python 合并不同文件夹下名称相同的文件

转载:https://blog.csdn.net/qq_42769683/article/details/104565285?utm_source=app&app_version=4.10.0&code=app_1562916241&uLinkId=usr1mkqgl919blen

1.目的:将不同文件夹下具有相同文件名称的文件进行合并,并保证文件中第一行标题不重复(按行整合)。

代码:

import pandas as pd
import os

def merge_data(id):
    base = 'path to Excel file/'  #大文件夹,注意最后的/
    dir_names = os.listdir(base) # 打开大文件夹后的各个小文件夹名dir_names
    df_all = pd.DataFrame()
    for dir in dir_names:  #遍历每一个小文件夹
        file_path = base+dir
        file_names = os.listdir(file_path)  #打开小文件夹后每一个excel文件的名称
        for file in file_names:  #遍历小文件夹里的每一个文件
            file_name = file_path+'/'+file
 
            if file.split('.')[0] == id:
                    df = pd.read_excel(file_name)
                   # df_all = df_all.append(df)
                    df_all = df_all.append(df,ignore_index=True)#索引重新排序
    return df_all

if __name__ == '__main__':
    
    base = 'path to excel file/'
    dir_names = os.listdir(base)
    ids = set()  # set()可以看做一个列表,这里面不包含重复的元素,不可以使用索引
    for dir in dir_names:
        file_path = base+dir
        file_names = os.listdir(file_path)
        for file_name in file_names:
            id = file_name.split('.')[0]  #id表示这一个股票的代码
            ids.add(id)  #把所有id放入ids中,这样不会有重复的id存在
    for id in ids:
        df = merge_data(id)
        base = 'path to save file/'  #存放合并后的文件路径
        id_path = base + id + '.xlsx'
        df.to_excel(id_path)

2.更新:将两个不同文件夹下同名文件合并,并且要求合并后的文件包括两个sheet。
代码

import pandas as pd
import os

from pandas import ExcelWriter
writer = ExcelWriter("path to save output excel/output.xlsx")

base = 'path to excel /'  #大文件夹,注意最后的/
dir_names = os.listdir(base) # 打开大文件夹后的各个小文件夹名dir_names
df_all = pd.DataFrame()

for dir in dir_names:  #遍历每一个小文件夹
    file_path = base+dir
    file_names = os.listdir(file_path)  #打开小文件夹后每一个excel文件的名称
    
    for file in file_names:  #遍历小文件夹里的每一个文件
        file_name = file_path+'/'+file
        
        df_excel = pd.read_excel(file_name)
        (_, f_name) = os.path.split(file_name)
        (f_short_name, _) = os.path.splitext(f_name)
        df_excel.to_excel(writer, f_short_name, index=False)

writer.save()

3.更新:将两个不同文件夹同名的文件合并,按列合并成一个sheet

import pandas as pd

import numpy as np

import os

import glob

from os.path import join

#%%
#合并不同文件夹下相同文件名Excel,按列合并

# out file path
outDir = os.path.abspath('save path') 

#one file
imageDir1 = os.path.abspath('file path1')

#define area file variable
image1 = [] #1.txt;2.txt
imgname1 = [] #1;2

#get all .txt file
imageList1 = glob.glob(os.path.join(imageDir1, '*.xlsx'))

#get filename (1.txt;2.txt)
for item in imageList1:
    image1.append(os.path.basename(item))

#get filename(1;2)
for item in image1:
    (temp1, temp2) = os.path.splitext(item)
    imgname1.append(temp1)

#second file 
imageDir2 = os.path.abspath('file path2')
image2 = []
imgname2 = []
imageList2 = glob.glob(os.path.join(imageDir2, '*.xlsx'))

for item in imageList2:
    image2.append(os.path.basename(item))

for item in image2:
    (temp1, temp2) = os.path.splitext(item)
    imgname2.append(temp1)

#f the first file name and sencond file name are the same, the two groups of data are merged 
for item1 in imgname1:
    for item2 in imgname2:
        if item1 == item2:
            
            dir1 = imageList1[imgname1.index(item1)]
            dir2 = imageList2[imgname2.index(item2)]
            
            data1 = pd.read_excel(dir1)
            area = data1
            # print(data)
            name1 = os.path.basename(dir1)
            
            data2 = pd.read_excel(dir2)
            height = data2[0:132].reset_index(drop=True)
            # print(data)
            name2 = os.path.basename(dir2)
            
            data = pd.concat([area,height],axis=1,ignore_index=True)
            pd_data = pd.DataFrame(data)

            pd_data.to_csv(os.path.join(outDir, name1.split('.')[0]+'.csv'))
print('done!')

参考资料:https://blog.csdn.net/weixin_43668299/article/details/97807698

4.快速将一个文件夹下所有csv文件合并为一个文件:copy *.csv all.csv需要手动删除表头。

    原文作者:RS&Hydrology
    原文地址: https://blog.csdn.net/qq_32649321/article/details/118416005
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