在 python 中使用 glob 读取多个文件时，如何匹配同一文件夹中具有相同起始模式的文件名模式？

Question

我有十个文件，名称如下：

加强剂量文件：

• sacoronavirus_total_number_vaccinated_type_2022-04-30.csv
• sacoronavirus_total_number_vaccinated_type_2022-05-31.csv
• sacoronavirus_total_number_vaccinated_type_2022-06-30.csv
• sacoronavirus_total_number_vaccinated_type_2022-07-31.csv
• sacoronavirus_total_number_vaccinated_type_2022-08-31.csv

完全疫苗接种文件：

• sacoronavirus_total_number_vaccinated_type_fully_vaccinated_2022-04-30.csv
• sacoronavirus_total_number_vaccinated_type_fully_vaccinated_2022-05-31.csv
• sacoronavirus_total_number_vaccinated_type_fully_vaccinated_2022-06-30.csv
• sacoronavirus_total_number_vaccinated_type_fully_vaccinated_2022-07-31.csv
• sacoronavirus_total_number_vaccinated_type_fully_vaccinated_2022-08-31.csv

它们必须放在同一个文件夹中，并且文件名不能更改。 当尝试将加强剂量文件加载到一个 dataframe 并将完全接种的文件加载到另一个时，在一个 python 脚本中，如下所示：

import pandas as pd
from pathlib import Path

path_name = r'C:/Users/Tim/Documents/UPDATED/data/'

def load_and_concat_files(path_name,month_year_col:str,filename_prefix:str,start_index:int,end_index:int):
    """
    It takes a path name, a column name, a filename prefix, a start index and an end index as parameters
    and returns a dataframe with the concatenated data from all the files in the path that match the
    filename prefix
    
    :param path_name: The path to the folder where the files are stored
    :type path_name: str
    :param month_year_col: The column name that will be added to the dataframe with the month and year
    of the file
    :type month_year_col: str
    :param filename_prefix: The prefix of the file name. For example, if the file name is
    "sales_jan_2020.csv", then the prefix is "sales_jan_"
    :type filename_prefix: str
    :param start_index: The index of the first character of the month-year string in the filename
    :type start_index: int
    :param end_index: the index of the last character of the month-year string in the filename
    :type end_index: int
    :return: A dataframe
    """
    try:
        files = Path(path_name).glob(filename_prefix + '*.csv')
        dfs = list()
        for f in files:
            data = pd.read_csv(f)
            data[month_year_col] = f.stem[start_index:end_index]
            dfs.append(data)
        df = pd.concat(dfs, ignore_index=True)
        return df
    except Exception as e:
        print(f"Exception on loading and concatenating files with error: {e}")

df_fully_vaccinated = load_and_concat_files(path_name,'mon_yr','sacoronavirus_total_number_vaccinated_type_fully_vaccinated',60,70)

df_booster_doses = load_and_concat_files(path_name,'mon_yr','sacoronavirus_total_number_vaccinated_type_',43,53)

创建df_fully_vaccinated创建正确/所需的 output，而df_booster_doses包括所有完全接种的文件。 显然，因为文件名的起始模式是相似的。 因为我无法更改文件名和路径，如何确保在加载每个 dataframe 的数据时，它只读取所需的数据？