如何索引熊猫列表中的元素？

Question

我用 np.shape(sample_size, ) 创建了一个 numpy 数组，然后我使用 pandas dataframe 来显示数据。 我想请问一下，如何为列表添加子列索引，例如：示例 1、示例 2、示例 3 等？

low = 0
high = 500
sample_size = 5

def get_numbers(low, high, sample_size):
    return random.sample(range(low, high), sample_size)

p_one = np.array(get_numbers(low, high, sample_size), dtype = int)
p_two = np.array(get_numbers(low, high, sample_size), dtype = int)
p_three = np.array(get_numbers(low, high, sample_size), dtype = int)
p_four = np.array(get_numbers(low, high, sample_size), dtype = int)
p_five = np.array(get_numbers(low, high, sample_size), dtype = int)

  for idn in range(0,n): #------------------n+1 for the last process step

        p = [p_one, p_two, p_three, p_four, p_five]

df_rawdata = pd.DataFrame(list(zip(p)),columns =['Processing'])

Answer 1

例子

我们需要代码中最小且可重现的示例

df = pd.DataFrame([[[1, 2, 3]], [[4, 5, 6]]], index=['step1', 'step2'], columns=['process'])

df

        process
step1   [1, 2, 3]
step2   [4, 5, 6]

---

代码

第一的。 将列表扩展到列

df1 = df['process'].apply(pd.Series).rename(columns=lambda x: f'sp {x+1}')

df1

        sp 1    sp 2    sp 3
step1   1       2       3
step2   4       5       6

第二。 制作多索引

out = pd.concat([df1], keys=['process'], axis=1)

out

              process
        sp 1    sp 2    sp 3
step1   1       2       3
step2   4       5       6

---

更新

或使用以下示例代码：

df = pd.DataFrame([[[1, 2], [3, 4]], [[5, 6], [7, 8]]], index=['step1', 'step2'], columns=['process1', 'process2'])

df

      process1  process2
step1   [1, 2]  [3, 4]
step2   [5, 6]  [7, 8]

out = (df.stack()
       .apply(pd.Series).rename(columns=lambda x: f'sp {x+1}')
       .unstack().swaplevel(0, 1, axis=1).sort_index(axis=1))

out

        process1        process2
        sp 1    sp 2    sp 1    sp 2
step1   1       2       3       4
step2   5       6       7       8