Question

我在熊猫数据框的一列中有一个列表。我想将“ price_label”列中的值附加到列表中。

我目前正在使用自己编写的函数来执行此操作，但这是最好的方法吗？我觉得我在这里想念什么吗？

我的方法有效，但效果不佳。有任何建议。

之前列出：

+---+--------------+-----------+-----------------------------------------+-------+------------------------------+-------------+
|   | prodref      | prodcateg | webproddesc                             | price | categories                   | price_label |
+---+--------------+-----------+-----------------------------------------+-------+------------------------------+-------------+
| 0 | 1004 10-14MM | STR4      | Gold Plated Bracelet 1004 - 10-14MM     | 5.98  | [Gold, Plated, Bracelet]     | 1           |
| 1 | 1004 16-22MM | STR4      | Gold Plated Bracelet 1004 - 16-22MM     | 5.98  | [Gold, Plated, Bracelet]     | 1           |
| 2 | 1007 10-14MM | STR4      | Bi-Colour Bracelet 1007 - 10-14MM       | 5.98  | [BiColour, Bracelet]         | 1           |
| 3 | 1007 16-22MM | STR4      | Bi-Colour Bracelet 1007 - 16-22MM       | 5.98  | [BiColour, Bracelet]         | 1           |
| 4 | 1010 10-14MM | STR4      | Stainless Steel Bracelet 1010 - 10-14MM | 5.98  | [Stainless, Steel, Bracelet] | 1           |
| 5 | 1010 16-22MM | STR4      | Stainless Steel Bracelet 1010 - 16-22MM | 5.98  | [Stainless, Steel, Bracelet] | 1           |
| 6 | W108/22      | STR1      | Grey Calf Watch Strap (S) - W108/22     | 4.18  | [Grey, Calf, Watch]          | 1           |
| 7 | W404/14      | STR1      | White Lizard Grain Strap (S) - W404/14  | 5.98  | [White, Lizard, Grain]       | 1           |
| 8 | W404/18      | STR1      | White Lizard Grain Strap (S) - W404/18  | 5.98  | [White, Lizard, Grain]       | 1           |
+---+--------------+-----------+-----------------------------------------+-------+------------------------------+-------------+

功能：

def appendPrice(vert):
    cat_list = vert["categories"]
    cat_list.append(vert["price_label"])
    return cat_list

test["categories"] = test.apply(lambda x:appendPrice(x),axis=1)

输出

+---+--------------+-----------+-----------------------------------------+-------+---------------------------------+-------------+
|   | prodref      | prodcateg | webproddesc                             | price | categories                      | price_label |
+---+--------------+-----------+-----------------------------------------+-------+---------------------------------+-------------+
| 0 | 1004 10-14MM | STR4      | Gold Plated Bracelet 1004 - 10-14MM     | 5.98  | [Gold, Plated, Bracelet, 1]     | 1           |
| 1 | 1004 16-22MM | STR4      | Gold Plated Bracelet 1004 - 16-22MM     | 5.98  | [Gold, Plated, Bracelet, 1]     | 1           |
| 2 | 1007 10-14MM | STR4      | Bi-Colour Bracelet 1007 - 10-14MM       | 5.98  | [BiColour, Bracelet, 1]         | 1           |
| 3 | 1007 16-22MM | STR4      | Bi-Colour Bracelet 1007 - 16-22MM       | 5.98  | [BiColour, Bracelet, 1]         | 1           |
| 4 | 1010 10-14MM | STR4      | Stainless Steel Bracelet 1010 - 10-14MM | 5.98  | [Stainless, Steel, Bracelet, 1] | 1           |
| 5 | 1010 16-22MM | STR4      | Stainless Steel Bracelet 1010 - 16-22MM | 5.98  | [Stainless, Steel, Bracelet, 1] | 1           |
| 6 | W108/22      | STR1      | Grey Calf Watch Strap (S) - W108/22     | 4.18  | [Grey, Calf, Watch, 1]          | 1           |
| 7 | W404/14      | STR1      | White Lizard Grain Strap (S) - W404/14  | 5.98  | [White, Lizard, Grain, 1]       | 1           |
| 8 | W404/18      | STR1      | White Lizard Grain Strap (S) - W404/18  | 5.98  | [White, Lizard, Grain, 1]       | 1           |
| 9 | W409/14      | STR1      | Pink Lizard Grain Strap (S) - W409/14   | 5.98  | [Pink, Lizard, Grain, 1]        | 1           |
+---+--------------+-----------+-----------------------------------------+-------+---------------------------------+-------------+

Answer 1

正如@ALollz所指出的那样，很少使用Series或DataFrame中的列表。

也就是说，如果您现在想要保留模式，则可以完全删除appendPrice：

test["categories"] = test.apply(lambda x: x.categories + [x.price_label],axis=1)

否则，您可以在尝试完成此操作时共享更多颜色。

Answer 2

假设

df["categories"]
0        [Gold, Plated, Bracelet]
1        [Gold, Plated, Bracelet]
2            [BiColour, Bracelet]
3            [BiColour, Bracelet]
4    [Stainless, Steel, Bracelet]
5    [Stainless, Steel, Bracelet]
6             [Grey, Calf, Watch]
7          [White, Lizard, Grain]
8          [White, Lizard, Grain]
Name: categories, dtype: object

和df['price_label']作为int，将执行以下操作：

df.apply(lambda row: row["categories"] + list(str(row["price_label"])), axis=1)
0        [Gold, Plated, Bracelet, 1]
1        [Gold, Plated, Bracelet, 1]
2            [BiColour, Bracelet, 1]
3            [BiColour, Bracelet, 1]
4    [Stainless, Steel, Bracelet, 1]
5    [Stainless, Steel, Bracelet, 1]
6             [Grey, Calf, Watch, 1]
7          [White, Lizard, Grain, 1]
8          [White, Lizard, Grain, 1]
dtype: object

Answer 3

这行代码将满足您的要求。它将price_label列中的值附加到每个类别行的末尾。对于较大的数据，此操作比df.apply（）更有效。

Person

Pandas Dataframe Apply-将lambda函数追加到列表

3 个答案: