假設(shè)我有以下數(shù)據(jù)幀：import pandas as pddf = pd.DataFrame()df['A'] = ('1/05/2019','2/05/2019','3/05/2019','4/05/2019','5/05/2019','6/05/2019','7/05/2019','8/05/2019','9/05/2019','10/05/2019','11/05/2019','12/05/2019','13/05/2019','14/05/2019','15/05/2019','16/05/2019','17/05/2019','18/05/2019','19/05/2019','20/05/2019')df['B'] = ('SIT','SCLOSE', 'SHODL', 'SHODL', 'SHODL', 'SHODL', 'SHODL', 'SELL','SIT','SIT','BCLOSE', 'BHODL', 'BHODL', 'BHODL', 'BHODL', 'BHODL', 'BHODL','BUY','SIT','SIT')df['C'] = (0.00,1.00,10.00, 5.00,6.00,-6.00, 6.00, 0.00,0.00,0.00,-8.00,33.00,-15.00,6.00,-1.00,5.00,10.00,0.00,0.00,0.00)df.loc[19, 'D'] = 100.0000可以看出，我從最后一行的100開始D列。我正在嘗試為列D編寫計算代碼，因此從底部行（第19行）開始，當B列上顯示買入或賣出時，D列上的數(shù)字被鎖定（例如100），并用于基于每個SHODL或BHODL的col C的計算，直到顯示BCLOSE或SCLOSE之后的行。鎖定的數(shù)字用于根據(jù)列 C 中的百分比計算運行余額。如您在第 16 行中看到的，C 列的“10”表示 10%。當 100 的 10% = 10 時，新的運行平衡為 110。第 15 行 C 列有 5%，因此將 5 添加到運行余額中以產(chǎn)生 115。下一行 14 列 C 具有 -1% 的變化，因為 100 的 1% 是 = 1，因此新的運行余額是 114，依此類推。以下是運行正確的代碼后，應(yīng)在數(shù)據(jù)幀的 col D 中返回的結(jié)果df['D'] = ('158.60','158.60', '157.30', '144.30', '137.80', '130.00', '137.80', '130.00','130.00','130.00','130.00', '138.00', '105.00', '120.00', '114.00', '115.00', '110.00','100.00','100.00','100.00')這種情況一直持續(xù)到 SCLOSE 或 BCLOSE 顯示為 BCLOSE 或 SCLOSE 行是計算運行余額的最后一行。如您所見，當顯示新的買入或賣出時，此過程將重新啟動。
3 回答
喵喵時光機
TA貢獻1846條經(jīng)驗 獲得超7個贊
下一個起始值取決于上一組的最后一個值，因此我認為它無法矢量化。它需要某種迭代過程。我想出了在groupby的組上進行迭代的解決方案。反轉(zhuǎn)并分配給 。處理每組組并將最終組列表分配給原始組dfdf1df1df
df1 = df[::-1]
s = df1.B.isin(['BCLOSE','SCLOSE']).shift(fill_value=False).cumsum()
grps = df1.groupby(s)
init_val= 100
l = []
for _, grp in grps:
s = grp.C * 0.01 * init_val
s.iloc[0] = init_val
s = s.cumsum()
init_val = s.iloc[-1]
l.append(s)
df['D'] = pd.concat(l)
Out[50]:
A B C D
0 1/05/2019 SIT 0.0 158.6
1 2/05/2019 SCLOSE 1.0 158.6
2 3/05/2019 SHODL 10.0 157.3
3 4/05/2019 SHODL 5.0 144.3
4 5/05/2019 SHODL 6.0 137.8
5 6/05/2019 SHODL -6.0 130.0
6 7/05/2019 SHODL 6.0 137.8
7 8/05/2019 SELL 0.0 130.0
8 9/05/2019 SIT 0.0 130.0
9 10/05/2019 SIT 0.0 130.0
10 11/05/2019 BCLOSE -8.0 130.0
11 12/05/2019 BHODL 33.0 138.0
12 13/05/2019 BHODL -15.0 105.0
13 14/05/2019 BHODL 6.0 120.0
14 15/05/2019 BHODL -1.0 114.0
15 16/05/2019 BHODL 5.0 115.0
16 17/05/2019 BHODL 10.0 110.0
17 18/05/2019 BUY 0.0 100.0
18 19/05/2019 SIT 0.0 100.0
19 20/05/2019 SIT 0.0 100.0
慕斯王
TA貢獻1864條經(jīng)驗 獲得超2個贊
下面的這篇文章應(yīng)該可以幫助你。它產(chǎn)生預期的輸出，并且速度相對較快，因為它避免了對數(shù)據(jù)幀行的直接迭代。
endpoints = [df.first_valid_index(), df.last_valid_index()]
# occurrences of 'BCLOSE' or 'SCLOSE'
breakpoints = df.index[(df.B =='BCLOSE') | (df.B == 'SCLOSE')][::-1]
# remove the endpoints of the dataframe that do not break the structure
breakpoints = breakpoints.drop(endpoints, errors='ignore')
PERCENTAGE_CONST = 100
top = 100 # you can specify any initial value here
for i in range(len(breakpoints) + 1):
prv = breakpoints[i - 1] - 1 if i else -1 # previous or first breakpoint
try:
nex = breakpoints[i] - 1 # next breakpoint
except IndexError:
nex = None # last breakpoint
# cumulative sum of appended to 'D' column
res = top + (df['C'][prv: nex: -1] * top / PERCENTAGE_CONST).cumsum()[::-1]
df.loc[res.index, 'D'] = res
# saving the value that will be the basis for percentage calculations
# for the next breakpoint
top = res.iloc[0]
紅糖糍粑
TA貢獻1815條經(jīng)驗 獲得超6個贊
我認為有一種更優(yōu)化和pythonic的方法可以解決這個問題。但是一個帶有迭代的解決方案：
df['D'] = pd.to_numeric(df['D'])
df['C'] = pd.to_numeric(df['C'])
D_val = None
for i in range(len(df)-1, 0, -1):
if df.loc[i, 'B'] == 'BUY':
D_val = df.loc[i, 'D']
continue
if D_val is None:
continue
df.loc[i, 'D'] = df.loc[i+1, 'D'] + (D_val * df.loc[i, 'C']/100)
每次遇到 in 時，您都會更新 .我們還可以有一個停止的條件，如OP所提到的，如 或 。BUYcolumn DD_valSCLOSEBCLOSE
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