Question

我是Python新手，有一个我不知道如何解决的问题。
我有以下代码：

#Calculating the closest distances
df_final=pd.DataFrame()

records = df_ipos.to_dict('records') #converting dataframe to a list of dictionaries

def return_closest(df,inp_record):
    """returns the closest euclidean distanced record"""
    
    filtered_records = df.to_dict('records')#converting dataframe to a list of dictionaries

    for record in filtered_records: #iterating through dictionaries

        params = ['z_LogSales','z_Leverage','z_LevR', 'z_ATR','z_ProfitMargin','z_RevenueGrowth']  #parameters  to calculate euclidean distance
        distance = []
        for param in params:
            d1,d2 = record.get(param,0),inp_record.get(param,0) # fetching value of these parameters. default is0 if not found
            if d1!=d1:  #checking isNan
                d1 = 0
            if d2!=d2:
                d2 = 0
            distance.append((d1 - d2)**2)
        euclidean = math.sqrt(sum(distance))
        record['Euclidean distance'] = round(euclidean,6) #assigning to a new key 
        
    distance_records = sorted(filtered_records,key = lambda x:x['Euclidean distance']) #sorting in increasing order
    
    return next(filter(lambda x:x['Euclidean distance'],distance_records),None) #returning the lowest value which is not zero. Default None

for record in records:
    ipo_year = record.get('IPO Year')
    MainSICCode_industry48 = record.get('MainSICCode_industry48')
    df = df_fundamentals[df_fundamentals['Year']==ipo_year]
    if df.shape[0]>0:
        df = df[df['MainSICCode_industry48_y']==MainSICCode_industry48] #filtering dataframe
        df.drop(df[df['Name'] == record.get('Name')].index, inplace = True) #Droping same row having same Name
        closest_record = return_closest(df,record)
        if closest_record:
            record['Closest Company'] = closest_record.get('Name')  #adding new columns
            record['Actual Distance'] = closest_record.get('Euclidean distance')
            df_dist = pd.DataFrame([record]) #changing list of dictionaries back to dataframe
            df_dist.replace(-np.inf, np.nan)
            df_final=df_final.append(df_dist)
df_final.drop_duplicates()
df_final2 = df_final

该代码计算公司样本与一组可比较公司之间的欧式距离，并最终选择与同一行业最接近的一家。唯一的问题是我在计算期间不知道如何处理NaN值。如果一个参数（参数）为NaN，我想跳过这家公司的信息，只计算其余参数上的距离。目前，NaN被视为0。有人知道我该如何实现。

Answer 1

如果param为NaN，则可以break的for循环退出循环，也可以continue进入下一个参数。

for param in params:
            d1,d2 = record.get(param,0),inp_record.get(param,0) # fetching value of these parameters. default is0 if not found
            if d1!=d1:  #checking isNan
                continue # continues the loop with the next param
            if d2!=d2:
                continue # continues the loop with the next param
            distance.append((d1 - d2)**2)
        euclidean = math.sqrt(sum(distance))

如何处理函数中的Nan值

1 个答案: