在DataFrame中将新的派生列从布尔值转换为整数

问题描述：

假设我有一个具有以下架构的DataFrame x:

Suppose I have a DataFrame x with this schema:

xSchema = StructType([ \
    StructField("a", DoubleType(), True), \
    StructField("b", DoubleType(), True), \
    StructField("c", DoubleType(), True)])

然后我有了DataFrame:

I then have the DataFrame:

DataFrame[a :double, b:double, c:double]

我想有一个整数派生的列.我可以创建一个布尔列:

I would like to have an integer derived column. I am able to create a boolean column:

x = x.withColumn('y', (x.a-x.b)/x.c > 1)

我的新架构是:

DataFrame[a :double, b:double, c:double, y: boolean]

但是，我希望列y的False包含0，True包含1.

However, I would like column y to contain 0 for False and 1 for True.

cast函数只能在列上运行，而不能在DataFrame上运行，而withColumn函数只能在DataFrame上运行.如何添加新列并将其同时转换为整数?

The cast function can only operate on a column and not a DataFrame and the withColumn function can only operate on a DataFrame. How to I add a new column and cast it to integer at the same time?

答

您使用的表达式求值到列，因此您可以像这样直接转换:

Expression you use evaluates to column so you can cast directly like this:

x.withColumn('y', ((x.a-x.b) / x.c > 1).cast('integer')) # Or IntegerType()

在DataFrame中将新的派生列从布尔值转换为整数

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