将Python列表转换为带索引的数据帧

Question

我有一个很大的python列表，其中包含许多格式的字符串

list = ['state1', 'town1','town2','town3', 'state2', 'town4', 'state3', 'town5','town6']

每个州有不同数量的城镇

如何将其嵌套，使其看起来像：

list = [['state1', 'town1','town2','town3'], ['state2', 'town4'],['state3', 'town5','town6']

然后把这个列表放到一个数据框中，州作为索引，城镇作为一列

Answer 1

那么，让我们首先来看一些列表示例：

state_lst = ['California', 'New Mexico', 'Arizona', 'etc.']
state_town_lst = ['California', 'San Francisco', 'Los Angeles', 'San Diego', 'New Mexico', 'Albuquerque', 'Santa Fe', 'Arizona', 'Tucson']
town_lst =[]

如你所见，加利福尼亚州应该有三个城市，新墨西哥州两个，亚利桑那州一个。因此，我们通过state_town_lst检查项目是否出现在state_lst中

for item in state_town_lst:
    if item in state_lst:
        state = item
        continue
    else:
        town = item
        
    town_item = (state, town)
    town_lst.append(town_item)
    
df = pd.DataFrame(town_lst, columns = ["State", "Town"])

这将为您提供：

    State       Town
0   California  San Francisco
1   California  Los Angeles
2   California  San Diego
3   New Mexico  Albuquerque
4   New Mexico  Santa Fe
5   Arizona     Tucson

Answer 2

让我们把清单看作：

lst = [['state', 'town','town','town'], ['state', 'town'],['state', 'town','town']]

要将其转换为以状态作为索引的数据帧，请执行以下操作：

df=pd.DataFrame(lst).set_index(0, drop=True)

输出：

0        1       2       3
            
state   town    town    town
state   town    None    None
state   town    town    None