我在下表中给出了一个映射:
Input Output
<4 0
5 0.4
6 0.5
7 0.65
8 0.75
9 0.85
>=10 1
到现在为止,我写了3个版本:
k1 <- function(h) {
if (h <= 4) { k <- 0
} else if (h == 5) { k <- 0.4
} else if (h == 6) { k <- 0.5
} else if (h == 7) { k <- 0.65
} else if (h == 8) { k <- 0.75
} else if (h == 9) { k <- 0.85
} else if (h >= 10) { k <- 1}
return(k)
}
第二:
k2 <- function(h) {
k <- 0
k[h == 5] <- 0.4
k[h == 6] <- 0.5
k[h == 7] <- 0.65
k[h == 8] <- 0.75
k[h == 9] <- 0.85
k[h >= 10] <- 1.0
return(k)
}
第三:
k3 <- function(h) {
k <- cut(h, breaks=c(0, 5, 6, 7, 8, 9, Inf), labels=c(0, 0.5, 0.65, 0.75, 0.85, 1), right=FALSE)
return(k)
}
我需要在两种不同的场景中使用该功能.首先,评估标量输入,然后评估值的向量.
对于标量输入:
h <- 5
microbenchmark(k1(h), k2(h), k3(h))
Unit: microseconds
expr min lq mean median uq max neval
k1(h) 1.208 1.5110 2.38264 1.8125 2.114 15.698 100
k2(h) 4.529 5.5855 8.71286 6.3400 7.849 73.053 100
k3(h) 52.224 54.0360 71.74953 68.9785 79.393 304.286 100
对于矢量输入:
h <- rep(5, 250)
microbenchmark(sapply(h, k1), k2(h), k3(h))
Unit: microseconds
expr min lq mean median uq max neval
sapply(h, k1) 595.592 617.327 641.8598 637.8535 654.9100 857.918 100
k2(h) 15.397 17.207 19.5470 18.1130 19.6225 49.508 100
k3(h) 110.486 116.070 131.3117 121.2020 140.6720 275.910 100
因此,k1对于标量输入是最快的,对于向量输入是k2.
你认为有可能提高速度吗？我无法相信这样一个笨拙的if/else代码在标量情况下应该是最快的.而且,我想要一个统一的功能,而不是两个单独的功能.
首先,为什么要在标量输入上优化几微秒？如果答案是"因为标量版本必须多次调用",或许是在一个循环中,而不是那个问题; 操作应该是矢量化的.(请注意,您k2可以在k1处理15个时间内处理250个输入).
无论如何,另一种选择是:
outputs <- c(0, .4, .5, .65, .75, .85, 1)
k4 <- function(h) {
output[pmin.int(pmax.int(h, 4), 10) - 3]
}
在我的系统中,这只是k2在矢量化情况下的关系,但它的速度大约k2是标量情况的两倍:
h <- 5
microbenchmark(k1(h), k2(h), k3(h), k4(h))
# Unit: nanoseconds
# expr min lq mean median uq max neval
# k1(h) 748 933.5 1314.29 1181.5 1655.0 3091 100
# k2(h) 4131 5424.5 6378.31 6236.5 7021.5 18140 100
# k3(h) 72149 74495.0 79796.22 75716.0 80936.5 176857 100
# k4(h) 1730 2259.5 3396.04 3338.5 3801.0 17001 100
h <- rep(5, 250)
microbenchmark(sapply(h, k1), k2(h), k3(h), k4(h))
# Unit: microseconds
# expr min lq mean median uq max neval
# sapply(h, k1) 311.099 327.5710 341.05200 335.9330 348.6405 405.830 100
# k2(h) 13.973 18.4965 20.64351 20.4160 22.4015 34.289 100
# k3(h) 117.401 125.0180 134.49228 129.2455 138.8240 241.896 100
# k4(h) 15.042 17.8870 20.33141 19.0690 20.4260 37.386 100
它也更简洁k2,更容易扩展到更多的整数输入.
最后,如果你愿意依赖Rcpp,你可以相对于k2和获得5倍的加速k4:
library(Rcpp)
cppFunction('NumericVector k5(IntegerVector h) {
int n = h.size();
NumericVector out(n);
for (int i = 0; i < n; ++i) {
int val = h[i];
if (val <= 4) out[i] = 0;
else if (val == 5) out[i] = .4;
else if (val == 6) out[i] = .5;
else if (val == 7) out[i] = .65;
else if (val == 8) out[i] = .75;
else if (val == 9) out[i] = .85;
else if (val >= 10) out[i] = 1;
}
return out;
}')
h <- rep(5, 250)
microbenchmark(sapply(h, k1), k2(h), k3(h), k4(h), k5(h))
# Unit: microseconds
# expr min lq mean median uq max neval
# sapply(h, k1) 382.383 410.7310 429.88844 423.7150 442.5765 501.400 100
# k2(h) 17.129 20.5865 23.95221 22.1340 23.7915 46.827 100
# k3(h) 123.519 127.6830 142.24084 140.5400 150.1525 218.919 100
# k4(h) 15.168 18.2705 20.45797 19.1985 20.6105 52.650 100
# k5(h) 2.988 4.9045 6.49218 5.9135 6.8455 33.219 100
(请参阅此Rcpp介绍的"矢量输入,矢量输出"部分,作为此类功能的指南).但请注意,它仍然比k1标量情况慢2倍!
findInterval 是R中最快的功能
out <- c(0, .4, .5, .65, .75, .85, 1)
k6 <- function(h){
ind <- findInterval(h, c(4, 5, 6, 7, 8, 9) +0.1) + 1
out[ind]
}
h <- rep(5, 250)
microbenchmark(k2(h), k4(h), k6(h), unit="relative")
Unit: relative
# expr min lq mean median uq max neval
# k2(h) 2.283983 2.347714 2.225037 2.392578 2.319125 1.184224 100
# k4(h) 1.830939 1.725286 1.699866 1.701196 1.599973 1.414026 100
# k6(h) 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 100