问题描述
我有一张客户(资产)和资产分配(资产)表
I've got a table of clients (coper) and asset allocation (asset)
A = [[1,2],[3,4],[5,6]]
idx = ['coper1','coper2','coper3']
cols = ['asset1','asset2']
df = pd.DataFrame(A,index = idx, columns = cols)
所以我的数据看起来像
asset1 asset2
coper1 1 2
coper2 3 4
coper3 5 6
,我想通过线性优化来运行它们(我有约束-像sum of all of asset_i <= amount_on_hand_i
和sum of coper_j = price_j
一样)
and I want to run them through a linear optimization (i've got constraints- somtehing like sum of all of asset_i <= amount_on_hand_i
and sum of coper_j = price_j
)
所以我必须将2D矩阵转换为1D向量.融化很容易
so I have to turn this 2D matrix into a 1D vector. Which is easy with melt
df2 = pd.melt(df,value_vars=['asset1','asset2'])
但是现在,当我尝试解开它时,我得到了一个6行数组,其中有很多空白!
But now, when I try to unmelt it, I get a 6-row array with lots of blanks!
df2.pivot(columns = 'variable', values = 'value')
variable asset1 asset2
0 1.0 NaN
1 3.0 NaN
2 5.0 NaN
3 NaN 2.0
4 NaN 4.0
5 NaN 6.0
在使用melt时,有什么方法可以保留索引的"coper"部分吗?
Is there any way to preserve the 'coper' part of my indexing while using melt?
推荐答案
您需要通过 reset_index
和参数id_vars
:
df2 = pd.melt(df.reset_index(), id_vars='index',value_vars=['asset1','asset2'])
print (df2)
index variable value
0 coper1 asset1 1
1 coper2 asset1 3
2 coper3 asset1 5
3 coper1 asset2 2
4 coper2 asset2 4
5 coper3 asset2 6
然后枢轴运行良好:
print(df2.pivot(index='index',columns = 'variable', values = 'value'))
variable asset1 asset2
index
coper1 1 2
coper2 3 4
coper3 5 6
使用 stack
:
Another possible solution with stack
:
df2 = df.stack().reset_index()
df2.columns = list('abc')
print (df2)
a b c
0 coper1 asset1 1
1 coper1 asset2 2
2 coper2 asset1 3
3 coper2 asset2 4
4 coper3 asset1 5
5 coper3 asset2 6
print(df2.pivot(index='a',columns = 'b', values = 'c'))
b asset1 asset2
a
coper1 1 2
coper2 3 4
coper3 5 6
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