使用Python从多个文本文件中提取列

Question

我有一个文件夹，里面有5个与各种网站有关的文本文件 -

标题的格式是这样的：

Rockspring_18_SW.417712.WRFc36.ET.2000-2050.txt

Rockspring_18_SW.417712.WRFc36.RAIN.2000-2050.txt

WICA.399347.WRFc36.ET.2000-2050.txt

WICA.399347.WRFc36.RAIN.2000-2050.txt

所以，基本上文件名的格式为 - （网站名称）。（网站编号）。（WRFc36）。（某些变量）。（2000-2050.txt

这些文本文件中的每一个都具有与其类似的格式，没有标题行：年月日值（每个文本文件中包含~18500行）

我希望Python搜索类似的文件名（站点名称和站点号匹配），并从其中一个文件中挑选第一到第三列数据并将其粘贴到新的txt文件中。 我还想复制并粘贴每个变量的第4列（雨，等等），并在新文件中按特定顺序粘贴它们。

我知道如何使用csv模块（并为空间分隔符定义新的方言）从ALL文件中获取数据并打印到新的文本文件，但我不知道如何为每个站点自动创建新文件姓名/号码，并确保我的变量以正确的顺序绘制出来 -

我想要使​​用的输出是每个站点的一个文本文件（不是5），具有以下格式（年，月，日，变量1，变量2，变量3，变量4，变量5），用于~18500行...

我确定我正在寻找一些非常简单的东西......这看起来很简陋......但是 - 任何帮助都会非常感激！

更新

========

我已更新代码以反映以下评论。
http://codepad.org/3mQEM75e

来自集合import defaultdict import glob import csv

#Create dictionary of lists--   [A] = [Afilename1, Afilename2, Afilename3...]
#                               [B] = [Bfilename1, Bfilename2, Bfilename3...] 
def get_site_files():
    sites = defaultdict(list)
    #to start, I have a bunch of files in this format ---
    #"site name(unique)"."site num(unique)"."WRFc36"."Variable(5 for each site name)"."2000-2050"
    for fname in glob.glob("*.txt"):
        #split name at every instance of "."
        parts = fname.split(".")
        #check to make sure i only use the proper files-- having 6 parts to name and having WRFc36 as 3rd part
        if len(parts)==6 and parts[2]=='WRFc36':
            #Make sure site name is the full unique identifier, the first and second "parts"
            sites[parts[0]+"."+parts[1]].append(fname)
    return sites

#hardcode the variables for method 2, below
Var=["TAVE","RAIN","SMOIS_INST","ET","SFROFF"]

def main():
    for site_name, files in get_site_files().iteritems():
        print "Working on *****"+site_name+"*****"
####Method 1- I'd like to not hardcode in my variables (as in method 2), so I can use this script in other applications.
        for filename in files:
            reader = csv.reader(open(filename, "rb"))
            WriteFile = csv.writer(open("XX_"+site_name+"_combined.txt","wb"))
            for row in reader:
                row = reader.next()
####Method 2 works (mostly), but skips a LOT of random lines of first file, and doesn't utilize the functionality built into my dictionary of lists...            
##        reader0 = csv.reader(open(site_name+".WRFc36."+Var[0]+".2000-2050.txt", "rb"))    #I'd like to copy ALL columns from the first file
##        reader1 = csv.reader(open(site_name+".WRFc36."+Var[1]+".2000-2050.txt", "rb"))    #    and just the fourth column from all the rest of the files
##        reader2 = csv.reader(open(site_name+".WRFc36."+Var[2]+".2000-2050.txt", "rb"))    #    (the columns 1-3 are the same for all files)
##        reader3 = csv.reader(open(site_name+".WRFc36."+Var[3]+".2000-2050.txt", "rb"))
##        reader4 = csv.reader(open(site_name+".WRFc36."+Var[4]+".2000-2050.txt", "rb"))
##        WriteFile = csv.writer(open("XX_"+site_name+"_COMBINED.txt", "wb"))               #creates new command to write a text file
##
##        for row in reader0:
##            row  = reader0.next()
##            row1 = reader1.next()
##            row2 = reader2.next()
##            row3 = reader3.next()
##            row4 = reader4.next()
##            WriteFile.writerow(row + row1 + row2 + row3 + row4)
##        print "***finished with site***"

if __name__=="__main__":
    main()

Answer 1

这是按网站分组的更简单的迭代文件的方法。

from collections import defaultdict
import glob

def get_site_files():
    sites = defaultdict(list)
    for fname in glob.glob('*.txt'):
        parts = fname.split('.')
        if len(parts)==6 and parts[2]=='WRFc36':
            sites[parts[0]].append(fname)
    return sites

def main():
    for site,files in get_site_files().iteritems():
        # you need to better explain what you are trying to do here!
        print site, files

if __name__=="__main__":
    main()

我仍然不理解你的剪切和粘贴专栏 - 你需要更清楚地解释你想要完成的事情。

Answer 2

至于获取文件名，我会使用如下内容：

import os

# Gets a list of all file names that end in .txt
# ON *nix
file_names = os.popen('ls *.txt').read().split('\n')

# ON Windows
file_names = os.popen('dir /b *.txt').read().split('\n')

然后要获取通常用句点分隔的元素，请使用：

# For some file_name in file_names
file_name.split('.')

然后，您可以继续进行比较并提取所需的列（使用open（file_name，'r'）或CSV解析器）

迈克尔G.