我如何在嵌套的data.table-data.table中的data.table中进行FAST / ADVANCE数据操作 [英] How can i do FAST/ADVANCE data manipulation in nested data.table - data.table within data.table
问题描述
我在R中有一个名为 route_data 的数据表。我需要创建一个嵌套的data.table leg_data 到 route_data 的每一行，并从 route_data
I have a data.table named route_data in R. I need to create a nested data.table leg_data to each row of route_data with information extracted from each row of route_data
route_data <- data.table(route = c("Seattle>NewDelhi>Patna>Motihari", "Seattle>NewDelhi>Motihari","Seattle>Hyderabad>NewDelhi>Patna>Motihari"),
travel_type = c("business_meeting", "casual_trip","office_meeting"),
leg1_time_hr = c(18.0,18.0,18.0),
leg2_time_hr = c(2,18,2.25),
leg3_time_hr = c(4.0,NA,1.75),
leg4_time_hr = c(NA,NA,4.0))
route_data
route travel_type leg1_time_hr leg2_time_hr leg3_time_hr leg4_time_hr
1: Seattle>NewDelhi>Patna>Motihari business_meeting 18 2.00 4.00 NA
2: Seattle>NewDelhi>Motihari casual_trip 18 18.00 NA NA
3: Seattle>Hyderabad>NewDelhi>Patna>Motihari office_meeting 18 2.25 1.75 4
我需要在<$中创建嵌套的 leg_data c $ c> route_data 例如在第一行中应如下所示：
I need to create a nested leg_data in route_datafor example in the first row that should look like this:
example_nested_data <- data.table(leg = c("Seattle>Hyderabad", "Hyderabad>NewDelhi","NewDelhi>Patna","Patna>Motihari"),
leg_num = c(1,2,3,4),
leg_transit_time_hr = c(18.0,2.25,1.75,4.0)
)
example_nested_data 在 route_data
leg leg_num leg_transit_time_hr
1: Seattle>Hyderabad 1 18.00
2: Hyderabad>NewDelhi 2 2.25
3: NewDelhi>Patna 3 1.75
4: Patna>Motihari 4 4.00
类似地，在 route_data 的第二行和第三行中p>
Similarly, in the second and third row of route_data
推荐答案
我将尝试自己回答。我看到警告消息，希望对任何限制都可以更好地理解它。但是，对我来说，它工作正常（忽略警告消息）。
I am going to try answering it myself. I am seeing Warning message in a hope to understand it better for any limitations. However, for me it works fine (ignoring the warning message).
另一方面，data.table打破了R的所有限制，阻止了R进行大数据处理。
On a side note, data.table breaks all the limitation of R that stops it to do Big Data processing, and lest i forget my own research would like it to be documented.
同时，让我们创建一个函数来中断行程：
Meanwhile let us create a function that breaks up the route in legs:
construct.legs <- function(ro) {
node_vector <- unlist(strsplit(ro, ">"))
d_nodes <- node_vector[!node_vector %in% node_vector[1]]
o_nodes <- node_vector[!node_vector %in% node_vector[length(node_vector)]]
legs <- paste(o_nodes,d_nodes, sep = ">")
}
现在创建对于每个包含该路段的路线，嵌套 leg_table 。当然可以使用上面定义的函数 construct.legs ：
Now create nested leg_table for each route containing legs of the route. Of course using the function construct.legs that was defined above:
route_data[, leg_data := .(list(data.table(leg = construct.legs(route)))), by = list(row.names(route_data))]
我们的 route_data 看起来像现在吗？
How does our route_data look like now?
route travel_type leg1_time_hr leg2_time_hr leg3_time_hr leg4_time_hr leg_data
1: Seattle>NewDelhi>Patna>Motihari business_meeting 18 2.00 4.00 NA <data.table>
2: Seattle>NewDelhi>Motihari casual_trip 18 18.00 NA NA <data.table>
3: Seattle>Hyderabad>NewDelhi>Patna>Motihari office_meeting 18 2.25 1.75 4 <data.table>
让我们看看如果<$ c $的第三行中嵌套的data.table是什么c> route_data
route_data$leg_data[3] #Access the leg_table like we do in data.frame. But this returns leg_data as a list
route_data$leg_data[[3]] #This returns leg_data as a data.table
route_data[3, leg_data] #Access the leg_table like we do in data.table. This returns leg_data as a list
route_data[3, leg_data[[1]]] #This returns leg_data as a data.table
data.table存储在 route_data
data.table stored in the 3rd row of route_data
leg
1: Seattle>Hyderabad
2: Hyderabad>NewDelhi
3: NewDelhi>Patna
4: Patna>Motihari
让我在 route_data 中添加行号，稍后我将在填充运输中使用嵌套表中的时间 leg_data
Let me add row number in route_data tha i will use later in populating transit time within nested table leg_data
route_data[, route_num := seq_len(.N)]
类似地在嵌套表 leg_Table
route_data[, leg_data := .(list(leg_data[[1]][, leg_num := seq_len(.N)])), by = list(row.names(route_data))]
您看到一条警告消息，指出存在无效的内部自我参照，该自我参照已通过浅层复制得到了修复。因此，到目前为止，我将忽略这一点。在这里，我需要有人的帮助，该人可以帮助我了解它是否有任何故障。无论如何，让我们继续。
You see a Warning message that says there was invalid internal self reference that has been fixed by shallow copying. So, i am going to ignore this as of now. I would need help here from someone who can help me understand if it breaks anything. Anyway, lets proceed.
为什么我们有 [[1]] ？这是为了确保sub_table值以data.table而不是list的形式返回。尝试运行 route_data [3，leg_data [[1]]] 和 route_data [3，leg_data] 来查看区别。
Why do we have [[1]]? This is to ensure that sub_table values are returned as data.table, not as list. Try running route_data[3, leg_data[[1]]] and route_data[3, leg_data] to see the difference.
现在终于在 route_data leg_data 中添加了运输时间$ c>
Now finally add the transit time in nested leg_data from route_data
route_data[, leg_data := .(list(leg_data[[1]][, leg_transit_time_hr := sapply(leg_num, function(x) {route_data[[route_num, 2+x, with = FALSE]]})])), by = list(row.names(route_data))]
我们在这里做什么？
我们只是循环输入了行号 route_num route_data 来确定要从 route_data 中提取的运输时间的右列。
We just looped in row number leg_num of leg_data via sapply by passing it as vector and utilized the row number route_num of route_data to identify right column of transit time to extract from the route_data.
我们为什么在 [[route_num，2 + x，= FALSE]]上放置双 [[]] / code>？
Why did we place double [[]] on the [[route_num, 2+x, with = FALSE]]?
大括号确保其返回的值是向量而不是数据表。
Double braces ensure it returns value as vector not as data.table
最后，让我们看一下嵌套的数据。 route_data 第三行的表 leg_data
And, finally, let's take a look into the nested data.table leg_data of 3rd row of route_data
route_data[3, leg_data[[1]]]
leg leg_num leg_transit_time_hr
1: Seattle>Hyderabad 1 18.00
2: Hyderabad>NewDelhi 2 2.25
3: NewDelhi>Patna 3 1.75
4: Patna>Motihari 4 4.00
让我们看看第二行嵌套表的外观：
Let's see how 2nd row nested table looks like:
leg leg_num leg_transit_time_hr
1: Seattle>NewDelhi 1 18
2: NewDelhi>Motihari 2 18
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