Python:如何填充依赖于前一个值(上一行)的Pandas列

Question

我正在构建一个金融应用程序。我的位置取决于上一个位置(上一行(和"信号"列(同一行(。

数据帧称为 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、更有效的方法可以做到这一点？

Answer 1

您可以移动position列并在列轴上使用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)

Answer 2

您可以在"信号"列上使用apply并使用变量来保存以前的值。

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)