R 加速应用

Question

我有以下循环脚本：

number_of_rows_similar_addresses <- as.data.table(cbind(
    distinct_similar_addresses,
    sapply(distinct_similar_addresses, function(x) {
        length(similar_addresses[Original_Address == x]$people_names) / length(unique(similar_addresses[Original_Address == x]$people_names))
    })
))

问题是它大大减慢了循环速度。

数据如下所示：

distinct_similar_addresses:

"U 2 5 TIMPERLEY ST NICHOLLS VIC"       
"U 1 3 TIMPERLEY ST NICHOLLS VIC"                            
"U 1 11 TIMPERLEY ST NICHOLLS VIC"                            
"U 1 33 TIMPERLEY ST NICHOLLS VIC"                           
"U 1 2 TIMPERLEY ST NICHOLLS VIC"                            
"U 1 3 TIMPERLEY ST NICHOLLS VIC"                            
"U 1 5 TIMPERLEY ST NICHOLLS VIC"

similar_addresses:

    people_names,Original_Address,Numbers,street_Name,street_type,post_code,suburb,PO,UID
Giuseppe Conte,U 1 3 TIMPERLEY ST NICHOLLS VIC,1,TIMPERLEY,ST,5469,NICHOLLS,,
Giuseppe Conte,U 1 3 TIMPERLEY ST NICHOLLS VIC,TIMPERLEY,ST,5469,NICHOLLS,,
Mario Pertini,U 2 5 TIMPERLEY ST NICHOLLS VIC,TIMPERLEY,ST,5469,NICHOLLS,,
Mario Pertini,U 2 5 TIMPERLEY ST NICHOLLS VIC,5,TIMPERLEY,ST,5469,NICHOLLS,,

脚本正在评估地址是指一个单元还是一个独立的房子。有什么方法可以更快地执行此任务？

我正在添加一个结果集和一个解释，以便它的作用变得更容易理解。

结果集：

   distinct_similar_addresses      V2
"U 2 5 TIMPERLEY ST NICHOLLS VIC"   2
"U 1 3 TIMPERLEY ST NICHOLLS VIC"   2

该代码只是计算与单行地址关联的姓名数。事实上，如果地址重复，则表示它指的是一个单元，否则就是一个单独的房子。

Answer 1

对于给您带来的数据不便，我们深表歉意，感谢 Roland 的帮助。

这就是解决方案

  x <- similar_addresses[, .N, by = Original_Address] %>% select('N')
  y <- similar_addresses[, length(unique(people_names)) , by = Original_Address] %>% select('V1')
  number_of_rows_similar_addresses <- cbind(unique(similar_addresses$Original_Address), x/y)

Answer 2

谢谢格雷戈尔，这可能更好：

 x <- similar_addresses[, .N, by = Original_Address]$N
 y <- similar_addresses[, length(unique(people_names)) , by = Original_Address]$V1
 number_of_rows_similar_addresses <- cbind(unique(similar_addresses$Original_Address), x/y)

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