python 列表成对生成字典 表格的列元素成对生成列表

列表成对生成字典list_1 = list(range(3)) list_2 = ['col' + str(i) for i in list_1] dict_1 = { key: value for key, value in zip(list_2, list_1) } dict_1

list_1 = list(range(3))
list_2 = ['col' + str(i) for i in list_1]
dict_1 = {
    key: value for key, value in zip(list_2, list_1)
}
dict_1

表格的列元素成对生成列表

'''工具_列元素成对'''
df_zip = pd.DataFrame(data=np.arange(0, 12).reshape(4, 3), columns=['col' + str(i) for i in range(3)])
def fun_zip(arg_df=df_zip):
    list_result = [
        (col_name, *row) for col_name, row in zip(
            arg_df.columns, 
            zip(*[arg_df.iloc[i] for i in range(arg_df.shape[0])])
        )
    ]
    return list_result

def fun_zip(arg_df=df_zip):
    list_result = [
        tuple(row) for row in zip(*[arg_df.iloc[i] for i in range(arg_df.shape[0])])
    ]
    return list_result
fun_zip()

    def fun_zip(
        self,
        df_arg=pd.DataFrame(),
        bool_arg=False
    ):
        # 每列元素打包
        zipped = zip(*[df_arg.iloc[i] for i in range(df_arg.shape[0])])
        # 是否包含标题
        if bool_arg:
            list_zip = [(col_name, *row) for col_name, row in zip(df_arg.columns, zipped)]
        else:
            list_zip = [tuple(row) for row in zipped]
        return list_zip