如果字典列表中存在键，则获取值

Question

我有一个类似python的列表

df.groupBy("geohash").agg(collect_list("timehash")).alias("timehash").show

//output
+-------+--------+
|geohash|timehash|
+-------+--------+
|      x|  [y, z]|
|      z|     [y]|
+-------+--------+

并希望像这样生成它

[{'month': 8, 'total': 31600.0}, {'month': 9, 'total': 2000.0}]

为此，我正在[ {'month': 1, 'total': 0}, {'month': 2, 'total': 0}, ... {'month': 8, 'total': 31600}, {'month': 9, 'total': 2000}, ... {'month': 12, 'total': 0} ] 范围内运行迭代

(1,13)

如何检查 i 在 month 中是否存在并获取字典？

Answer 1

您可以将字典列表转换为直接月份：

monthly_totals = {item['month']: item['total'] for item in data_list}

并使用带有dict.get的简单列表理解来处理缺失值：

new_list = [{'month': i, 'total': monthly_totals.get(i, 0)} for i in range(1, 13)]

Answer 2

创建一个包含默认值的新列表，然后从原始列表中更新所需的值

>>> lst = [{'month': 8, 'total': 31600.0}, {'month': 9, 'total': 2000.0}]
>>> new_lst = [dict(month=i, total=0) for i in range(1,13)]
>>> for d in lst:
...     new_lst[d['month']-1] = d
... 
>>> pprint(new_lst)
[{'month': 1, 'total': 0},
 {'month': 2, 'total': 0},
 {'month': 3, 'total': 0},
 {'month': 4, 'total': 0},
 {'month': 5, 'total': 0},
 {'month': 6, 'total': 0},
 {'month': 7, 'total': 0},
 {'month': 8, 'total': 31600.0},
 {'month': 9, 'total': 2000.0},
 {'month': 11, 'total': 0},
 {'month': 12, 'total': 0}]

Answer 3

exist_lst =  [{'month': 8, 'total': 31600.0}, {'month': 9, 'total': 2000.0}]

new_lst = []

for i in range(1,13):
    found = False
    for dict_item in exist_lst:
        if dict_item['month'] == i:
            new_lst.append(dict_item)
            found = True

    if not found: 
        new_lst.append({'month': i, 'total': 0}) # default_dict_item

print(new_lst)