Question

我正在尝试使用大小相同的各种熊猫数据帧构建数据库。我想为数据框中的每个点分配所有其他点。

我得到这样的东西：

数据帧1：

```{r}
valid_markdown <- your_function()
```

(ref:my_hack) `r valid_markdown`

```{block2, type="block"}
(ref:my_hack)

```

数据框2：

   1  2  3  4  
 1 A  B  C  D  
 2 A  B  C  D  
 3 A  B  C  D

最终数据帧（类似这样）

   1  2  3  4  
 1 A  D  G  J
 2 B  E  H  K
 3 C  F  I  L

Answer 1

您可以添加两个数据帧中的值，以及两者之间的分隔符，并使用str.split遍历列以创建列表。最后使用pd.concat将结果转换为数据框：

df = df1 + ',' + df2
pd.concat([df[col].str.split(',') for col in df.columns], axis=1)

     1       2       3       4
1  [A, A]  [B, D]  [C, G]  [D, J]
2  [A, B]  [B, E]  [C, H]  [D, K]
3  [A, C]  [B, F]  [C, I]  [D, L]

或使用DataFrame.combine：

df1.combine(df2, func=lambda a,b: (a + ',' + b).str.split(','))

     1       2       3       4
1  [A, A]  [B, D]  [C, G]  [D, J]
2  [A, B]  [B, E]  [C, H]  [D, K]
3  [A, C]  [B, F]  [C, I]  [D, L]

Answer 2

检查

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如果我是我，我将关注多个索引而不是列表类型的列

pd.concat([df1,df2],1).groupby(level=0,axis=1).agg(lambda x : x.values.tolist())
Out[529]: 
        1       2       3       4
1  [A, A]  [B, D]  [C, G]  [D, J]
2  [A, B]  [B, E]  [C, H]  [D, K]
3  [A, C]  [B, F]  [C, I]  [D, L]

Answer 3

效率不高，但是您可以将所有单元格转换为列表，然后只需添加如下数据框即可：

bluetoothAdapter.getBluetoothLeScanner()

输出

import pandas as pd
import numpy as np
np.random.seed(42)

df1 = pd.DataFrame(
    [list('ABCD') for _ in range(5)]
)
df2 = pd.DataFrame(
    [np.random.choice(list('ABCD'), size=4).tolist() for _ in range(5)]
)

df1 = df1.apply(lambda x: [[v] for v in x])
df2 = df2.apply(lambda x: [[v] for v in x])

df1 + df2

如何将多个相同大小的熊猫数据框组合到一个内聚数据库中？

3 个答案: