英文:
rolling function using previous row values [R]
问题
我明白你的需求，你需要翻译代码和相关注释，但不需要翻译其他部分。以下是代码和注释的中文翻译：
# 我有一些地点（`a,b,c`），土壤具有一定的最大容量（`100,200`）。对于每个时间段（数据集中的每一行），我已经计算了水含量的变化（`change`）。# 设置随机种子set.seed(2023)db <- data.table(ID=c(rep("a",3),rep("b",3),rep("c",3)),capacity=c(rep(100,3),rep(200,6)),prc=round(runif(9,1,100),0),evpt=round(runif(9,1,100),0))# 计算水含量的变化db[prc>=evpt, change := prc - evpt]db[prc<evpt, change := exp(-((evpt - prc) / capacity))]# 在时间0时，地点存储最大的`capacity`水量。然后，如果`prc` > `evpt`，`change`是添加到先前含量的水，但不超过最大容量。如果`prc` < `evpt`，`change`是先前水含量的乘数，从而减少含量。# 草稿函数new_content <- function(change, capacity, prc, evpt) {if (prc > evpt) {min(change + previous_content, capacity)} else {previous_content * change}}# 然后，我应该应用该函数到每个地点，以获得一个新列(`content`)，显示水含量的时间变化。如下面在Excel中计算的输出：# 输出# > db# ID capacity prc evpt change content# 1: a 100 39 20 19.0000000 100.0000# 2: a 100 74 35 39.0000000 100.0000# 3: a 100 67 19 48.0000000 100.0000# 4: b 200 49 78 0.8650223 173.0045# 5: b 200 41 70 0.8650223 149.6527# 6: b 200 94 75 19.0000000 168.6527# 7: c 200 74 51 23.0000000 200.0000# 8: c 200 67 71 0.9801987 196.0397# 9: c 200 77 73 4.0000000 200.0000# 我无法解决的问题是如何定义变量`previous_content`，即一开始等于最大容量，然后等于前一行的内容。# 我查阅了`data.table::frollapply`和`zoo:rollapply`的文档，但甚至无法为此起草一个初步的草稿。有什么帮助吗？谢谢！
希望这能满足你的需求。如果你有其他问题，请随时提出。
英文:
I have some sites (a,b,c) where the soil has a certain maximum capacity (100,200). For each period of time (every row in my dataset), I have calculated the change in water content (change).
set.seed(2023)db <- data.table(ID=c(rep("a",3),rep("b",3),rep("c",3)),capacity=c(rep(100,3),rep(200,6)),prc=round(runif(9,1,100),0),evpt=round(runif(9,1,100),0))db[prc>=evpt,change:=prc-evpt]db[prc<evpt,change:=exp(-((evpt-prc)/capacity))]ID capacity prc evpt change1: a 100 39 20 19.00000002: a 100 74 35 39.00000003: a 100 67 19 48.00000004: b 200 49 78 0.86502235: b 200 41 70 0.86502236: b 200 94 75 19.00000007: c 200 74 51 23.00000008: c 200 67 71 0.98019879: c 200 77 73 4.0000000
At time 0 the site hold the maximum capacity of water. Then, if prc >evpt, change is the water added to the previous content, without exceeding the maximum capacity. If prc < evpt, change is a multiplier of the previous water content, thus reducing the content.
# draft functionnew_content <- function(change,capacity,prc,evpt){if (prc>evpt) {min(change+previous_content,capacity)} else {previous_content*change}}
Then I should apply the function for each site to get a new column (content) showing the temporal variation of water content. As in the following output calculated in Excel:
> dbID capacity prc evpt change content1: a 100 39 20 19.0000000 100.00002: a 100 74 35 39.0000000 100.00003: a 100 67 19 48.0000000 100.00004: b 200 49 78 0.8650223 173.00455: b 200 41 70 0.8650223 149.65276: b 200 94 75 19.0000000 168.65277: c 200 74 51 23.0000000 200.00008: c 200 67 71 0.9801987 196.03979: c 200 77 73 4.0000000 200.0000
The problem I cannot solve is how to define the variable previous_content, i.e. at the beginning equal to the maximum capacity and then to the content of the previous row.
