我正在尝试检查何时以及是否满足条件的多个列值0。我们的Spark数据框的列从1到11,需要检查其值。目前,我的代码如下:
df3 =df3.withColumn('Status', when((col("1") ==0)|(col("2") ==0)|(col("3") ==0)| (col("4") ==0) |(col("5") ==0)|(col("6") ==0)|(col("7") ==0)| (col("8") ==0)|(col("9") ==0)|(col("10") ==0)| (col("11") ==0) ,'Incomplete').otherwise('Complete'))
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我如何仅通过使用for循环而不是那么多or条件来实现此目的
我提出了一个更pythonic的解决方案。使用functools.reduce和operator.or_。
import operator
import functools
colnames = [str(i+1) for i in range(11)]
df1 = spark._sc.parallelize([
[it for it in range(11)],
[it for it in range(1,12)]]
).toDF((colnames))
df1.show()
+---+---+---+---+---+---+---+---+---+---+---+
| 1| 2| 3| 4| 5| 6| 7| 8| 9| 10| 11|
+---+---+---+---+---+---+---+---+---+---+---+
| 0| 1| 2| 3| 4| 5| 6| 7| 8| 9| 10|
| 1| 2| 3| 4| 5| 6| 7| 8| 9| 10| 11|
+---+---+---+---+---+---+---+---+---+---+---+
cond_expr = functools.reduce(operator.or_, [(f.col(c) == 0) for c in df1.columns])
df1.withColumn('test', f.when(cond_expr, f.lit('Incomplete')).otherwise('Complete')).show()
+---+---+---+---+---+---+---+---+---+---+---+----------+
| 1| 2| 3| 4| 5| 6| 7| 8| 9| 10| 11| test|
+---+---+---+---+---+---+---+---+---+---+---+----------+
| 0| 1| 2| 3| 4| 5| 6| 7| 8| 9| 10|Incomplete|
| 1| 2| 3| 4| 5| 6| 7| 8| 9| 10| 11| Complete|
+---+---+---+---+---+---+---+---+---+---+---+----------+
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这样,您无需定义任何函数,评估字符串表达式或使用python lambdas。希望这可以帮助。
您可以使用下面的代码来收集条件并将它们连接到单个字符串中,然后调用eval.
代码
cond ='|'.join('(col("'+str(_)+'")==0)' for _ in range(1, 12))
cond = '('+cond+')'
print(cond)
#((col("1")==0)|(col("2")==0)|(col("3")==0)|(col("4")==0)|(col("5")==0)|(col("6")==0)|(col("7")==0)|(col("8")==0)|(col("9")==0)|(col("10")==0)|(col("11")==0))
df3 = df3.withColumn('Status', when(eval(cond),'Incomplete').otherwise('Complete'))
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