group_a = df.groupby('Category').get_group('A') print(group_a) 问题2:分组时遇到KeyError错误怎么办? 解决方法:确保用于分组的列名在 DataFrame 中存在且拼写正确。 代码语言:txt 复制 # 确保列名正确 if 'Category' in df.columns: grouped = df.groupby('Category')['Value'].mean() ...
6000,4500,5500]}df=pd.DataFrame(data)# 定义自定义函数计算工资差异defsalary_diff(group):returngroup['salary']-group['salary'].mean()# 使用apply()方法添加工资差异列df['salary_diff']=df.groupby('department')['salary'].apply
columns.values] grouped_df = grouped_df.reset_index() grouped_df 实例7 遍历分组 代码语言:javascript 代码运行次数:0 运行 AI代码解释 for key,group_df in df.groupby('product'): print("the group for product '{}' has {} rows".format(key,len(group_df))) 代码语言:javascript 代码运行次数...
'B','C','C'],'product':['X','Y','X','Y','X','Y'],'sales':[100,150,200,120,80,250]}df=pd.DataFrame(data)# 对'category'和'product'列进行分组,然后计算sales的总和和平均值result=df.groupby(['category','product'])['sales'].agg(['sum','mean'])print(result)...
默认情况下,groupby 总是在 row 方向切割。可以指定在 columns 方向切割。 首先定义处理列索引的函数: def deal_column_name(col_name): print(f'### {col_name} ###') if ord(col_name) <= 66: return 'AB' else: return 'CD' 在调用 groupby 时指定沿 columns 方向切割: >> df.groupby(deal_...
我想创建两个新列。当小时=16时,一列将存储索引。当湿度达到最大值时,另一列将存储索引。对于每个日期,这两个操作都需要单独完成。我可以使用groupby和transform函数找到每个日期的最大湿度,如下所示: >>> df["max_humidity"] = "" >>> df["max_humidity"] = df["humidity"].groupby(df["dates"]).tr...
df_group=df.groupby("Product_Category")df_group.ngroups--Output5 小组内计数(Group Sizes) 统计分组内每个小组数据的个数,可以使用.size() df.groupby("Product_Category").size()# 筛选分组数等于nums的index,转化为listli_select=temp_df_30.groupby('codes').size().to_frame().rename(columns={0:...
You don't need to accept the names that GroupBy gives to the columns; notably(尤其)lambdafunctions have the name<lambdawhich makes them hard to identify(you can see for yourself by looking at a function's __ name__ attribute.) Thus, if you pass a list of(name, function)tuples, the...
Pandas按groupby上的列标准化 给定一个pandas数据帧,例如 import pandas as pd df = pd.DataFrame({'id': ['id1','id1','id2','id2'] , 'x': [1,2,3,4], 'y': [10,20,30,40]}) 每个数字列可以标准化为单位间隔[0,1] columns = ['x', 'y']...
df.groupby(['key1','key2']).mean() You may have noticed in the first casedf.groupby('key1').mean()that there is no key2 columns in the result. Because df['key2'] is not numeric data, it is said to be a nuisance column, which is therefore excluded from the result. By defaul...