如何將Python串列轉換為PandasDataFrame：

我有以下已簡化的串列：

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

請注意我的串列如何具有“select”和“from”陳述句。

我試圖實作的輸出是一個 DataFrame 或者讓我們說 Excel 輸出：

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

有沒有辦法達到這個結果？我已經嘗試了很多東西，但它不起作用.. :(

uj5u.com熱心網友回復：

類似于下面的內容（使用簡單的“狀態機”）

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)

輸出

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

uj5u.com熱心網友回復：

有很多方法可以做到這一點。這種方法獲取所有 'from' 的索引，并使用前面拆分 2 個空格，np.split以便每個新陣列的開始都是一個 'select'。最后一個是空的，所以我們將放棄它。

然后，您可以通過對每個陣列進行切片來構建一個 dict，并從中生成一個資料幀。

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')

輸出

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

uj5u.com熱心網友回復：

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

uj5u.com熱心網友回復：

制作一個通用且可重用的split功能，就像這個問題的答案中的功能一樣。然后更容易從每個拆分組中產生對。

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

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