【发布时间】:2020-01-09 20:15:01
【问题描述】:
我的python方法如下;
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)
下面的例子;
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]
你能帮我解决这个问题吗?我无法将方法应用于 spark 数据帧。
【问题讨论】:
标签: python function dataframe pyspark apache-spark-sql