databricks udf 广播字典值返回字典列表；无法访问该列表中字典的值

Question

下面的代码是基于udf dictionary broadcast 博客发布的。在下面的代码中，广播字典应该返回一个字典列表，即预期从广播字典返回 {'key':[{'key':'value','key2':'value'},{'key':'value ','key2':'值'}]

当我尝试从列表中的字典访问任何键时，不会返回任何值。我知道这一点是因为我收到错误“分配前引用的局部变量 'tmp'”。此变量应包含由列表中每个字典中定义的操作产生的计算值。谁能告诉我为什么或如何调试这个或为什么字典键没有返回值？我在返回广播密钥的代码中添加了打印语句，但它从未出现在输出或驱动程序日志中。

广播词典

calc dict {'name': 'Lvl1_Name_Score', 'operations': [{'operator': 'multi', 'value': 1.5}, {'operator': 'div', 'value': 1}]}

def apply_weights_a(calc_broadcasted,key): 
    def apply_weights(col_name):
        calc_operations = calc_broadcasted.value.get(key)
        results = 0
        print(f'calc_operations {calc_operations}')
        for op in calc_operations:
            print(f'op {op}')
            # for i,op in enumerate(calc['dependencies']
            if op['operator'] == 'div': tmp = col_name / float(op['value'])
            elif op['operator'] == 'multi': tmp = col_name * float(op['value'])
            elif op['operator'] == 'sub':tmp = col_name - float(op['value'])
            elif op['operator'] == 'add': tmp = col_name + float(op['value'])
        # sdf = sdf['tmp'].apply(lambda x: 0 if x < 0 else x)
        results = results + tmp
        return results
    return udf(apply_weights)


def multi_apply_weights(col_names,weights): 
    def inner(sdf):
        for col_name in col_names:
            #print(f"weights {weights}")
            size = len(col_name)  #get col_name without _c
            calc = [mylist for mylist in weights if mylist['name'] == col_name[:size-2] ]
            #print(f"calc {calc}")
            calc_dict = calc[0]
            #print(f"calc dict {calc[0]}")
            b = spark.sparkContext.broadcast(calc_dict)
            sdf = sdf.withColumn(
                col_name,
                apply_weights_a(b,'operations')(col_name)
            )
        return sdf
    return inner

sdf = multi_apply_weights(weight_list,config['algorithmScores'])(sdf)
                                    

此 spark 代码是转换此 pandas 函数的结果，该函数在一组列上执行 1 到多个不同的计算...

def apply_weights(df, weights):
    for calc in weights:
        col = calc['name']
        for op in calc['operations']:
            if op['operator'] == 'div':
                df[col + '_w'] = df[col]/float(op['value'])
            elif op['operator'] == 'multi':
                df[col+ '_w'] = df[col]*float(op['value'])
            elif op['operator'] == 'sub':
                df[col+ '_w'] = df[col]-float(op['value'])
            elif op['operator'] == 'add':
                df[col+ '_w'] = df[col]+float(op['value'])
        df[col] = df[col].apply(lambda x: 0 if x < 0 else x)

    return df

Answer 1

我发现我可以通过做一个小的改动来运行原始的 python 代码。这适用于我的 4 个功能中的 3 个。在我的最后一个函数上使用这种方法时，它导致我的集群崩溃并出现错误“火花驱动程序已意外停止并正在重新启动。您的笔记本将自动重新附加'。我已经向 microsoft 开了一张票，以获得额外的帮助。

def apply_weights(kdf, weights) -> ks.DataFrame[zip(kdf.dtypes, kdf.columns)]:

我还解决了广播字典的问题，方法是在广播之前添加一个键。我把广播改​​成

brdcst = dict({"rules":newcolrules})
b = spark.sparkContext.broadcast(brdcst) 

calcs = b.value.get('rules') #test broadcasted dictionary
print(f"calcs : {calcs}") #test broadcasted dictionary