I checked the documentation of both data.table::frollapply and zoo:rollapply but I was not able even to write a starting draft for that. Any help? Thanks!
答案1
得分: 1
Assuming the input shown in the Note at the end content2 generated using Reduce is the same as content:
db[, content2 := Reduce(function(x, i) with(.SD[i, ], {if (prc > evpt) min(change+x, capacity)else x * change}), 1:.N, init = capacity[1], acc = TRUE)[-1], by = ID]db## ID capacity prc evpt change content content2## 1: a 100 39 20 19.0000000 100.0000 100.0000## 2: a 100 74 35 39.0000000 100.0000 100.0000## 3: a 100 67 19 48.0000000 100.0000 100.0000## 4: b 200 49 78 0.8650223 173.0045 173.0045## 5: b 200 41 70 0.8650223 149.6527 149.6527## 6: b 200 94 75 19.0000000 168.6527 168.6527## 7: c 200 74 51 23.0000000 200.0000 200.0000## 8: c 200 67 71 0.9801987 196.0397 196.0397## 9: c 200 77 73 4.0000000 200.0000 200.0000
In the example in the question capacity is always constant within ID, and change is multiplied if it is <= 1 and added otherwise. If those are true generally, we can simplify the above:
db[, content3 := Reduce(function(x, change)if (change > 1) min(change + x, capacity[1]) else x * change,change, init = capacity[1], acc = TRUE)[-1],by = ID]
Note
library(data.table)Lines <- "ID capacity prc evpt change contenta 100 39 20 19.0000000 100.0000a 100 74 35 39.0000000 100.0000a 100 67 19 48.0000000 100.0000b 200 49 78 0.8650223 173.0045b 200 41 70 0.8650223 149.6527b 200 94 75 19.0000000 168.6527c 200 74 51 23.0000000 200.0000c 200 67 71 0.9801987 196.0397c 200 77 73 4.0000000 200.0000"db <- fread(Lines)
英文:
Assuming the input shown in the Note at the end content2 generated using Reduce is the same as content:
db[, content2 := Reduce(function(x, i) with(.SD[i, ], {if (prc > evpt) min(change+x, capacity)else x * change}), 1:.N, init = capacity[1], acc = TRUE)[-1], by = ID]db## ID capacity prc evpt change content content2## 1: a 100 39 20 19.0000000 100.0000 100.0000## 2: a 100 74 35 39.0000000 100.0000 100.0000## 3: a 100 67 19 48.0000000 100.0000 100.0000## 4: b 200 49 78 0.8650223 173.0045 173.0045## 5: b 200 41 70 0.8650223 149.6527 149.6527## 6: b 200 94 75 19.0000000 168.6527 168.6527## 7: c 200 74 51 23.0000000 200.0000 200.0000## 8: c 200 67 71 0.9801987 196.0397 196.0397## 9: c 200 77 73 4.0000000 200.0000 200.0000
In the example in the question capacity is always constant within ID and change is multiplied if it is <= 1 and added otherwise. If those are true generally we can simplify the above:
db[, content3 := Reduce(function(x, change)if (change > 1) min(change + x, capacity[1]) else x * change,change, init = capacity[1], acc = TRUE)[-1],by = ID]
Note
library(data.table)Lines <- "ID capacity prc evpt change contenta 100 39 20 19.0000000 100.0000a 100 74 35 39.0000000 100.0000a 100 67 19 48.0000000 100.0000b 200 49 78 0.8650223 173.0045b 200 41 70 0.8650223 149.6527b 200 94 75 19.0000000 168.6527c 200 74 51 23.0000000 200.0000c 200 67 71 0.9801987 196.0397c 200 77 73 4.0000000 200.0000"db <- fread(Lines)
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