Python中用字典替换DataFrame空值失败，求解决方法

原代码存在的问题 #

• 字典迭代错误：for k,v in dic应改为for k,v in dic.items()，否则会抛出ValueError（字典默认迭代键，无法直接解包为键值对）
• 缺失值判断错误：pandas中的缺失值为np.nan，不能用== None判断，需用pd.isna()或np.isnan()
• 链式索引修改问题：df['Weight'][i]属于链式索引，修改操作可能无法作用于原DataFrame，应使用df.loc[i, 'Weight']

更高效的解决方案（无需循环） #

对于大型DataFrame，循环遍历效率极低，推荐使用pandas内置分组填充方法：

方法1：groupby.transform + fillna

按Name分组，用每组的有效值（与原逻辑一致取max）填充缺失值：

import pandas as pd
import numpy as np

lst1 = ["AA","BB","CC","AA","BB","CC","AA","BB","CC"]
lst2 = [12,np.nan,14,12,15,14,np.nan,np.nan,14]
df = pd.DataFrame(list(zip(lst1,lst2)), columns = ['Name','Weight'])

# 按Name分组，用组内最大值填充缺失
df['Weight'] = df.groupby('Name')['Weight'].transform(lambda x: x.fillna(x.max()))
print(df)

方法2：字典映射填充

若坚持使用字典逻辑，可结合map和fillna避免循环：

# 生成每个Name对应的Weight字典
weight_dict = df.groupby('Name')['Weight'].max().to_dict()
# 用字典映射填充缺失值
df['Weight'] = df['Weight'].fillna(df['Name'].map(weight_dict))
print(df)

修复后的循环代码（不推荐用于大型DataFrame） #

若必须使用循环，修正后代码如下：

import pandas as pd
import numpy as np

lst1 = ["AA","BB","CC","AA","BB","CC","AA","BB","CC"]
lst2 = [12,np.nan,14,12,15,14,np.nan,np.nan,14]
df = pd.DataFrame(list(zip(lst1,lst2)), columns = ['Name','Weight'])

df_2 = df.groupby('Name')['Weight'].max()
dic = df_2.to_dict()

# 修正循环逻辑
for k, v in dic.items():
    for i in range(len(df)):
        if pd.isna(df.loc[i, 'Weight']) and df.loc[i, 'Name'] == k:
            df.loc[i, 'Weight'] = v
print(df)

内容的提问来源于stack exchange，提问作者Mariana Arismendi