Python 从上面的行中提取差异比率,并使用多索引将值存储在另一列中
我想知道如何获得具有多索引列的两行之间的差异比率,并将它们存储在特定列中 我有一个数据框,看起来像这样Python 从上面的行中提取差异比率,并使用多索引将值存储在另一列中,python,pandas,rows,multi-index,difference,Python,Pandas,Rows,Multi Index,Difference,我想知道如何获得具有多索引列的两行之间的差异比率,并将它们存储在特定列中 我有一个数据框,看起来像这样 >>>df A B C total diff total diff total diff 2020-08-15 100 0 200 0 20 0 df_new
>>>df
A B C
total diff total diff total diff
2020-08-15 100 0 200 0 20 0
df_new
A B C
total diff total diff total diff
2020-08-16 200 - 50 - 30 -
每天,我都会增加一行。新的一排看起来像这样
>>>df
A B C
total diff total diff total diff
2020-08-15 100 0 200 0 20 0
df_new
A B C
total diff total diff total diff
2020-08-16 200 - 50 - 30 -
对于diff
列,我想从上一行中取比率,取total
的值。因此,公式将是([total of today]-[total of the day])/[total of the day]
A B C
total diff total diff total diff
2020-08-15 100 0 200 0 20 0
2020-08-16 200 1.0 50 -0.75 30 0.5
我知道如何添加新行
day = dt.today()
df.loc[day.strftime("%Y-%m-%d"), :] = df_new.squeeze()
但我不知道如何才能得到两行多索引列之间的差异。。。任何帮助都将不胜感激!谢谢。使用
shift
计算结果并更新原始df:
s = df.filter(like="total").rename(columns={"total":"diff"}, level=1)
res = ((s - s.shift(1))/s.shift(1))
df.update(res)
print (df)
A B C
total diff total diff total diff
2020-08-15 100 0.0 200 0.00 20 0.0
2020-08-16 200 1.0 50 -0.75 30 0.5
使用
shift
计算结果并更新原始df:
s = df.filter(like="total").rename(columns={"total":"diff"}, level=1)
res = ((s - s.shift(1))/s.shift(1))
df.update(res)
print (df)
A B C
total diff total diff total diff
2020-08-15 100 0.0 200 0.00 20 0.0
2020-08-16 200 1.0 50 -0.75 30 0.5
可以使用和更新多索引值
#df
# A B C
# total diff total diff total diff
#0 100 0 200 0 20 0
#df2
# A B C
# total diff total diff total diff
#0 200.0 NaN 50.0 NaN 30.0 NaN
# Take last row of current DataFrame i.e. `df`
curr = df.iloc[-1].xs('total', level=1) #Get total values
# Take total values of new DataFrame you get everyday i.e. `df2`
new = df2.iloc[0].xs('total',level=1)
# Calculate diff values
diffs = new.sub(curr).div(curr) # This is equal to `(new-curr)/curr`
idx = pd.IndexSlice
x = pd.concat([df, df2]).reset_index(drop=True)
x.loc[x.index[-1], idx[:,'diff']] = diffs.tolist()
x
A B C
total diff total diff total diff
0 100.0 0.0 200.0 0.00 20.0 0.0
1 200.0 1.0 50.0 -0.75 30.0 0.5
如果不想创建新的数据帧(x
),请使用DataFrame.append
追加值
一切都是一样的,直到步骤idx=pd.indexlice
,不要创建x
,而是将值附加到df
df2.loc[:, idx[:,'diff']] = diffs.tolist()
df.append(df2)
A B C
total diff total diff total diff
0 100.0 0.0 200.0 0.00 20.0 0.0
0 200.0 1.0 50.0 -0.75 30.0 0.5
可以使用和更新多索引值
#df
# A B C
# total diff total diff total diff
#0 100 0 200 0 20 0
#df2
# A B C
# total diff total diff total diff
#0 200.0 NaN 50.0 NaN 30.0 NaN
# Take last row of current DataFrame i.e. `df`
curr = df.iloc[-1].xs('total', level=1) #Get total values
# Take total values of new DataFrame you get everyday i.e. `df2`
new = df2.iloc[0].xs('total',level=1)
# Calculate diff values
diffs = new.sub(curr).div(curr) # This is equal to `(new-curr)/curr`
idx = pd.IndexSlice
x = pd.concat([df, df2]).reset_index(drop=True)
x.loc[x.index[-1], idx[:,'diff']] = diffs.tolist()
x
A B C
total diff total diff total diff
0 100.0 0.0 200.0 0.00 20.0 0.0
1 200.0 1.0 50.0 -0.75 30.0 0.5
如果不想创建新的数据帧(x
),请使用DataFrame.append
追加值
一切都是一样的,直到步骤idx=pd.indexlice
,不要创建x
,而是将值附加到df
df2.loc[:, idx[:,'diff']] = diffs.tolist()
df.append(df2)
A B C
total diff total diff total diff
0 100.0 0.0 200.0 0.00 20.0 0.0
0 200.0 1.0 50.0 -0.75 30.0 0.5