如何提取r/sparklyr数据集中没有空值的列名？

Question

我只想提取 r 的大型数据集中没有空值的列名。

如果我的 table 有 4 列（id、Price、Product、Status），其中 Price 和 Status 列有一些空值，而列 id 和 Product 没有空值。然后我希望我的输出为：id, Product

Answer 1

data <- data.frame(ID = c(1,2,3,4),
                   Price = c(50, NA, 10, 20),
                   Product = c("A", "B", "C", "D"),
                   Status = c("Complete", NA, "Complete", "Incomplete"))

names(apply(data, 2, anyNA)[apply(data, 2, anyNA) == FALSE])

Answer 2

如果您需要一个确切的答案，您必须先扫描整个数据集，以计算缺失值：

library(dplyr)

df <- copy_to(sc, tibble(
  id = 1:4,  Price = c(NA, 3.20, NA, 42),
  Product = c("p1", "p2", "p3", "p4"),
  Status = c(NA, "foo", "bar", NA)))

null_counts <- df %>% 
    summarise_all(funs(sum(as.numeric(is.na(.)), na.rm=TRUE))) %>% 
    collect() 

null_counts

# A tibble: 1 x 4
     id Price Product Status
  <dbl> <dbl>   <dbl>  <dbl>
1     0     2       0      2

确定哪些列的缺失计数为零：

cols_without_nulls <- null_counts %>% 
  select_if(funs(. == 0)) %>% 
  colnames()

cols_without_nulls

[1] "id"      "Product"

并将这些用于 select

df %>% select(one_of(cols_without_nulls))

# Source: spark<?> [?? x 2]
     id Product
  <int> <chr>  
1     1 p1     
2     2 p2     
3     3 p3     
4     4 p4

存在更短的变体：

df %>% select_if(funs(sum(as.numeric(is.na(.)), na.rm=TRUE) == 0))

Applying predicate on the first 100 rows
# Source: spark<?> [?? x 2]
     id Product
  <int> <chr>  
1     1 p1     
2     2 p2     
3     3 p3     
4     4 p4

但如您所见，它只会对数据进行采样。

如何提取r/sparklyr数据集中没有空值的列名？

How to extract the column names which doesn't have any null values in a dataset in r/sparklyr?

r

sparklyr