假设它存储在Pandas的Dataframe（名为df）中：

no_of_weeks = len(set(df['Week']))
check = pd.DataFrame(df.groupby('City').Week.apply(lambda x: x.count() == no_of_weeks))
always_recorded = pd.DataFrame({'City': check.index.values, 'always_recorded': check['Week']})
always_recorded.index = [*range(1, len(always_recorded)+1)]
df2 = pd.merge(df, always_recorded, how = "left", on = "City")
subset = df2[df2.always_recorded == True]

假设每个城市在给定日期只能出现一次，一种选择是使用关系划分技术：

select city
from mytable
group by city
having count(*) = (select count(distinct week) from mytable)

这将为您提供为所有日期显示的所有城市。
如果某个特定的城市在某个特定的日期出现不止一次，那么我们可以稍微改变一下 having 条款：

having count(distinct week) = (select count(distinct week) from mytable)

一个可能的解决方案是计算每个城市的周数，并将其与当前周数相匹配。
在sql中，可以执行以下操作

select 
   city, 
   count(week) over (partition by city) as appearance_num
from table_name