Question

我将dadaframe命名为＆＃34;数据＆＃34;并希望简单地按照以下代码进行调整。

花了很多时间跑。解决这个问题有什么好的解决方案吗？

for i in range(0,len(data)):
    if data['Theta'][i]<=np.pi/8 and data['Theta'][i]>=0: 
       data['Theta'][i]=0
    elif data['Theta'][i]<=3*np.pi/8 and data['Theta'][i]>1*np.pi/8: 
         data['Theta'][i] = np.pi/4
    elif data['Theta'][i]<=np.pi and data['Theta'][i]>3*np.pi/8: 
         data['Theta'][i] = np.arctan(2)
    elif data['Theta'][i]<=13*np.pi/8 and data['Theta'][i]>np.pi: 
         data['Theta'][i] = np.arctan(-2)
    elif data['Theta'][i] <= 15 * np.pi/8 and data['Theta'][i] >13*np.pi/8:
         data['Theta'][i] = -1*np.pi/4
    elif data['Theta'][i] >=15 * np.pi / 8 and data['Theta'][i] <= 2*np.pi:
         data['Theta'][i] = 0

Answer 1

如果为每个条件创建掩码，速度会更快：

data.loc[(data['Theta']<=np.pi/8) & (data['Theta']>=0), 'Theta'] = 0 data.loc[(data['Theta']<=3*np.pi/8) & (data['Theta']>1*np.pi/8), 'Theta'] = np.pi/4 
data.loc[(data['Theta']<=np.pi) & (data['Theta']>3*np.pi/8), 'Theta'] = np.arctan(2)
data.loc[(data['Theta']<=3*np.pi/8) & (data['Theta']>np.pi), 'Theta'] = np.arctan(-2) 
data.loc[(data['Theta']<=15*np.pi/8) & (data['Theta']>13*np.pi/8), 'Theta'] = -1*np.pi/4 
data.loc[(data['Theta']>=15*np.pi/8) & (data['Theta']<2*np.pi), 'Theta'] = 0

所以这使用loc和一个掩码，我们将df中的所有行与条件进行比较，这将返回一个掩盖整个df的布尔掩码，以便只更新满足条件的行。此外，我们必须使用&而不是and，因为您要比较数组而不是标量值。由于运算符优先级，我们还需要将条件包装在括号中。

Answer 2

你可以定义一个你想要的功能，然后将它应用于所有行，它应该很快。

def f(x):
    if x <= np.pi / 8 and x >=0:
        x = 0
    elif x <= 3 * np.pi / 8 and x > 1 * np.pi / 8:
        x = np.pi / 4
    elif x <= np.pi and x > 3 * np.pi / 8:
        x = np.arctan(2)
    elif x <= 13 * np.pi / 8 and x > np.pi:
        x = np.arctan(-2)
    elif x <= 15 * np.pi / 8 and x > 13 * np.pi / 8:
        x = -1 * np.pi / 4
    elif x >= 15 * np.pi / 8 and x <= 2 * np.pi:
         x = 0
    return x

data["Theta"] = data["Theta"].apply(f)

如何简单地遵循代码，Dataframe，迭代，for

2 个答案: