用pandas函数代替下列语句： 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 r=len(data2_0_time3)-1-i 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 for j in range(1,r): 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

可以使用以下代码来代替原先的语句：

data2_0_time3.loc[i,['speed_391_24','speed_391_23','speed_391_22']] = 0
k = 0
r = len(data2_0_time3) - 1 - i
for j in range(1, 4):
    if not np.isnan(data2_0_time3.iloc[i-j, 0]):
        data2_0_time3.loc[i,['speed_391_24','speed_391_23','speed_391_22']] += data2_0_time3.iloc[i-j:i-j+1,:3]
        k += 1
for j in range(1, r):
    if not np.isnan(data2_0_time3.iloc[i+j, 0]):
        data2_0_time3.loc[i,['speed_391_24','speed_391_23','speed_391_22']] += data2_0_time3.iloc[i+j:i+j+1,:3]
        k += 1
data2_0_time3.loc[i,['speed_391_24','speed_391_23','speed_391_22']] /= k

这段代码使用了 `loc` 和 `iloc` 方法来进行索引和赋值，对于 `iloc` 方式，可以直接通过整数索引进行数据访问。使用了逻辑判断 `not np.isnan` 来判断指定位置的值是否为 NaN，并在判断为 True 时累计计算其它行的数值。最后将累计值除以 k 得到平均值，再使用 `/=` 符号将计算结果重新赋值给 `data2_0_time3` 中指定的行和列。