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我有数据“ li”，我想运行算法 FPGrowth，但我不知道如何

set.seed(123)

# make fake data
li <- list()
for(i in 1:10) li[[i]] <- make.unique(letters[sample(1:26,sample(5:20,1),rep = T)])
 

require(sparklyr)
sc <- spark_connect(master = "local",version = "3.0.1")


df <- copy_to(sc,  **....??????what should be here??????...**  )
            fp_growth_model <- ml_fpgrowth(df)

这里有一个类似的答案，但它不起作用，我得到了错误

sc <- spark_connect(master = "local", version = "2.3")
tb <- tibble::tibble(items=c("a b c", "a b", "c f g", "b c"))

df <- copy_to(sc, tb)  %>% 
 mutate(items = split(items, "\\\\s+"))

Error in mutate(., items = split(items, "\\\\s+")) : 
  could not find function "mutate"

/// plyr::mutate

df <- copy_to(sc, tb)  %>% 
  plyr::mutate(items = split(items, "\\\\s+"))

Error in sdf_import.default(x, sc, name, memory, repartition, overwrite,  : 
  table tb already exists (pass overwrite = TRUE to overwrite)

/// SparkR::mutate

df <- copy_to(sc, tb)  %>% 
  SparkR::mutate(items = split(items, "\\\\s+"))

Error in sdf_import.default(x, sc, name, memory, repartition, overwrite,  : 
  table tb already exists (pass overwrite = TRUE to overwrite)
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1 回答 1

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上述答案中的代码示例有效。你得到两个错误，第一个是因为mutate没有加载。第二个是因为对象tb已经加载到 Spark 中。

尝试从新会话中运行以下代码：

library(tidyverse)
library(sparklyr)

sc <- spark_connect(master = "local")

tb <- tibble::tibble(items=c("a b c", "a b", "c f g", "b c"))

df <- copy_to(sc, tb)  %>% 
  mutate(items = split(items, "\\\\s+"))

fp_growth_model <- ml_fpgrowth(df)

ml_association_rules(fp_growth_model)
ml_freq_itemsets(fp_growth_model)

要使用您的数据集执行 FP-growth li，您需要更改格式。该函数ml_fpgrowth需要一个带有包含序列的列表列的 SparkDataFrame。您不能将带有列表的 R DataFrame 直接传输到 Spark。首先，您创建一个 SparkDataFrame，其中序列为字符串，然后使用mutate和split函数生成列表。

这是应用于您的数据的代码。

> tb_li <- tibble(items=sapply(li, function(x) paste(x, collapse=" ")))
> tb_li
# A tibble: 10 x 1
   items                                            
   <chr>                                            
 1 o s n c j r v k e t n.1 v.1 y z e.1 s.1 y.1 y.2 i
 2 c h z g j i s d n q k g.1 u l o j.1 m            
 3 i i.1 j w u g u.1 f y b e                        
 4 l m r a y y.1 f u o i o.1 z                      
 5 p t f k h v v.1 g p.1 q v.2 r q.1 b d m          
 6 v s y t v.1 y.1 n y.2 w                          
 7 h p l y n c n.1                                  
 8 g c w v z o u e h s j r j.1 l b j.2 v.1          
 9 l t n q n.1 v c h n.2 s o x q.1 w k g o.1 w.1 z  
10 n g j e f p x u w k

将数据传输到 Spark 并生成列表：

> df_li <- copy_to(sc, tb_li, overwrite = TRUE) %>%
+   mutate(items = split(items, "\\\\s+"))
> df_li
# Source: spark<?> [?? x 1]
   items      
   <list>     
 1 <list [19]>
 2 <list [17]>
 3 <list [11]>
 4 <list [12]>
 5 <list [16]>
 6 <list [9]> 
 7 <list [7]> 
 8 <list [17]>
 9 <list [19]>
10 <list [10]>

数据已准备好供模型使用，如上面的示例。

> fp_growth_model_li <- ml_fpgrowth(df_li)

> ml_association_rules(fp_growth_model_li)
# Source: spark<?> [?? x 4]
   antecedent consequent confidence  lift
   <list>     <list>          <dbl> <dbl>
 1 <list [4]> <list [1]>          1     2
 2 <list [3]> <list [1]>          1     2
 3 <list [3]> <list [1]>          1     2
 4 <list [3]> <list [1]>          1     2
 5 <list [5]> <list [1]>          1     2
 6 <list [5]> <list [1]>          1     2
 7 <list [3]> <list [1]>          1     2
 8 <list [3]> <list [1]>          1     2
 9 <list [3]> <list [1]>          1     2
10 <list [3]> <list [1]>          1     2
# ... with more rows

> ml_freq_itemsets(fp_growth_model_li)
# Source: spark<?> [?? x 2]
   items       freq
   <list>     <dbl>
 1 <list [1]>     3
 2 <list [2]>     3
 3 <list [3]>     3
 4 <list [2]>     3
 5 <list [1]>     5
 6 <list [2]>     3
 7 <list [3]>     3
 8 <list [3]>     3
 9 <list [4]>     3
10 <list [2]>     4
# ... with more rows
于 2021-01-23T22:03:16.293 回答