用pandas函数代替下列语句：for i in range(len(data2_0_time3)): if (np.isnan(data2_0_time3.iat[i,0])==True): if 3<=i<=len(data2_0_time3)-4: data2_0_time3.loc[i,'speed_391_24']=0 data2_0_time3.loc[i,'speed_391_23']=0 data2_0_time3.loc[i,'speed_391_22']=0 k=0 for j in range(1,4): if np.isnan(data2_0_time3.iat[i-j,0])==False: data2_0_time3.loc[i,'speed_391_24']=data2_0_time3.iat[i,0]+data2_0_time3.iat[i-j,0] data2_0_time3.loc[i,'speed_391_23']=data2_0_time3.iat[i,1]+data2_0_time3.iat[i-j,1] data2_0_time3.loc[i,'speed_391_22']=data2_0_time3.iat[i,2]+data2_0_time3.iat[i-j,2] k=k+1 if np.isnan(data2_0_time3.iat[i+j,0])==False: data2_0_time3.loc[i,'speed_391_24']=data2_0_time3.iat[i,0]+data2_0_time3.iat[i+j,0] data2_0_time3.loc[i,'speed_391_23']=data2_0_time3.iat[i,1]+data2_0_time3.iat[i+j,1] data2_0_time3.loc[i,'speed_391_22']=data2_0_time3.iat[i,2]+data2_0_time3.iat[i+j,2] k=k+1 data2_0_time3.loc[i,'speed_391_24']=data2_0_time3.iat[i,0]/k data2_0_time3.loc[i,'speed_391_23']=data2_0_time3.iat[i,1]/k data2_0_time3.loc[i,'speed_391_22']=data2_0_time3.iat[i,2]/k

可以使用下列代码代替原始的for循环代码：

mask = np.isnan(data2_0_time3.iloc[:, 0])
mask[3:-4] &= True
for i in range(1, 4):
    mask[3:-4] &= ~np.isnan(data2_0_time3.iloc[:, i])
data2_0_time3.loc[mask, 'speed_391_24'] = data2_0_time3.loc[mask, 0]
data2_0_time3.loc[mask, 'speed_391_23'] = data2_0_time3.loc[mask, 1]
data2_0_time3.loc[mask, 'speed_391_22'] = data2_0_time3.loc[mask, 2]
for i in range(1, 4):
    data2_0_time3.loc[mask, 'speed_391_24'] += data2_0_time3.shift(-i)[mask][0]
    data2_0_time3.loc[mask, 'speed_391_23'] += data2_0_time3.shift(-i)[mask][1]
    data2_0_time3.loc[mask, 'speed_391_22'] += data2_0_time3.shift(-i)[mask][2]
data2_0_time3.loc[mask, 'speed_391_24'] /= (k + i)
data2_0_time3.loc[mask, 'speed_391_23'] /= (k + i)
data2_0_time3.loc[mask, 'speed_391_22'] /= (k + i)

此代码将创建一个掩码（即一个布尔数组），以标识哪些行需要更新“speed_391_XX”列。接下来，通过使用函数`shift`和`loc`将该列中的值从相邻的3行中收集到目标行。最后，它更新列的平均值为除以观察到的非空值数量。