Python 将相同字典列表转换为数据帧

我有这样一份清单：

[{'FirstOfficer': '1'}, {'SecondOfficer': '2'}, {'ThirdOfficer': '3'},{'FirstOfficer': '4'}, {'SecondOfficer': '5'}, {'ThirdOfficer': '6'},{'FirstOfficer': '7'}, {'SecondOfficer': '8'}, {'ThirdOfficer': '9'},{'FirstOfficer': '10'}, {'SecondOfficer': '11'}, {'ThirdOfficer': '12'}]

   FirstOfficer SecondOfficer ThirdOfficer
0             1           NaN          NaN
1           NaN             2          NaN
2           NaN           NaN            3
3             4           NaN          NaN
4           NaN             5          NaN
5           NaN           NaN            6
6             7           NaN          NaN
7           NaN             8          NaN
8           NaN           NaN            9
9            10           NaN          NaN
10          NaN            11          NaN
11          NaN           NaN           12

我想将其转换为数据帧，但得到的数据帧如下：

[{'FirstOfficer': '1'}, {'SecondOfficer': '2'}, {'ThirdOfficer': '3'},{'FirstOfficer': '4'}, {'SecondOfficer': '5'}, {'ThirdOfficer': '6'},{'FirstOfficer': '7'}, {'SecondOfficer': '8'}, {'ThirdOfficer': '9'},{'FirstOfficer': '10'}, {'SecondOfficer': '11'}, {'ThirdOfficer': '12'}]

   FirstOfficer SecondOfficer ThirdOfficer
0             1           NaN          NaN
1           NaN             2          NaN
2           NaN           NaN            3
3             4           NaN          NaN
4           NaN             5          NaN
5           NaN           NaN            6
6             7           NaN          NaN
7           NaN             8          NaN
8           NaN           NaN            9
9            10           NaN          NaN
10          NaN            11          NaN
11          NaN           NaN           12

列名称可以是任何名称，因此我无法硬编码它

预期数据帧为：

   FirstOfficer SecondOfficer ThirdOfficer
0             1           2          3
1             4           5          6
2             7           8          9
3            10          11         12

有人能给我一个解决办法吗

---

任何帮助都将不胜感激。

使用

defaultdict

存储按字典键列出的值：

from collections import defaultdict

d = defaultdict(list)
for x in L:
    a, b = tuple(x.items())[0]
    d[a].append(b)
print (d)


df = pd.DataFrame(d)
print (df)
  FirstOfficer SecondOfficer ThirdOfficer
0            1             2            3
1            4             5            6
2            7             8            9
3           10            11           12

输出：

  FirstOfficer SecondOfficer ThirdOfficer
0            1             2            3
1            4             5            6
2            7             8            9
3           10            11           12

---

一种方法是预处理您的列表

Ex:

import pandas as pd

lst = [{'FirstOfficer': '1'}, {'SecondOfficer': '2'}, {'ThirdOfficer': '3'},{'FirstOfficer': '4'}, {'SecondOfficer': '5'}, {'ThirdOfficer': '6'},{'FirstOfficer': '7'}, {'SecondOfficer': '8'}, {'ThirdOfficer': '9'},{'FirstOfficer': '10'}, {'SecondOfficer': '11'}, {'ThirdOfficer': '12'}]

data = []
for i in range(0, len(lst), 3):
    temp = []
    for d in lst[i:i+3]:
        for _, v in d.items():
            temp.append(v)
    data.append(temp)

df = pd.DataFrame(data, columns=["FirstOfficer", "SecondOfficer", "ThirdOfficer"]) 
print(df)

  FirstOfficer SecondOfficer ThirdOfficer
0            1             2            3
1            4             5            6
2            7             8            9
3           10            11           12

输出：

import pandas as pd

lst = [{'FirstOfficer': '1'}, {'SecondOfficer': '2'}, {'ThirdOfficer': '3'},{'FirstOfficer': '4'}, {'SecondOfficer': '5'}, {'ThirdOfficer': '6'},{'FirstOfficer': '7'}, {'SecondOfficer': '8'}, {'ThirdOfficer': '9'},{'FirstOfficer': '10'}, {'SecondOfficer': '11'}, {'ThirdOfficer': '12'}]

data = []
for i in range(0, len(lst), 3):
    temp = []
    for d in lst[i:i+3]:
        for _, v in d.items():
            temp.append(v)
    data.append(temp)

df = pd.DataFrame(data, columns=["FirstOfficer", "SecondOfficer", "ThirdOfficer"]) 
print(df)

  FirstOfficer SecondOfficer ThirdOfficer
0            1             2            3
1            4             5            6
2            7             8            9
3           10            11           12

---

如果性能不是问题，您可以使用：

df=pd.DataFrame(l).apply(lambda x: pd.Series(x.dropna().values))
print(df)

---

---

嗨，我修改了这个问题。请看一看。@jezrael想不出用这个的方法：(