Question

我有10个不同的子目录，每个目录中有相同的文件名（每个目录20个文件），第0列是每个文件中的索引列。

例如

     **strong text**DIRECTORY  A
    - data_20170101_k.csv
    - data_20170102_k.csv
    - data_20170102_k.csv
    - data_20170103_k.csv
    - data_20170104_k.csv
    - data_20170105_k.csv
    .....
    .....
    - data_20170120_k.csv  



    **DIRECTORY  B**
    - data_20170101_k.csv
    - data_20170102_k.csv
    - data_20170102_k.csv
    - data_20170103_k.csv
    - data_20170104_k.csv
    - data_20170105_k.csv
    .....
    .....
    - data_20170120_k.csv                




    **DIRECTORY  C**
    - data_20170101_k.csv
    - data_20170102_k.csv
    - data_20170102_k.csv
    - data_20170103_k.csv
    - data_20170104_k.csv
    - data_20170105_k.csv
    .....
    .....
    - data_20170120_k.csv                


   Each of the above files contains 6 columns and index_col = 0  with NO
   column headers

   **DIRECTORY  FILES_MERGED**
   - data_20170101_k.csv
   - data_20170102_k.csv
   - data_20170102_k.csv
   - data_20170103_k.csv
   - data_20170104_k.csv
   - data_20170105_k.csv
   .....
   .....
   - data_20170120_k.csv

我想将所有文件与EACH子目录中的SAME NAME合并使用SAME NAME将1个文件保存到一个NEW子目录中例如目录FILES_MERGED与INDEX =第0列。合并文件只有一个索引列，每个文件有1,2,3,4,5列每个目录中的相同名称

我已将csv文件读入pandas数据帧

   df= pd.read_csv(filename, sep=",", header = None, usecols=[0, 1, 2, 3, 4, 5])

这是dataframe的格式

我最初的原始数据框：

             0       1        2        3        4     5
   0  1451606820  1.0862  1.08630  1.08578  1.08578  25
   1  1451608800  1.0862  1.08630  1.08578  1.08610  10
   2  1451608860  1.0862  1.08620  1.08578  1.08578  16
   3  1451610180  1.0862  1.08630  1.08578  1.08578  27
   4  1451610480  1.0858  1.08590  1.08560  1.08578  21
   5  1451610540  1.0857  1.08578  1.08570  1.08578   2
   6  1451610600  1.0857  1.08578  1.08570  1.08578   2
   7  1451610720  1.0857  1.08578  1.08570  1.08578   2
   8  1451610780  1.0857  1.08578  1.08570  1.08578   2

   Column '0' = Datetime in Epoch time 
   Columns 1,2,3,4,5 are values

Answer 1

有很多方法可以做到这一点，留在熊猫我做了以下几点。

使用文件结构

root/  
├── dir1/  
│   ├── data_20170101_k   
│   ├── data_20170102_k    
│   ├── ...  
├── dir2/    
│   ├── data_20170101_k    
│   └── data_20170101_k  
│   └── ...   
└── ...

这段代码可行，但解释时有点冗长，但您可以缩短实施时间。

import glob
import pandas as pd

CONCAT_DIR = "/FILES_CONCAT/"

# Use glob module to return all csv files under root directory. Create DF from this.
files = pd.DataFrame([file for file in glob.glob("root/*/*")], columns=["fullpath"])

#    fullpath
# 0  root\dir1\data_20170101_k.csv
# 1  root\dir1\data_20170102_k.csv
# 2  root\dir2\data_20170101_k.csv
# 3  root\dir2\data_20170102_k.csv

# Split the full path into directory and filename
files_split = files['fullpath'].str.rsplit("\\", 1, expand=True).rename(columns={0: 'path', 1:'filename'})

#    path       filename
# 0  root\dir1  data_20170101_k.csv
# 1  root\dir1  data_20170102_k.csv
# 2  root\dir2  data_20170101_k.csv
# 3  root\dir2  data_20170102_k.csv

# Join these into one DataFrame
files = files.join(files_split)

#    fullpath                       path        filename
# 0  root\dir1\data_20170101_k.csv  root\dir1   data_20170101_k.csv
# 1  root\dir1\data_20170102_k.csv  root\dir1   data_20170102_k.csv
# 2  root\dir2\data_20170101_k.csv  root\dir2   data_20170101_k.csv
# 3  root\dir2\data_20170102_k.csv  root\dir2   data_20170102_k.csv

# Iterate over unique filenames; read CSVs, concat DFs, save file
for f in files['filename'].unique():
    paths = files[files['filename'] == f]['fullpath'] # Get list of fullpaths from unique filenames
    dfs = [pd.read_csv(path, header=None) for path in paths] # Get list of dataframes from CSV file paths
    concat_df = pd.concat(dfs) # Concat dataframes into one
    concat_df.to_csv(CONCAT_DIR + f) # Save dataframe

Answer 2

这可以在shell中以非常简单的方式实现：

find . -name "*.csv" | xargs cat > mergedCSV

（注意：请勿在扩展名中使用.csv，因为这会导致与find不一致。完成此命令后，可以将文件重命名为.csv

在10个不同的子目录中合并多个具有相同名称的csv文件

2 个答案: