Pyspark 圆函数问题 [英] Trouble With Pyspark Round Function

问题描述

在 pyspark 中使用 round 函数时遇到一些问题 - 我有下面的代码块，我试图将 new_bid 列四舍五入到小数点后两位，并将该列重命名为bid 之后 - 我正在导入 pyspark.sql.functions AS func 以供参考，并使用其中包含的 round 函数:

Having some trouble getting the round function in pyspark to work - I have the below block of code, where I'm trying to round the new_bid column to 2 decimal places, and rename the column as bid afterwards - I'm importing pyspark.sql.functions AS func for reference, and using the round function contained within it:

output = output.select(col("ad").alias("ad_id"),
                       col("part").alias("part_id"),
                       func.round(col("new_bid"), 2).alias("bid"))

这里的 new_bid 列是 float 类型 - 生成的数据框没有像我试图做的那样将新命名的 bid 列四舍五入到 2 个小数位，而是它仍然是小数点后 8 或 9 位.

the new_bid column here is of type float - the resulting dataframe does not have the newly named bid column rounded to 2 decimal places as I am trying to do, rather it is still 8 or 9 decimal places out.

我尝试了各种方法，但似乎无法使生成的数据帧具有四舍五入的值 - 任何指针都将不胜感激！谢谢！

I've tried various things but can't seem to get the resulting dataframe to have the rounded value - any pointers would be greatly appreciated! Thanks!

推荐答案

这里有几种方法可以处理一些玩具数据:

Here are a couple of ways to do it with some toy data:

spark.version
# u'2.2.0'

import pyspark.sql.functions as func

df = spark.createDataFrame(
        [(0.0, 0.2, 3.45631),
         (0.4, 1.4, 2.82945),
         (0.5, 1.9, 7.76261),
         (0.6, 0.9, 2.76790),
         (1.2, 1.0, 9.87984)],
         ["col1", "col2", "col3"])

df.show()
# +----+----+-------+ 
# |col1|col2|   col3|
# +----+----+-------+
# | 0.0| 0.2|3.45631| 
# | 0.4| 1.4|2.82945|
# | 0.5| 1.9|7.76261| 
# | 0.6| 0.9| 2.7679| 
# | 1.2| 1.0|9.87984| 
# +----+----+-------+

# round 'col3' in a new column:
df2 = df.withColumn("col4", func.round(df["col3"], 2)).withColumnRenamed("col4","new_col3")
df2.show()
# +----+----+-------+--------+ 
# |col1|col2|   col3|new_col3|
# +----+----+-------+--------+
# | 0.0| 0.2|3.45631|    3.46|
# | 0.4| 1.4|2.82945|    2.83|
# | 0.5| 1.9|7.76261|    7.76|
# | 0.6| 0.9| 2.7679|    2.77|
# | 1.2| 1.0|9.87984|    9.88|
# +----+----+-------+--------+

# round & replace existing 'col3':
df3 = df.withColumn("col3", func.round(df["col3"], 2))
df3.show()
# +----+----+----+ 
# |col1|col2|col3| 
# +----+----+----+ 
# | 0.0| 0.2|3.46| 
# | 0.4| 1.4|2.83| 
# | 0.5| 1.9|7.76| 
# | 0.6| 0.9|2.77| 
# | 1.2| 1.0|9.88| 
# +----+----+----+ 

这是个人品味，但我不喜欢 col 或 alias - 我更喜欢 withColumn 和 withColumnRenamed 代替.不过，如果您想坚持使用 select 和 col，以下是您应该如何调整自己的代码片段:

It's a personal taste, but I am not a great fan of either col or alias - I prefer withColumn and withColumnRenamed instead. Nevertheless, if you would like to stick with select and col, here is how you should adapt your own code snippet:

from pyspark.sql.functions import col

df4 = df.select(col("col1").alias("new_col1"), 
                col("col2").alias("new_col2"), 
                func.round(df["col3"],2).alias("new_col3"))
df4.show()
# +--------+--------+--------+ 
# |new_col1|new_col2|new_col3| 
# +--------+--------+--------+
# |     0.0|     0.2|    3.46|
# |     0.4|     1.4|    2.83|
# |     0.5|     1.9|    7.76|
# |     0.6|     0.9|    2.77|
# |     1.2|     1.0|    9.88|
# +--------+--------+--------+

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