熊猫循环优化

Question

是否有更好的方法（在性能方面）在熊猫中执行以下循环（假设df为DataFrame）？for i in range(len(df)):    if df['signal'].iloc[i] == 0:   # if the signal is negative        if df['position'].iloc[i - 1] - 0.02 < -1:   # if the row above - 0.1 < -1 set the value of current row to -1            df['position'].iloc[i] = -1        else:   # if the new col value above -0.1 is > -1 then subtract 0.1 from that value            df['position'].iloc[i] = df['position'].iloc[i - 1] - 0.02    elif df['signal'].iloc[i] == 1:     # if the signal is positive        if df['position'].iloc[i - 1] + 0.02 > 1:     # if the value above + 0.1 > 1 set the current row to 1            df['position'].iloc[i] = 1        else:   # if the row above + 0.1 < 1 then add 0.1 to the value of the current row            df['position'].iloc[i] = df['position'].iloc[i - 1] + 0.02我将不胜感激，因为我刚开始走熊猫路，很显然，可能会错过一些关键的事情。源CSV数据：Date,sp500,sp500 MA,UNRATE,UNRATE MA,signal,position2000-01-01,,,4.0,4.191666666666665,1,02000-01-02,,,4.0,4.191666666666665,1,02000-01-03,102.93,95.02135,4.0,4.191666666666665,1,02000-01-04,98.91,95.0599,4.0,4.191666666666665,1,02000-01-05,99.08,95.11245000000001,4.0,4.191666666666665,1,02000-01-06,97.49,95.15450000000001,4.0,4.191666666666665,1,02000-01-07,103.15,95.21575000000001,4.0,4.191666666666665,1,02000-01-08,103.15,95.21575000000001,4.0,4.191666666666665,1,02000-01-09,103.15,95.21575000000001,4.0,4.191666666666665,1,0更新下面的所有答案（在撰写本文时）产生的position0.02常数与我的幼稚循环方法不同。换句话说，我要寻找这样会给一个解决方案0.02，0.04，0.06，0.08等为position列。

Answer 1

感谢您添加数据和示例输出。首先，我很确定您不能对它进行矢量化处理，因为每个计算都取决于上一个的输出。所以这是我所能做到的最好的。

您的方法大约0.116999在我的机器上几秒钟

这个大约在0.0039999几秒钟之内

没有向量化，但是速度得到了很好的提高，因为为此使用列表并将其添加回末尾的数据帧更快。

def myfunc(pos_pre, signal):

    if signal == 0:  # if the signal is negative

        # if the new col value above -0.2 is > -1 then subtract 0.2 from that value

        pos = pos_pre - 0.02

        if pos < -1:  # if the row above - 0.2 < -1 set the value of current row to -1

            pos = -1

    elif signal == 1:

        # if the row above + 0.2 < 1 then add 0.2 to the value of the current row

        pos = pos_pre + 0.02

        if pos > 1:  # if the value above + 0.1 > 1 set the current row to 1

            pos = 1

    return pos

''' set first position value because you aren't technically calculating it correctly in your method since there is no 

position minus 1... IE: it will always be 0.02'''

new_pos = [0.02]

# skip index zero since there is no position 0 minus 1

for i in range(1, len(df)):

    new_pos.append(myfunc(pos_pre=new_pos[i-1], signal=df['signal'].iloc[i]))

df['position'] = new_pos

输出：

df.position

0    0.02

1    0.04

2    0.06

3    0.08

4    0.10

5    0.12

6    0.14

7    0.16

8    0.18

Answer 2

不要使用循环。熊猫专门从事矢量化运算，例如signal == 0：

pos_shift = df['position'].shift() - 0.02

m1 = df['signal'] == 0

m2 = pos_shift < -1

df.loc[m1 & m2, 'position'] = -1

df['position'] = np.where(m1 & ~m2, pos_shift, df['position'])

您可以为编写类似的内容signal == 1。

Answer 3

有很多更好的方法，但是这种方法也应该起作用：

df['previous'] = df.signal.shift()

def get_signal_value(row):

    if row.signal == 0:

        compare = row.previous - 0.02

        if compare < -1:

            return -1

        else:

            return compare

    elif row.signal == 1: 

        compare = row.previous + 0.01

        if compare > 1:

            return 1

        else:

            return compare

df['new_signal'] = df.apply(lambda row: get_signal_value(row), axis=1)