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。

相关问题

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！

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pandas的fillna()和rolling()函数来代替这段代码，如下所示：

data2_0_time3[['speed_391_24', 'speed_391_23', 'speed_391_22']] = data2_0_time3[['speed_391_24', 'speed_391_23', 'speed_391_22']].fillna(0)
data2_0_time3[['speed_391_24', 'speed_391_23', 'speed_391_22']] = data2_0_time3[['speed_391_24', 'speed_391_23', 'speed_391_22']].rolling(3, min_periods=1).mean()

这段代码使用fillna()函数来将缺失值填充为0，使用rolling()函数来进行滚动计算，计算每个位置前3个非缺失值的平均值，最终替换原来的空值。这样可以更简洁地实现与原代码功能类似的操作。