#>    janitor_tably 6005.490 6439.6870 8137.82606 7497.1985 8283.3670 53352.680
#>   collapse_fnobs   14.591   21.9790   32.63891   27.2530   32.6465   417.987
#>    base_tabulate    1.879    4.3310    5.68916    5.5990    6.6210    16.789
#>      dplyr_count 1832.648 1969.8005 2546.17131 2350.0450 2560.3585  7210.992
#>  neval
#>    100
#>    100
#>    100
#>    100
#>    100
#>    100
#>    100
#>    100

df <- data.frame(X1 = sample(1:5, 1000, replace = TRUE))
microbenchmark::microbenchmark(tabulate_rcpp = {tabulate_rcpp(df$X1, max(df$X1))},
base_table = {base::table(factor(df$X1, 1:max(df$X1)))},
stats_aggregate = {stats::aggregate(. ~ X1, cbind(df, n = 1), sum)},
graphics_hist = {hist(df$X1, plot = FALSE, right = FALSE)[c("breaks", "counts")]},
janitor_tably = {adorn_totals(tabyl(df, X1))},
collapse_fnobs = {fnobs(df, df$X1)},
base_tabulate = {tabulate(df$X1)},
dplyr_count = {count(df, X1)})
#> Unit: microseconds
#>             expr      min        lq       mean    median        uq       max
#>    tabulate_rcpp    4.847    8.8465   10.92661   10.3105   12.6785    28.407
#>       base_table   83.736  107.2040  121.77962  118.8450  129.9560   184.427
#>  stats_aggregate 1027.918 1155.9205 1338.27752 1246.6205 1434.8990  2085.821
#>    graphics_hist  209.273  237.8265  274.60654  258.9260  300.3830   523.803
#>    janitor_tably 5988.085 6497.9675 7833.34321 7593.3445 8422.6950 13759.142
#>   collapse_fnobs   26.085   38.6440   51.89459   47.8250   57.3440   333.034
#>    base_tabulate    4.501    6.7360    8.09408    8.2330    9.2170    11.463
#>      dplyr_count 1852.290 2000.5225 2374.28205 2145.9835 2516.7940  4834.544
#>  neval
#>    100
#>    100
#>    100
#>    100
#>    100
#>    100
#>    100
#>    100

^{由reprex包(v2.0.0)创建于2021-08-01}

Answer 3

我们也可以使用janitor::tabyl:

library(janitor)
df %>%
tabyl(X1) %>%
adorn_totals()
X1   n percent
1   9    0.09
2   6    0.06
3  15    0.15
4  13    0.13
5  11    0.11
6   9    0.09
7   7    0.07
8   9    0.09
9  11    0.11
10  10    0.10
Total 100    1.00

Answer 4

这不是你想要的，但也许你可以使用这个:

library(dplyr)
set.seed(8192)
df <- data.frame(X1 = sample(1:10, 100, replace = TRUE))
df %>% 
count(X1)

返回

X1  n
1   1  9
2   2  6
3   3 15
4   4 13
5   5 11
6   6  9
7   7  7
8   8  9
9   9 11
10 10 10

如果需要计算更多的数字(包括丢失的数字)，可以使用

library(tidyr)
library(dplyr)
df2 <- data.frame(X1 = 1:12)
df %>% 
count(X1) %>% 
right_join(df2, by="X1") %>% 
mutate(n = replace_na(n, 0L))

,

X1  n
1   1  9
2   2  6
3   3 15
4   4 13
5   5 11
6   6  9
7   7  7
8   8  9
9   9 11
10 10 10
11 11  0
12 12  0

Answer 5

使用aggregate的基本R选项(从@Martin Gal借用df)

> aggregate(. ~ X1, cbind(df, n = 1), sum)
X1  n
1   1  9
2   2  6
3   3 15
4   4 13
5   5 11
6   6  9
7   7  7
8   8  9
9   9 11
10 10 10

---

另一个选择是使用hist

> hist(df$X1, plot = FALSE, right = FALSE)[c("breaks", "counts")]
$breaks
[1]  1  2  3  4  5  6  7  8  9 10
$counts
[1]  9  6 15 13 11  9  7  9 21

Answer 6

还有一个:summarytools

数据来自Martin Gal!非常感谢:

library(summarytools)
set.seed(8192)
df <- data.frame(X1 = sample(1:10, 100, replace = TRUE))
summarytools::freq(df$X1, cumul=FALSE)

输出:

Freq   % Valid   % Total
----------- ------ --------- ---------
1      9      9.00      9.00
2      6      6.00      6.00
3     15     15.00     15.00
4     13     13.00     13.00
5     11     11.00     11.00
6      9      9.00      9.00
7      7      7.00      7.00
8      9      9.00      9.00
9     11     11.00     11.00
10     10     10.00     10.00
<NA>      0                0.00
Total    100    100.00    100.00

Answer 7

如果需要table()更快的替代方案，包括交叉制表，v1.8.0(2022年5月)以来可用的collapse::qtab()是一个可靠且明显更快的替代方案。fcount()也可用于单变量情况，并返回data.frame。

library(collapse) # > v1.8.0, and > 1.9.0 for fcount()
library(microbenchmark)
v = sample(10000, 1e6, TRUE)
microbenchmark(qtab(v, sort = FALSE), fcount(v), tabulate(v), times = 10)
Unit: milliseconds
expr      min       lq     mean   median       uq      max neval
qtab(v, sort = FALSE) 1.911707 1.945245 2.002473 1.963654 2.027942 2.207891    10
fcount(v) 1.885549 1.906746 1.978894 1.932310 2.103997 2.138027    10
tabulate(v) 2.321543 2.323716 2.333839 2.328206 2.334499 2.372506    10
v2 = sample(10000, 1e6, TRUE)
microbenchmark(qtab(v, v2), qtab(v, v2, sort = FALSE), table(v, v2), times = 10)
Unit: milliseconds
expr       min        lq      mean   median        uq      max neval
qtab(v, v2)  45.61279  51.14840  74.16168  60.7761  72.86385 157.6501    10
qtab(v, v2, sort = FALSE)  41.30812  49.66355  57.02565  51.3568  54.69859 118.1289    10
table(v, v2) 281.60079 282.85273 292.48119 286.0535 288.19253 349.5513    10

话虽这么说，tabulate()在C代码中是最快的。但它有一个明确的警告，即它根本不散列值，而是确定最大值并分配该长度的结果向量，将其用作表来计数值。想想看:

v[10] = 1e7L # Adding a random large value here
length(tabulate(v))
[1] 10000000
length(table(v))
[1] 10001
length(qtab(v))
[1] 10001

所以你得到一个包含699万个零的结果向量，你的性能就会下降

microbenchmark(qtab(v, sort = FALSE), fcount(v), tabulate(v), times = 10)
Unit: milliseconds
expr      min       lq     mean   median       uq       max neval
qtab(v, sort = FALSE) 1.873249 1.900473 1.966721 1.923064 2.064186  2.126588    10
fcount(v) 1.829338 1.850330 1.926676 1.880199 2.021013  2.057667    10
tabulate(v) 4.207789 4.357439 5.066296 4.417012 4.558216 10.347744    10

考虑到这一点，qtab()实际上对每个值进行哈希并实现这种性能是相当了不起的。