我的python方法在下面;
def leadtime_crossdock_calc(slt, wlt, dow, freq):
temp_lt = [0, 0, 0, 0, 0, 0, 0]
remaining = []
for i in range(0, 7):
remaining.append((dow[i:] + dow[:i]).index(1))
for i in range(7):
if freq[i] == 1:
supplier_lt = int(slt[i])
warehouse_lt = int(wlt[(i + supplier_lt) % 7])
waiting = int(remaining[(i + supplier_lt + warehouse_lt) % 7])
temp_lt[i] = supplier_lt + warehouse_lt + waiting
for i in range(7):
if temp_lt[i] == 0:
temp_lt[i] = next((value for index, value in enumerate(temp_lt[i:] + temp_lt[:i]) if value), None)
return ''.join(str(x) for x in temp_lt)
下面是例子;
leadtime_crossdock_calc([0,2,0,2,0,3,0],[1,1,1,1,1,1,1],[0,0,1,0,1,0,1],[0,1,0,1,0,1,0])
'3333443'
问题是,我有一个如下所示的spark数据框;
Product Store slt wlt dow freq
A B [0,2,0,2,0,3,0] [1,1,1,1,1,1,1] [0,0,1,0,1,0,1] [0,1,0,1,0,1,0]
我想使用上述方法为数据框中的每个新行创建一个新列;
Product Store slt wlt dow freq result
A B [0,2,0,2,0,3,0] [1,1,1,1,1,1,1] [0,0,1,0,1,0,1] [0,1,0,1,0,1,0] [3,3,3,3,4,4,3]
您能帮我吗?我无法将方法应用于spark数据框。
答案 0 :(得分:0)
您可以使用User Defined Functions or UDF 首先在spark上注册您的UDF,指定返回函数类型。您可以使用以下内容:
from pyspark.sql.types import StringType, col
leadtime_udf = spark.udf.register("leadtime_udf", leadtime_crossdock_calc, StringType())
然后,您可以将该UDF应用到DataFrame(或在Spark SQL中)
df.select("*", leadtime_udf(col(slt), ... , col(freq)))
希望这会有所帮助