Question

我有4个列表，我想转换为熊猫数据框

carriersID=[1,2,3,5,6,7,8,9]
destinationId=[2,5,4,4,5,7,8,7]
departureDate=[1,2,3,4,5,7,8]
prices=[755,800,500,400,152,444,784,954,120]

和我想要的输出是这样的：

      carrierId     DestinationID    DeparturDate     Prices
1      2                 2               1              755
2      5                 5               2              800 
3      4                 4               3              500
4      4                 4               4              400
...  ...                 ...             ...           ....

Answer 1

使用移调T：

df = pd.DataFrame([carriersID, destinationId, departureDate, prices]).T
df.columns = ['carriersID', 'destinationId', 'departureDate', 'prices']
df


+---+-------------+---------------+---------------+--------+
|   |  carriersID | destinationId | departureDate | prices |
+---+-------------+---------------+---------------+--------+
| 0 | 1.0         | 2.0           | 1.0           |  755.0 |
| 1 | 2.0         | 5.0           | 2.0           |  800.0 |
| 2 | 3.0         | 4.0           | 3.0           |  500.0 |
| 3 | 5.0         | 4.0           | 4.0           |  400.0 |
| 4 | 6.0         | 5.0           | 5.0           |  152.0 |
| 5 | 7.0         | 7.0           | 7.0           |  444.0 |
| 6 | 8.0         | 8.0           | 8.0           |  784.0 |
| 7 | 9.0         | 7.0           | NaN           |  954.0 |
| 8 | NaN         | NaN           | NaN           |  120.0 |
+---+-------------+---------------+---------------+--------+

Answer 2

使用字典的另一种方法

>>> import pandas as pd
>>> pd.DataFrame.from_dict({
        'carrierId'    : carriersID, 
        'DestinationID': destinationId,
        'DeparturDate' : departureDate,
        'Prices'       : prices
    }, orient='index').transpose()

Answer 3

将列表转换为字典。然后将字典传递到pandas.DataFrame，这将获得键作为列名和值行（您甚至不需要使用pandas.from_dict）。请注意，由于列表的长度不相等，因此在创建DataFrame时需要将其考虑在内并遍历字典。

import pandas as pd

carriersID = [1, 2, 3, 5, 6, 7, 8, 9]
destinationId = [2, 5, 4, 4, 5, 7, 8, 7]
departureDate = [1, 2, 3, 4, 5, 7, 8]
prices = [755, 800, 500, 400, 152, 444, 784, 954, 120]

my_dict = {'carriersID':carriersID, 'destinationId':destinationId, 'departureDate':departureDate, 'prices':prices}
df = pd.DataFrame(dict([(k, pd.Series(v)) for k,v in my_dict.items()]))
print(df)

输出

   carriersID  departureDate  destinationId  prices
0         1.0            1.0            2.0     755
1         2.0            2.0            5.0     800
2         3.0            3.0            4.0     500
3         5.0            4.0            4.0     400
4         6.0            5.0            5.0     152
5         7.0            7.0            7.0     444
6         8.0            8.0            8.0     784
7         9.0            NaN            7.0     954
8         NaN            NaN            NaN     120

将列表转换为熊猫数据框python

3 个答案: