Python：如何填充取决于前一个值（前一行）的Pandas列？

我建立一个财务应用程序。我的位置取决于以前的位置（上排），也对“信号”列（同一行）。

DataFrame被称为SPY。

position_arr = [] 
position = 0 
for row in SPY['signal']: 
    if row=='BUY' and position == 0: 
     position = 1 
    elif row=='SELL' and position == 0: 
     position = -1 
    elif row=='CLOSE SELL' and position == -1: 
     position = 0 
    elif row=='CLOSE BUY' and position == 1: 
     position = 0 
    position_arr.append(position) 

SPY['position']=position_arr

有没有更好的，更有效的方式来做到这一点？

来源

2017-02-12 Tomas Silva Ebensperger

您可以使用“信号”列apply，并使用一个变量来保持前值。

prev_val = None #if you don't know what should be the first value 

def check_condition(current_val): 
    global prev_val 
    val = 0 
    if prev_val is not None: 
     if current_val == 'BUY' and prev_val == 0: 
      val = 1 
     elif current_val == 'SELL' and prev_val == 0: 
      val = -1 
     elif current_val == 'CLOSE SELL' and prev_val == -1: 
      val = 0 
     elif current_val == 'CLOSE BUY' and prev_val == 1: 
      val = 0 
    else: # handle the first row case separately 
     val = 0 # TODO: what is the value for the first row? 
    prev_val = val 
    return val 

df['position'] = df['signal'].apply(check_condition)

来源

2017-02-12 08:11:19 AndreyF

你可以转移position列，列轴上使用apply()：

def apply_func(row): 
    if row['signal']=='BUY' and row['pos_shifted'] == 0: 
     position = 1 
    elif row['signal']=='SELL' and row['pos_shifted'] == 0: 
     position = -1 
    elif row['signal']=='CLOSE SELL' and row['pos_shifted'] == -1: 
     position = 0 
    elif row['signal']=='CLOSE BUY' and row['pos_shifted'] == 1: 
     position = 0 
    return position 

SPY['pos_shift'] = SPY['position'].shift() 
SPY['position'] = SPY.apply(apply_func, axis=1)

来源

2017-02-12 12:12:35 Toterich

Python：如何填充取决于前一个值（前一行）的Pandas列？

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