简体   繁体   English

Pyspark - 计算每个数据框列中空值的数量

[英]Pyspark - Calculate number of null values in each dataframe column

I have a dataframe with many columns.我有一个包含许多列的数据框。 My aim is to produce a dataframe thats lists each column name, along with the number of null values in that column.我的目标是生成一个列出每个列名称的数据框,以及该列中空值的数量。

Example:例子:

+-------------+-------------+
| Column_Name | NULL_Values |
+-------------+-------------+
|  Column_1   |      15     |
|  Column_2   |      56     |
|  Column_3   |      18     |
|     ...     |     ...     |
+-------------+-------------+

I have managed to get the number of null values for ONE column like so:我已经设法获得一列的空值数量,如下所示:

df.agg(F.count(F.when(F.isnull(c), c)).alias('NULL_Count'))

where c is a column in the dataframe.其中c是数据框中的一列。 However, it does not show the name of the column.但是,它不显示列的名称。 The output is:输出是:

+------------+
| NULL_Count |
+------------+
|     15     |
+------------+

Any ideas?有任何想法吗?

You can use a list comprehension to loop over all of your columns in the agg , and use alias to rename the output column:您可以使用列表推导循环遍历agg中的所有列,并使用alias重命名输出列:

import pyspark.sql.functions as F

df_agg = df.agg(*[F.count(F.when(F.isnull(c), c)).alias(c) for c in df.columns])

However, this will return the results in one row as shown below:但是,这将在一行中返回结果,如下所示:

df_agg.show()
#+--------+--------+--------+
#|Column_1|Column_2|Column_3|
#+--------+--------+--------+
#|      15|      56|      18|
#+--------+--------+--------+

If you wanted the results in one column instead, you could union each column from df_agg using functools.reduce as follows:如果您希望将结果放在一列中,您可以使用functools.reducedf_agg每一列df_agg ,如下所示:

from functools import reduce
df_agg_col = reduce(
    lambda a, b: a.union(b),
    (
        df_agg.select(F.lit(c).alias("Column_Name"), F.col(c).alias("NULL_Count")) 
        for c in df_agg.columns
    )
)
df_agg_col.show()
#+-----------+----------+
#|Column_Name|NULL_Count|
#+-----------+----------+
#|   Column_1|        15|
#|   Column_2|        56|
#|   Column_3|        18|
#+-----------+----------+

Or you can skip the intermediate step of creating df_agg and do:或者您可以跳过创建df_agg的中间步骤并执行以下操作:

df_agg_col = reduce(
    lambda a, b: a.union(b),
    (
        df.agg(
            F.count(F.when(F.isnull(c), c)).alias('NULL_Count')
        ).select(F.lit(c).alias("Column_Name"), "NULL_Count")
        for c in df.columns
    )
)

Scala alternative could be Scala 替代方案可能是

case class Test(id:Int, weight:Option[Int], age:Int, gender: Option[String])

val df1 = Seq(Test(1, Some(100), 23, Some("Male")), Test(2, None, 25, None), Test(3, None, 33, Some("Female"))).toDF()

df1.show()

+---+------+---+------+
| id|weight|age|gender|
+---+------+---+------+
|  1|   100| 23|  Male|
|  2|  null| 25|  null|
|  3|  null| 33|Female|
+---+------+---+------+

val s = df1.columns.map(c => sum(col(c).isNull.cast("integer")).alias(c))

val df2 = df1.agg(s.head, s.tail:_*)

val t = df2.columns.map(c => df2.select(lit(c).alias("col_name"), col(c).alias("null_count")))

val df_agg_col = t.reduce((df1, df2) => df1.union(df2))

df_agg_col.show()

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM