问题描述
我有一个 dataframe，其中包含多个“堆栈”及其相应的“长度”。
df = pd.DataFrame({'stack-1-material': ['rock','paper','scissors','rock'],'stack-2-material': ['rock','rock','scissors'],'stack-1-length': [3,1,2,3],'stack-2-length': [3,3,2]})
stack-1-material stack-2-material stack-1-length stack-2-length
0 rock rock 3 3
1 paper paper 1 1
2 paper rock 1 3
3 scissors paper 2 1
4 rock scissors 3 2
我正在尝试为每种材料创建一个单独的列，以跟踪长度的累积总和，而不管它们是哪个“堆栈”。我试过使用 groupby 但只能将累积总和放入一列。这是我要找的：
stack-1-material stack-2-material stack-1-length stack-2-length rock_cumsum paper_cumsum scissors_cumsum
0 rock rock 3 3 6 0 0
1 paper paper 1 1 6 2 0
2 paper rock 1 3 9 3 0
3 scissors paper 2 1 9 4 2
4 rock scissors 3 2 12 4 4
解决方法
您可以使用列材料作为列长度的掩码，然后使用 sum 沿列和 cumsum，用于每个材料。
#separate material and length
material = df.filter(like='material').to_numpy()
lentgh = df.filter(like='length')
# get all unique material
l_mat = np.unique(material)
# iterate over nique materials
for mat in l_mat:
df[f'{mat}_cumsum'] = lentgh.where(material==mat).sum(axis=1).cumsum()
print(df)
stack-1-material stack-2-material stack-1-length stack-2-length \
0 rock rock 3 3
1 paper paper 1 1
2 paper rock 1 3
3 scissors paper 2 1
4 rock scissors 3 2
rock_cumsum paper_cumsum scissors_cumsum
0 6.0 0.0 0.0
1 6.0 2.0 0.0
2 9.0 3.0 0.0
3 9.0 4.0 2.0
4 12.0 4.0 4.0
首先，反转您的列名，以便我们可以使用 wide_to_long 来重塑 DataFrame。
然后取材料中的 cumsum 并确定每行每种材料的最大值。然后我们可以 reshape this 和 ffill 并用 0 替换剩余的 NaN 并连接回原始。
df.columns = ['-'.join(x[::-1]) for x in df.columns.str.rsplit('-',n=1)]
res = (pd.wide_to_long(df.reset_index(),stubnames=['material','length'],i='index',j='whatever',suffix='.*')
.sort_index(level=0))
# material length
#index whatever
#0 -stack-1 rock 3
# -stack-2 rock 3
#1 -stack-1 paper 1
# -stack-2 paper 1
#2 -stack-1 paper 1
# -stack-2 rock 3
#3 -stack-1 scissors 2
# -stack-2 paper 1
#4 -stack-1 rock 3
# -stack-2 scissors 2
res['csum'] = res.groupby('material')['length'].cumsum()
res = (res.groupby(['index','material'])['csum'].max()
.unstack(-1).ffill().fillna(0,downcast='infer')
.add_suffix('_cumsum'))
df = pd.concat([df,res],axis=1)
material-stack-1 material-stack-2 length-stack-1 length-stack-2 paper_cumsum rock_cumsum scissors_cumsum
0 rock rock 3 3 0 6 0
1 paper paper 1 1 2 6 0
2 paper rock 1 3 3 9 0
3 scissors paper 2 1 4 9 2
4 rock scissors 3 2 4 12 4