舍入双精度值并转换为整数 [英] Round double values and cast as integers

问题描述

我在 PySpark 中有一个数据框，如下所示.

I have a data frame in PySpark like below.

import pyspark.sql.functions as func

df = sqlContext.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))
df2.show()

+----+----+-------+----+
|col1|col2|   col3|col4|
+----+----+-------+----+
| 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|
+----+----+-------+----+

在上面的数据框中col4是double.现在我想将 col4 转换为 Integer

In the above data frame col4 is double. Now I want to convert col4 as Integer

df2 = df.withColumn("col4", func.round(df["col3"], 2).cast('integer'))

+----+----+-------+----+
|col1|col2|   col3|col4|
+----+----+-------+----+
| 0.0| 0.2|3.45631|   3|
| 0.4| 1.4|2.82945|   2|
| 0.5| 1.9|7.76261|   7|
| 0.6| 0.9| 2.7679|   2|
| 1.2| 1.0|9.87984|   9|
+----+----+-------+----+

但我想将 col4 值四舍五入到最接近的

But I want to round the col4 values to nearest

预期结果

+----+----+-------+----+
|col1|col2|   col3|col4|
+----+----+-------+----+
| 0.0| 0.2|3.45631|   3|
| 0.4| 1.4|2.82945|   3|
| 0.5| 1.9|7.76261|   8|
| 0.6| 0.9| 2.7679|   3|
| 1.2| 1.0|9.87984|  10|
+----+----+-------+----+

我该怎么做?

推荐答案

您应该使用 round 函数，然后转换为整数类型.但是，不要对 round 函数使用第二个参数.通过在那里使用 2，它将四舍五入到 2 个小数位，cast 到整数将 向下舍入到最接近的数字.

You should use the round function and then cast to integer type. However, do not use a second argument to the round function. By using 2 there it will round to 2 decimal places, the cast to integer will then round down to the nearest number.

改为使用:

df2 = df.withColumn("col4", func.round(df["col3"]).cast('integer'))

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