在 R 中向量化 for 循环
[英]Vectorizing a for loop in R
问题描述
我试图通过受益于 R 的矢量化来改进我的代码，比如使用更多的应用系列函数而不是 for 循环，因为我使用的数据集达到 300K 记录，我希望能够减少脚本的时间跑步。
我已经准备了一个 repex 以及实际的 for 循环，我只是不知道是否可以将其转换为非循环结构。
它是这样的：
df <- structure(list(time = structure(c(1500697800, 1500698100, 1500698400,
1500698700, 1500699000, 1500699300, 1500699600, 1500699900, 1500700200,
1500700500, 1500700800, 1500701100, 1500701400, 1500701700, 1500702000,
1500702300, 1500702600, 1500702900, 1500703200, 1500703500, 1500703800,
1500704100, 1500704400, 1500704700, 1500705000, 1500705300, 1500705600,
1500705900, 1500706200, 1500706500, 1500706800, 1500707100, 1500707400,
1500707700, 1500708000, 1500708300, 1500708600, 1500708900, 1500709200,
1500709500, 1500709800, 1500710100, 1500710400, 1500710700, 1500711000,
1500711300, 1500711600, 1500711900, 1500712200, 1500712500, 1500712800,
1500713100, 1500713400, 1500713700, 1500714000, 1500714300, 1500714600,
1500714900, 1500715200, 1500715500, 1500715800, 1500716100, 1500716400,
1500716700, 1500717000, 1500717300, 1500717600, 1500717900, 1500718200,
1500718500, 1500718800, 1500719100, 1500719400, 1500719700, 1500720000,
1500720300, 1500720600, 1500720900, 1500721200, 1500721500, 1500721800,
1500722100, 1500722400, 1500722700, 1500723000, 1500723300, 1500723600,
1500723900, 1500724200, 1500724500, 1500724800, 1500725100, 1500725400,
1500725700, 1500726000, 1500726300, 1500726600, 1500726900, 1500727200,
1500727500, 1500727800, 1500728100, 1500728400, 1500728700, 1500729000,
1500729300, 1500729600, 1500729900, 1500730200, 1500730500, 1500730800,
1500731100, 1500731400, 1500731700, 1500732000, 1500732300, 1500732600,
1500732900, 1500733200, 1500733500, 1500733800, 1500734100, 1500734400,
1500734700, 1500735000, 1500735300, 1500735600, 1500735900, 1500736200,
1500736500, 1500736800, 1500737100, 1500737400, 1500737700, 1500738000,
1500738300, 1500738600, 1500738900, 1500739200, 1500739500, 1500739800,
1500740100, 1500740400, 1500740700, 1500741000), class = c("POSIXct",
"POSIXt"), tzone = "UTC"), rate = c(8021.22624828867, 8022.17252092756,
4026.57093082574, 0, 0, 0, 0, 0, 0, 0, 0, 1092.48742657481, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2352.47712160156, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), is.rate = c("OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF",
"OFF", "OFF", "OFF", "OFF", "OFF", "OFF", "OFF")), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -145L))
快速解释数据：它有一个时间变量、一些速率和一个标志，当速率不为 0 --> ON 时。
for 循环的想法是它会接收高于 0 的速率值，并且从时间的角度来看，它将在下一小时“拖尾”is.rate 标志。 我知道这听起来很复杂，但是一旦你在 repex 上运行 for 循环，它应该是有意义的。
谈到 for 循环，这里是：
for (i in which(temp_df$rate != 0)) {
temp_df$is.rate[i:(i + 12)] <- "ON" # 12 in this case is a factor of lag-time. Since data is in 5 min intervals, this means the next hour
}
我很想尝试优化此代码，最好完全删除for循环并使用类似的东西来应用家庭功能，但我无法真正看到代码结构。
2 个解决方案
解决方案1
1 2020-01-16 22:52:39
我认为您正在寻找在rate > 0时设置"ON"并在接下来的 11 行滞后。
我上面的评论没有包含align="right" ，这是获得我认为你想要的逻辑所必需的。 尝试这个：
zoo::rollapply(df$rate > 0, 12, any, align = "right", partial = TRUE)
# [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
# [13] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE
# [25] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# [37] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# [49] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# [61] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# [73] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# [85] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# [97] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# [109] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# [121] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE TRUE TRUE
# [133] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE
# [145] FALSE
ifelse(zoo::rollapply(df$rate > 0, 12, any, align = "right", partial = TRUE), "YES", "NO")
# [1] "YES" "YES" "YES" "YES" "YES" "YES" "YES" "YES" "YES" "YES" "YES" "YES"
# [13] "YES" "YES" "YES" "YES" "YES" "YES" "YES" "YES" "YES" "YES" "YES" "NO"
导致此数据：
print(df, n=26)
# # A tibble: 145 x 3
# time rate is.rate
# <dttm> <dbl> <chr>
# 1 2017-07-22 04:30:00 8021. YES
# 2 2017-07-22 04:35:00 8022. YES
# 3 2017-07-22 04:40:00 4027. YES
# 4 2017-07-22 04:45:00 0 YES
# 5 2017-07-22 04:50:00 0 YES
# 6 2017-07-22 04:55:00 0 YES
# 7 2017-07-22 05:00:00 0 YES
# 8 2017-07-22 05:05:00 0 YES
# 9 2017-07-22 05:10:00 0 YES
# 10 2017-07-22 05:15:00 0 YES
# 11 2017-07-22 05:20:00 0 YES ### counting rows from last non-zero rate
# 12 2017-07-22 05:25:00 1092. YES 1
# 13 2017-07-22 05:30:00 0 YES 2
# 14 2017-07-22 05:35:00 0 YES 3
# 15 2017-07-22 05:40:00 0 YES 4
# 16 2017-07-22 05:45:00 0 YES 5
# 17 2017-07-22 05:50:00 0 YES 6
# 18 2017-07-22 05:55:00 0 YES 7
# 19 2017-07-22 06:00:00 0 YES 8
# 20 2017-07-22 06:05:00 0 YES 9
# 21 2017-07-22 06:10:00 0 YES 10
# 22 2017-07-22 06:15:00 0 YES 11
# 23 2017-07-22 06:20:00 0 YES 12
# 24 2017-07-22 06:25:00 0 NO
# 25 2017-07-22 06:30:00 0 NO
# 26 2017-07-22 06:35:00 0 NO
# # ... with 119 more rows
解决方案2
0 2020-01-17 01:03:15
我认为您需要做的是找出rate != 0索引，在这些索引和inds + 12之间创建一个序列，并将这些索引的is.rate分配给"ON" 。
inds <- which(temp_df$rate != 0)
temp_df$is.rate[unique(c(mapply(`:`, inds, inds + 12)))] <- "ON"
它提供与for循环相同的输出。
在R中向量化for循环
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