如何将 Python 列表转换为 pandas DataFrame：

Question

我有下面的列表，我已经简化了：

my_list = ['select', 'fruit1', 'fruit2, 'fruit3', 'from', 'basket1',
           'select', 'fruit4', 'from', 'basket2',
           'select', 'fruit5', 'fruit6' 'from', 'basket3', ..... so on]

请注意我的列表如何包含“选择”和“来自”语句。

我想要实现的 output 是 DataFrame 或者说 Excel Z78E6221F6393D14CE5668：

Fruit number      Basket number
fruit1            basket1
fruit2            basket1
fruit3            basket1
fruit4            basket2
fruit5            basket3
fruit6            basket3
.                 .
.                 .
.                 .
.                 .

有没有办法达到这个结果？ 我已经尝试了很多东西，但它不会工作.. :(

Answer 1

类似下面的东西（使用一个简单的“状态机”）

import pandas as pd
lst = ['select', 'fruit1', 'fruit2', 'fruit3', 'from', 'basket1',
       'select', 'fruit4', 'from', 'basket2',
       'select', 'fruit5', 'fruit6', 'from', 'basket3']

data = []
fruits = []
state = 'select'
for word in lst:
  if word == 'select':
    state = 'select'
    continue
  if word == 'from':
    state = 'basket'
    continue
  if state == 'select':
    fruits.append(word)
  if state == 'basket':
    for f in fruits:
      data.append({'fruit':f,'basket':word})
    fruits = []

df = pd.DataFrame(data)
print(df)

output

    fruit   basket
0  fruit1  basket1
1  fruit2  basket1
2  fruit3  basket1
3  fruit4  basket2
4  fruit5  basket3
5  fruit6  basket3

Answer 2

有很多方法可以做到这一点。 这种方法获取所有“来自”的索引，并使用np.split将 2 个空格向前拆分，以便每个新数组的开头都是一个“选择”。 最后一个是空的，所以我们将删除它。

然后你可以通过分割每个数组来构建一个字典，并从中制作一个 dataframe 。

import numpy as np
import pandas as pd
my_list = ['select', 'fruit1', 'fruit2', 'fruit3', 'from', 'basket1',
           'select', 'fruit4', 'from', 'basket2',
          'select', 'fruit5', 'fruit6', 'from', 'basket3']

f = [i+2 for i, x in enumerate(my_list) if x == "from"][:-1]
s = np.split(my_list,f)

df = pd.DataFrame([{'basket':q[-1],'fruits':q[1:-2]} for q in s])
df = df.explode('fruits')

Output

    basket  fruits
0  basket1  fruit1
0  basket1  fruit2
0  basket1  fruit3
1  basket2  fruit4
2  basket3  fruit5
2  basket3  fruit6

Answer 3

data = {'Select' : {'Fruit_Number': 
['fruit1','fruit2','fruit3']},'From' : {'Basket_Number': 
['basket1','basket2','basket3']}}

data2 = data['Select']
data3 = data['From']

df2 = pd.DataFrame.from_dict(data2)
df3 = pd.DataFrame.from_dict(data3)

l = [df2,df3]
df_all = pd.concat(l,axis=1)


      Fruit_Number Basket_Number
0       fruit1       basket1
1       fruit2       basket2
2       fruit3       basket3

Answer 4

制作一个通用且可重复使用的split function ，就像这个问题的答案中的那样。 然后更容易从每个拆分组中产生对。

def split(sequence, sep):
    group = []
    for item in sequence:
        if item == sep:
            yield group
            group = []
        else:
            group.append(item)
    yield group
    
def parse_select(tokens):
    for group in split(tokens, "select"):
        for item in group[:-2]:
            yield item, group[-1]
        
import pandas as pd
print(pd.DataFrame(parse_select(my_list)))

或者：

def parse_select(tokens):
    for group in split(tokens, "select"):
        if group:
            items, (basket,) = split(group, "from")
            for item in items:
                yield item, basket