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
时间: 2023-06-08 09:06:15 浏览: 44
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。
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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。
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:
```python
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。