Python program to simply add a column level to a pandas dataframe # Importing pandas packageimportrandomimportpandasaspd# Creating a Dictionaryd={'A':[iforiinrange(25,35)],'B':[iforiinrange(35,45)] }# Creating a DataFramedf=pd.DataFrame(d,index=['a','b','c','d','e','f','...
dtype: float64'''#index返回一个RangeIndex对象,用来描述索引的取值范围#默认索引print(s.index)#RangeIndex(start=0, stop=5, step=1)#自定义索引print(s1.index)#Index(['a', 'b', 'c', 'd', 'e'], dtype='object')#通过.index.values 获取索引列表print(s.index.values)#[0 1 2 3 4]pri...
Parameters: axis : {0 or ‘index’, 1 or ‘columns’}, default 0 0 or ‘index’ for row-wise, 1 or ‘columns’ for column-wise level : int or level name, default None If the axis is a MultiIndex (hierarchical), count along a particular level, collapsing into a DataFrame numeric_...
add(other[, axis, level, fill_value])获取DataFrame和other的加法，逐元素执行（二进制运算符add）。
df.info() <class 'pandas.core.frame.DataFrame'> RangeIndex: 6040 entries, 0 to 6039 Data columns (total 5 columns): UserID 6040 non-null int64 Gender 6040 non-null object Age 6040 non-null int64 Occupation 6040 non-null int64 Zip-code 6040 non-null object dtypes: int64(3), object(2...
# df.columns是一个Index对象,也可使用.str # 成员资格:.isin() df.columns=df.columns.str.upper() print(df) 2.字符串常用方法 # 字符串常用方法(1) -lower,upper,len,startswith,endswith s= pd.Series(['A','b','bbhello','123',np.nan]) ...
4.MultiIndex可在 column 上设置 indexs 的多层索引 我们可以使用MultiIndex.from_product()函数创建一个...
index values on the otheraxes are still respected in the join.keys : sequence, default NoneIf multiple levels passed, should contain tuples. Constructhierarchical index using the passed keys as the outermost level.levels : list of sequences, default NoneSpecific levels (unique values) to use ...
df.set_index(['Gender','School']).groupby(level=1,axis=0).get_group('S_1').head() 1. 2. groupby对象的特点: 查看所有可调用的方法 分组对象的head 和first 分组依据 groupby的[]操作 连续型变量分组 a). 查看所有可调用的方法 由此可见,groupby对象可以使用相当多的函数,灵活程度很高 ...
IIUC,您需要添加一个助手列: (df.assign(group=df.groupby(level=0).cumcount()) .set_index('group', append=True)[0] # 0 is the name of the column here .unstack('group')) or: (df.reset_index() .assign(group=lambda d: d.groupby('index').cumcount()) .pivot('index', 'group', ...