python - 从 Pandas 数据框构建二维数组

Question

我有一个 Pandas 数据框:

import pandas as pd
import numpy as np

df = pd.DataFrame(columns=['Text','Selection_Values'])
df["Text"] = ["Hi", "this is", "just", "a", "single", "sentence.", "This", np.nan, "is another one.","This is", "a", "third", "sentence","."]
df["Selection_Values"] = [0,0,0,0,0,1,0,0,1,0,0,0,0,0]
print(df)

输出:

               Text  Selection_Values
0                Hi                 0
1           this is                 0
2              just                 0
3                 a                 0
4            single                 0
5         sentence.                 1
6              This                 0
7               NaN                 0
8   is another one.                 1
9           This is                 0
10                a                 0
11            third                 0
12         sentence                 0
13                .                 0

现在，我想根据 Selection Value 列将 Text 列重新组合到一个二维数组中。出现在 0(第一个整数，或 1 之后)和 1(包括)之间的所有单词都应放入二维数组中.数据集的最后一句话可能没有结束 1。这可以按照这个问题中的解释来完成:Regroup pandas column into 2D list based on another column

[["Hi this is just a single sentence."],["This is another one"], ["This is a third sentence ."]]

我想更进一步，提出以下条件:如果一个列表中有超过 max_number_of_cells_per_list 个非 NaN 单元格，那么这个列表应该分成大致相等的最多包含 +/- 1 个 max_number_of_cells_per_list 单元格元素的部分。

假设:max_number_of_cells_per_list = 2，那么预期的输出应该是:

 [["Hi this is"], ["just a"], ["single sentence."],["This is another one"], ["This is"], ["a third sentence ."]]

示例:

基于“Selection_Values”列，可以将单元格重新组合到以下二维列表中，使用:

[[s.str.cat(sep=' ')] for s in np.split(df.Text, df[df.Selection_Values == 1].index+1) if not s.empty]

输出(原始列表):

[["Hi this is just a single sentence."],["This is another one"], ["This is a third sentence ."]]

让我们看一下这些列表中的单元格数量:

如您所见，列表 1 有 6 个单元格，列表 2 有 2 个单元格，列表 3 有 5 个单元格。

现在，我想要实现的是:如果列表中的单元格数量超过一定数量，则应将其拆分，以便每个结果列表具有 +/-1 所需数量的单元格.

例如 max_number_of_cells_per_list = 2

修改列表:

你有办法做到这一点吗？

编辑: 重要说明:原始列表中的单元格不应放入相同的列表中。

编辑 2:

               Text  Selection_Values  New
0                Hi                 0  1.0
1           this is                 0  0.0
2              just                 0  1.0
3                 a                 0  0.0
4            single                 0  1.0
5         sentence.                 1  0.0
6              This                 0  1.0
7               NaN                 0  0.0
8   is another one.                 1  1.0
9           This is                 0  0.0
10                a                 0  1.0
11            third                 0  0.0
12         sentence                 0  0.0
13                .                 0  NaN

Answer 1

IIUC，你可以这样做:

n=2 #change this as you like for no. of splits
s=df.Text.dropna().reset_index(drop=True)
c=s.groupby(s.index//n).cumcount().eq(0).shift().shift(-1).fillna(False)

---

[[i] for i in s.groupby(c.cumsum()).apply(' '.join).tolist()]

---

[['Hi this is'], ['just a'], ['single sentence.'], 
    ['This is another one.'], ['This is a'], ['third sentence .']]

编辑:

d=dict(zip(df.loc[df.Text.notna(),'Text'].index,c.index))
ser=pd.Series(d)
df['new']=ser.reindex(range(ser.index.min(),
                        ser.index.max()+1)).map(c).fillna(False).astype(int)
print(df)

---

               Text  Selection_Values  new
0                Hi                 0    1
1           this is                 0    0
2              just                 0    1
3                 a                 0    0
4            single                 0    1
5         sentence.                 1    0
6              This                 0    1
7               NaN                 0    0
8   is another one.                 1    0
9           This is                 0    1
10                a                 0    0
11            third                 0    1
12         sentence                 0    0
13                .                 0    0