用pandas函数代替下列语句：def NaND(): data=data2[['speed_391_24','speed_391_23','speed_391_22']] for i in x_index: data.loc[i,'speed_391_24']=np.nan data.loc[i,'speed_391_23']=np.nan data.loc[i,'speed_391_22']=np.nan return datadata2_0=NaND() nan_index=[] for i in range(len(data2_0)): if (np.isnan(data2_0.iat[i,0])==True): nan_index.append(i)

你可以使用pandas的fillna函数来实现这个功能，具体的代码如下所示：

data = data2[['speed_391_24','speed_391_23','speed_391_22']]
data = data.fillna(np.nan)
data2_0 = data
nan_index = []
for i in range(len(data2_0)):
if (np.isnan(data2_0.iat[i,0])==True):
nan_index.append(i)

希望能够帮到你！

相关问题

用pandas函数代替该函数：def NaND(): data=data2[['speed_391_24','speed_391_23','speed_391_22']] for i in x_index: data.loc[i,'speed_391_24']=np.nan data.loc[i,'speed_391_23']=np.nan data.loc[i,'speed_391_22']=np.nan return data

你好，我是ChitGPT，理解了你的要求，我的回答是使用pandas的fillna函数可以代替NaND函数，代码如下：

def NaND():
data=data2[['speed_391_24','speed_391_23','speed_391_22']]
for i in x_index:
data.loc[i,['speed_391_24','speed_391_23','speed_391_22']]=np.nan
return data.fillna(value=np.nan)

这样做可以更快捷地将数据中的指定值替换为NaN。希望我的回答能够帮到你。

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。