我有疑問如何優化我的代碼，實際上只有回圈。我用來計算最多兩行的解決方案，或者有時是行和數量的最大值。
我嘗試使用 .loc 和 .clip 更改我的代碼，但是當它是關于 max 或 min 多次出現時，我在邏輯運算式方面遇到了一些麻煩。
它看著開頭：
def Calc(row):
if row['Forecast'] == 0:
return max(row['Qty'],0)
elif row['def'] == 1:
return 0
elif row['def'] == 0:
return round(max(row['Qty'] - ( max(row['Forecast_total']*14,(row['Qty_12m_1'] row['Qty_12m_2'])) * max(1, (row['Total']/row['Forecast'])/54)),0 ))
df['Calc'] = df.apply(Calc, axis=1)
我設法使用我指出的函式來改變它，但我有一個問題如何寫這個 max(max())
df.loc[(combined_sf2['Forecast'] == 0),'Calc'] = df.clip(0,None)
df.loc[(combined_sf2['def'] == 1),'Calc'] = 0
df.loc[(combined_sf2['def'] == 0),'Calc'] = round(max(df['Qty']- (max(df['Forecast_total']
*14,(df['Qty_12m_1'] df['Qty_12m_2']))
*max(1, (df['Total']/df['Forecast'])/54)),0))
前兩個功能有效，最后一個無效。
id Forecast def Calc Qty Forecast_total Qty_12m_1 Qty_12m_2 Total
31551 0 0 0 2 0 0 0 95
27412 0,1 0 1 3 0,1 11 0 7
23995 0,1 0 0 4 0 1 0 7
27411 5,527 1 0,036186 60 0,2 64 0 183
28902 5,527 0 0,963814 33 5,327 277 0 183
23954 5,527 0 0 6 0 6 0 183
23994 5,527 0 0 8 0 0 0 183
31549 5,527 0 0 6 0 1 0 183
31550 5,527 0 0 6 0 10 0 183
uj5u.com熱心網友回復：
使用numpy.select并改為max使用numpy.maximum：
m1 = df['Forecast'] == 0
m2 = df['def'] == 1
m3 = df['def'] == 0
s1 = df['Qty'].clip(lower=0)
s3 = round(np.maximum(df['Qty'] - (np.maximum(df['Forecast_total']*14,(df['Qty_12m_1'] df['Qty_12m_2'])) * np.maximum(1, (df['Total']/df['Forecast'])/54)),0 ))
df['Calc2'] = np.select([m1, m2, m3], [s1, 0, s3], default=None)
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