从现有的列名在 Pandas DataFrame 中创建一个新列

Question

我想解构一个 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。 展示比解释更容易：

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)

从这个 DataFrame 我希望得到以下列表：

[['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]]

有没有方便的方法来做到这一点？

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]]

通过以下，元素的顺序将与问题中的输出相匹配。

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]]

Answer 2

真正的熊猫风格：

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

Answer 3

我更喜欢堆叠而不是卸载然后交换级别：

>>> 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)
>>> 

Answer 4

您希望数据采用的结构非常混乱，因此鉴于您想要的数据，这可能是最好的方法。

# 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

Answer 5

您可以使用

data = []

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

data

您可以避免牺牲您请求输出的顺序的第二个循环，因为它是结构如何开始的一个完整分解。

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]]