用pandas函数代替下列语句: data2_0_time5.loc[i,'speed_391_24']=0 data2_0_time5.loc[i,'speed_391_23']=0 data2_0_time5.loc[i,'speed_391_22']=0 k=0 r=i for j in range(1,r+1): if np.isnan(data2_0_time5.iat[i-j,0])==False: data2_0_time5.loc[i,'speed_391_24']=data2_0_time5.iat[i,0]+data2_0_time5.iat[i-j,0] data2_0_time5.loc[i,'speed_391_23']=data2_0_time5.iat[i,1]+data2_0_time5.iat[i-j,1] data2_0_time5.loc[i,'speed_391_22']=data2_0_time5.iat[i,2]+data2_0_time5.iat[i-j,2] k=k+1 for j in range(1,6): if np.isnan(data2_0_time5.iat[i+j,0])==False: data2_0_time5.loc[i,'speed_391_24']=data2_0_time5.iat[i,0]+data2_0_time5.iat[i+j,0] data2_0_time5.loc[i,'speed_391_23']=data2_0_time5.iat[i,1]+data2_0_time5.iat[i+j,1] data2_0_time5.loc[i,'speed_391_22']=data2_0_time5.iat[i,2]+data2_0_time5.iat[i+j,2] k=k+1 data2_0_time5.loc[i,'speed_391_24']=data2_0_time5.iat[i,0]/k data2_0_time5.loc[i,'speed_391_23']=data2_0_time5.iat[i,1]/k data2_0_time5.loc[i,'speed_391_22']=data2_0_time5.iat[i,2]/k
时间: 2023-06-08 11:08:07 浏览: 48
可以使用pandas的滚动函数来代替这段代码,具体代码如下:
```python
data2_0_time5.loc[i, ['speed_391_24', 'speed_391_23', 'speed_391_22']] = 0
# 设置初始值为0
rolling_window = data2_0_time5.iloc[max(0, i-5):i, [0, 1, 2]].fillna(method='ffill').fillna(method='bfill')
# 获取最近的五个样本,并用前向填充和后向填充来处理缺失值
data2_0_time5.loc[i, ['speed_391_24', 'speed_391_23', 'speed_391_22']] = rolling_window.mean().values
# 将计算出的均值填充到指定位置
```
其中,rolling_window会返回一个5x3的DataFrame,包括最近五个样本的第1、2、3列的数据,也就是`['speed_391_24', 'speed_391_23', 'speed_391_22']`这三列数据。fillna() 用于填充缺失值,ffill表示前向填充,bfill表示后向填充。最后通过mean()计算均值,并将结果填充到指定位置。
相关问题
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=i for j in range(1,r+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 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 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
可以使用以下代码替代:
for j in range(1,r+1):
if not np.isnan(data2_0_time3.iloc[i-j,0]):
data2_0_time3.loc[i,'speed_391_24']+=data2_0_time3.iloc[i-j,0]
data2_0_time3.loc[i,'speed_391_23']+=data2_0_time3.iloc[i-j,1]
data2_0_time3.loc[i,'speed_391_22']+=data2_0_time3.iloc[i-j,2]
k+=1
if k>0:
data2_0_time3.loc[i,'speed_391_24']=data2_0_time3.loc[i,'speed_391_24']/k
data2_0_time3.loc[i,'speed_391_23']=data2_0_time3.loc[i,'speed_391_23']/k
data2_0_time3.loc[i,'speed_391_22']=data2_0_time3.loc[i,'speed_391_22']/k
这段代码使用了 for 循环和 iloc 函数来遍历数据,并计算出 speed_391_24、speed_391_23 和 speed_391_22 的平均值。注意,当 k=0 时,不能除以 0,我们需要进行判断。
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=i for j in range(1,r+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 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 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来的语句:
```
data2_0_time3.loc[i,['speed_391_24','speed_391_23','speed_391_22']]=[0,0,0]
for j in range(1,r+1):
if not np.isnan(data2_0_time3.iat[i-j,0]):
data2_0_time3.loc[i,['speed_391_24','speed_391_23','speed_391_22']]=[
data2_0_time3.iat[i,0]+data2_0_time3.iat[i-j,0],
data2_0_time3.iat[i,1]+data2_0_time3.iat[i-j,1],
data2_0_time3.iat[i,2]+data2_0_time3.iat[i-j,2]
]
else:
break
for j in range(1,4):
if not np.isnan(data2_0_time3.iat[i+j,0]):
data2_0_time3.loc[i,['speed_391_24','speed_391_23','speed_391_22']]=[
data2_0_time3.iat[i,0]+data2_0_time3.iat[i+j,0],
data2_0_time3.iat[i,1]+data2_0_time3.iat[i+j,1],
data2_0_time3.iat[i,2]+data2_0_time3.iat[i+j,2]
]
else:
break
k = j-1 if j>1 else 1
data2_0_time3.loc[i,['speed_391_24','speed_391_23','speed_391_22']]=[
data2_0_time3.iat[i,0]/k,
data2_0_time3.iat[i,1]/k,
data2_0_time3.iat[i,2]/k
]
```
这个代码将先将 'speed_391_24'、'speed_391_23'、'speed_391_22' 这三列设置为0,然后通过两个循环计算这三列的值,并最终将这三列的值除以 k 得到平均值。