为了清晰和简洁而重写 tidyverse 函数

Question

我使用chatgpt进行一些编码（羞愧我），我有这个功能我不明白：

escalc.df <- allmetadata %>%
  group_by(!!sym(testvariable)) %>%
  summarise(group = first(!!sym(testvariable)),
         ai = sum(!!sym(hydratelevel) == "Present" & modelpresence > 0),
         bi = sum(!!sym(hydratelevel) == "Present" & modelpresence == 0),
         ci = sum(!!sym(hydratelevel) == "Absent" & modelpresence > 0),
         di = sum(!!sym(hydratelevel) == "Absent" & modelpresence == 0))

我不知道从哪里开始，因为我不完全理解这个功能。有人可以向我解释这段代码，甚至分享一个缩短/更简化的版本吗？

我期望 df

allmetadata

按

testvariable

分组（之前设置以匹配特定的

allmetadata

列），然后我想对

allmetadata$modelpresence

中每个变量 > 或 = 为 0 时的实例求和allmetadata$hydratelevel

。这些值应输出到具有四列的新 df 

escalc.df

 中：

$ai

、

$bi

、

$ci

 和 

$di

。

例如，

testvariable

可以是

$surfacelith

，

hydratelevel

可以是

$AreaKnownHydrate

。

> dput(head(allmetadata))
structure(list(feature.id = c("AB094456", "AB094457", "AB094458", 
"AB094459", "AB094460", "AB094461"), seq = c("cct", "cct", "cct", 
"cct", "cct", "cct"), author = c("Inagaki", "Inagaki", "Inagaki", 
"Inagaki", "Inagaki", "Inagaki"), yearPub = c(2003L, 2003L, 2003L, 
2003L, 2003L, 2003L), yearCollected = c(2001L, 2001L, 2001L, 
2001L, 2001L, 2001L), ocean = c("Pacific", "Pacific", "Pacific", 
"Pacific", "Pacific", "Pacific"), region = c("SeaOkhotsk", "SeaOkhotsk", 
"SeaOkhotsk", "SeaOkhotsk", "SeaOkhotsk", "SeaOkhotsk"), location = c("ShiretokoPeninsula", 
"ShiretokoPeninsula", "ShiretokoPeninsula", "ShiretokoPeninsula", 
"ShiretokoPeninsula", "ShiretokoPeninsula"), waterType = c("marine", 
"marine", "marine", "marine", "marine", "marine"), methaneForm = c("HYD", 
"HYD", "HYD", "HYD", "HYD", "HYD"), waterDepth = c(1225, 1225, 
1225, 1225, 1225, 1225), sedDepth = c("UNK", "UNK", "UNK", "UNK", 
"UNK", "UNK"), latitude = c(44.5275, 44.5275, 44.5275, 44.5275, 
44.5275, 44.5275), longitude = c(145.0041, 145.0041, 145.0041, 
145.0041, 145.0041, 145.0041), sedProfile = c("UNK", "UNK", "UNK", 
"UNK", "UNK", "UNK"), sampleType = c("sediment", "sediment", 
"sediment", "sediment", "sediment", "sediment"), porosity = c(81.5954, 
81.5954, 81.5954, 81.5954, 81.5954, 81.5954), surfaceTOC = c(2.1923, 
2.1923, 2.1923, 2.1923, 2.1923, 2.1923), surfacelith = c("clay", 
"clay", "clay", "clay", "clay", "clay"), locationUSGSdatabase = c(NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_
), LatUSGSdatabase = c(NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_), LongUSGSdatabase = c(NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_), AreaKnownHydrate = c("Absent", 
"Absent", "Absent", "Absent", "Absent", "Absent"), ExactHydratePresent = c("UNK", 
"UNK", "UNK", "UNK", "UNK", "UNK"), hydInfoSource = c("UNK", 
"UNK", "UNK", "UNK", "UNK", "UNK"), modelpresence = c(0, 0, 0, 
0, 0, 0)), row.names = c(NA, 6L), class = "data.frame")

Answer 1

您需要定义一个函数然后调用它。 例如：

myFun <- function(testvariable, hydratelevel) {
   escalc.df <- allmetadata %>%
   group_by(!!sym(testvariable)) %>%
   summarise(group = first(!!sym(testvariable)),
             ai = sum(!!sym(hydratelevel) == "Present" & modelpresence > 0),
             bi = sum(!!sym(hydratelevel) == "Present" & modelpresence == 0),
             ci = sum(!!sym(hydratelevel) == "Absent" & modelpresence > 0),
             di = sum(!!sym(hydratelevel) == "Absent" & modelpresence == 0))
}
returned.df <- myFun("surfacelith", "AreaKnownHydrate")
returned.df