如何更新列一个动态由其他列根据电池的价值观的另一个栏

0

的问题

我们需要更新列A,无论是通过列B或C柱或柱D。 其列为可用会决定通过列X其中包含的元数值作为"列B"或者"列C"或"D栏"

这需要在pyspark数据帧

apache-spark-sql pyspark python-3.x
2021-11-23 15:00:21
1

最好的答案

0

你可以使用 when 要实现这一点。

变量 possible_values 内部 derive_column_A() 控制所有可能列价值观。 从此我们动态地构造条件,以选择适当列。


from pyspark.sql import functions as F

data = [("B1", "C1", "D1", "column C"),
        ("B2", "C2", "D2", "column D"), 
        ("B3", "C3", "D3", "column B"), 
        ("B4", "C4", "D4", "column D")]

df = spark.createDataFrame(data, ("B", "C", "D", "X"))

def derive_column_A():
    possible_values = ["column B", "column C", "column D"]
    column_mapping = [{col, col.split(" ")[1]} for col in possible_values]
    condition = F
    for possible_value in possible_values:
        condition = condition.when(F.col("X") == possible_value, F.col(possible_value.split(" ")[1]))
    return condition


df.withColumn("A", derive_column_A()).show()

输出

+---+---+---+--------+---+
|  B|  C|  D|       X|  A|
+---+---+---+--------+---+
| B1| C1| D1|column C| C1|
| B2| C2| D2|column D| D2|
| B3| C3| D3|column B| B3|
| B4| C4| D4|column D| D4|
+---+---+---+--------+---+
2021-11-23 15:48:59

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