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 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
时间: 2023-06-08 16:06:33 浏览: 44
可以使用pandas的rolling函数来代替循环,代码如下:
data2_0_time3.loc[i,['speed_391_24','speed_391_23','speed_391_22']] = [0,0,0]
if i >= 3:
rollingdata = data2_0_time3.iloc[i-3:i,:3]
rollingdata = rollingdata.dropna()
rollingcnt = len(rollingdata)
if rollingcnt > 0:
data2_0_time3.loc[i,['speed_391_24','speed_391_23','speed_391_22']] = rollingdata.mean()
其中,rolling函数会在数据窗口中滑动,计算滑动窗口内的均值。如果数据窗口中包含NaN值,则在均值计算中除去该值。最终,使用mean函数计算每个窗口的均值,并在原始数据中更新对应的值。