Python产品问题
我仍在努力处理以下代码:Python产品问题,python,pandas,dataframe,Python,Pandas,Dataframe,我仍在努力处理以下代码: xa= [0, 0, 0, 0, 65, 67, 69, 75, 0, 0, 0] xb = [.3, .3, .3,.3, .3, .3, .3, .3, .3, .3, .3] ideal = [0, 0, 0, 0, 65, 67, 69, 75, 67.5, 60.75, 54.675] df = pd.DataFrame({'a':xa, 'b':xb, 'i':ideal}) mask=(df['a']<51) & (df['b']>
xa= [0, 0, 0, 0, 65, 67, 69, 75, 0, 0, 0]
xb = [.3, .3, .3,.3, .3, .3, .3, .3, .3, .3, .3]
ideal = [0, 0, 0, 0, 65, 67, 69, 75, 67.5, 60.75, 54.675]
df = pd.DataFrame({'a':xa, 'b':xb, 'i':ideal})
mask=(df['a']<51) & (df['b']>0)
df['c'] = df['a'].where(mask,0.9).groupby(~mask.cumsum()).cumprod()
print(df)
xa=[0,0,0,65,67,69,75,0,0,0,0]
xb=[.3、.3、.3、.3、.3、.3、.3、.3、.3、.3、.3]
理想=[0,0,0,0,65,67,69,75,67.5,60.75,54.675]
数据帧({'a':xa,'b':xb,'i':理想})
掩码=(df['a']0)
df['c']=df['a'].where(mask,0.9).groupby(~mask.cumsum()).cumprod()
打印(df)
我希望“c”列变成“理想”列。这只是我的100K+行完整数据集的一个示例
“面具”是这样计算的:
什么时候'a'{i}0?那么对还是错
“c”列的计算方式如下:
当'mask'{i}=FALSE时,则'c'{i}='a'{i}否则'c'{i}=0.9*'c'{i-1}
所以我希望(有一天)“c”变成“理想的”…如果我理解正确的话
mask=(df['a']<51) & (df['b']>0)
df['c'] = np.where(mask, df['i'].shift() * 0.9, df['i'])
df.fillna(0, inlace = True)
a b i c
0 0 0.3 0.000 0.000
1 0 0.3 0.000 0.000
2 0 0.3 0.000 0.000
3 0 0.3 0.000 0.000
4 65 0.3 65.000 65.000
5 67 0.3 67.000 67.000
6 69 0.3 69.000 69.000
7 75 0.3 75.000 75.000
8 0 0.3 67.500 67.500
9 0 0.3 60.750 60.750
10 0 0.3 54.675 54.675
mask=(df['a']0)
df['c']=np.where(掩码,df['i'].shift()*0.9,df['i']))
df.fillna(0,嵌入=真)
a b i c
0 0 0.3 0.000 0.000
1 0 0.3 0.000 0.000
2 0 0.3 0.000 0.000
3 0 0.3 0.000 0.000
4 65 0.3 65.000 65.000
5 67 0.3 67.000 67.000
6 69 0.3 69.000 69.000
7 75 0.3 75.000 75.000
8 0 0.3 67.500 67.500
9 0 0.3 60.750 60.750
10 0 0.3 54.675 54.675
我相信这可以解决您的问题:
# First calculate the column as if there is no decay
mask=(df['a']<51) & (df['b']>0)
df['c'] = df['a'].where(~mask)
df['c'].fillna(method='ffill', inplace=True)
df['c'].fillna(0, inplace=True)
# Check how many rows since the mask has changed from True to False or v.v.
df['ones'] = 1
df['power'] = df['ones'].groupby((mask != mask.shift()).cumsum()).transform('cumsum')
# For the values in the mask, apply the decay
df['c'] = np.where(mask, 0.9 ** df['power']*df['c'], df['c'])
print(df)
主要的技巧是定义一列,该列定义0.9相乘的次数,另一列向前填充,以检查如果没有衰减,数字会是多少。希望这有帮助 我不想使用理想或“我”。我把它放在那里只是为了给你展示我需要的结果;你可以自由地尝试“a”和“b”来代替。你的逻辑不是面向数据分析的,而是通用的逻辑和数学,而且实际上是有效的!我喜欢你的方法。谢谢!
a b i c power ones
0 0 0.3 0.000 0.000 1 1
1 0 0.3 0.000 0.000 2 1
2 0 0.3 0.000 0.000 3 1
3 0 0.3 0.000 0.000 4 1
4 65 0.3 65.000 65.000 1 1
5 67 0.3 67.000 67.000 2 1
6 69 0.3 69.000 69.000 3 1
7 75 0.3 75.000 75.000 4 1
8 0 0.3 67.500 67.500 1 1
9 0 0.3 60.750 60.750 2 1
10 0 0.3 54.675 54.675 3 1