从现有的列名中在Pandas DataFrame中创建一个新的列

Question

我想要分解一份pandas DataFrame，使用列标题作为新的数据列，并创建一个包含行索引和列的所有组合的列表。更容易通过示例演示：

index_col = ["store1", "store2", "store3"]
cols = ["January", "February", "March"]
values = [[2,3,4],[5,6,7],[8,9,10]]
df = pd.DataFrame(values, index=index_col, columns=cols)

我希望从这个DataFrame中获得以下列表：

[['store1', 'January', 2],
 ['store1', 'February', 3],
 ['store1', 'March', 4],
 ['store2', 'January', 5],
 ['store2', 'February', 6],
 ['store2', 'March', 7],
 ['store3', 'January', 8],
 ['store3', 'February', 9],
 ['store3', 'March', 10]]

有没有方便的方法做到这一点？

- Håkon

Answer 1

df.unstack().swaplevel().reset_index().values.tolist()
#OR
df.reset_index().melt(id_vars="index").values.tolist()
# [['store1', 'January', 2],
#  ['store2', 'January', 5],
#  ['store3', 'January', 8],
#  ['store1', 'February', 3],
#  ['store2', 'February', 6],
#  ['store3', 'February', 9],
#  ['store1', 'March', 4],
#  ['store2', 'March', 7],
#  ['store3', 'March', 10]]

使用以下代码，元素的顺序将与问题中的输出匹配。

df.transpose().unstack().reset_index().values.tolist()
# [['store1', 'January', 2],
#  ['store1', 'February', 3],
#  ['store1', 'March', 4],
#  ['store2', 'January', 5],
#  ['store2', 'February', 6],
#  ['store2', 'March', 7],
#  ['store3', 'January', 8],
#  ['store3', 'February', 9],
#  ['store3', 'March', 10]]

- d.b

---

这绝对是最好的（虽然我的也不错，但这更加简洁，甚至更符合“真正的Pandas风格”）。 - user17242583

需要保持相同的顺序吗？ - Larry the Llama

@LarrytheLlama，如果顺序很重要，您可以执行 df.unstack().swaplevel().reset_index().sort_values("level_0").values.tolist() - d.b

Answer 2

真正的Pandas风格：

lst = [[*k, v] for k, v in df.unstack().swaplevel().to_dict().items()]

- user17242583

---

美妙的嵌套列表推导和多个Pandas方法的使用！ - Celius Stingher

Answer 3

我更喜欢堆叠而不是解除堆叠，然后交换层级：

>>> df.stack().reset_index().to_numpy()
array([['store1', 'January', 2],
       ['store1', 'February', 3],
       ['store1', 'March', 4],
       ['store2', 'January', 5],
       ['store2', 'February', 6],
       ['store2', 'March', 7],
       ['store3', 'January', 8],
       ['store3', 'February', 9],
       ['store3', 'March', 10]], dtype=object)
>>> 

或者使用 melt 和 ignore_index=False：

>>> df.melt(ignore_index=False).reset_index().to_numpy()
array([['store1', 'January', 2],
       ['store2', 'January', 5],
       ['store3', 'January', 8],
       ['store1', 'February', 3],
       ['store2', 'February', 6],
       ['store3', 'February', 9],
       ['store1', 'March', 4],
       ['store2', 'March', 7],
       ['store3', 'March', 10]], dtype=object)
>>> 

- U13-Forward

---

这也是一个不错的选择！ - user17242583

Answer 4

你想要的数据结构非常混乱，因此考虑到你所需的数据，这可能是最佳方法。

# Results
res = []

# Nested loop: first for length of index col, then next for cols
for i in range(len(index_col)):
    for j in range(len(cols)):
        # Format of data
        res.append([index_col[i], cols[j], values[i][j]])

# Return results
print(res)
return res

- Larry the Llama

Answer 5

您可以使用以下方法迭代数据框中的项目：

data = []

for col, row in df.items():
    for ind, val in row.reset_index().values:
        data.append([ind, col, val])

data

您可以避免第二个循环，以牺牲输出顺序为代价，因为这会完全分解结构的起始方式。

- fam-woodpecker

Answer 6

temp = df.stack()

[[*ent, val] for ent, val in zip(temp.index, temp)]

[['store1', 'January', 2],
 ['store1', 'February', 3],
 ['store1', 'March', 4],
 ['store2', 'January', 5],
 ['store2', 'February', 6],
 ['store2', 'March', 7],
 ['store3', 'January', 8],
 ['store3', 'February', 9],
 ['store3', 'March', 10]]

- sammywemmy