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我有一个包含 5 个文本文件的文件夹，这些文件与各个站点有关——

标题格式如下：

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，基本上文件名遵循以下格式-（站点名称）。（站点编号）。（WRFc36）。（一些变量）。（2000-2050.txt

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

我希望 Python 搜索相似的文件名（站点名称和站点编号匹配），并从其中一个文件中挑选出第一到第三列数据并将其粘贴到新的 txt 文件中。我还想复制并粘贴站点（雨等）的每个变量的第 4 列，并将它们以特定顺序粘贴到新文件中。

我知道如何使用 csv 模块（并为空格分隔符定义新方言）从所有文件中获取数据并打印到新的文本文件，但我不确定如何为每个站点自动创建新文件名称/编号并确保我的变量以正确的顺序绘制 -

我要使用的输出是每个站点的一个文本文件（不是 5 个），格式如下（年、月、日、变量 1、变量 2、变量 3、变量 4、变量 5）约 18500 行...

我确定我在这里查看的是非常简单的东西……这似乎很简陋……但是 - 任何帮助将不胜感激！

更新

========

我已经更新了代码以反映下面的评论。
http://codepad.org/3mQEM75e

从集合导入 defaultdict 导入 glob 导入 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()
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2 回答 2

2

这是一种更简单的遍历文件的方法，按站点分组。

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()

我仍然不明白您的剪切和粘贴列-您需要更清楚地解释您要完成的工作。

于 2012-08-13T23:57:00.683 回答
1

就获取文件名而言，我将使用以下内容：

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')

然后要获取通常由句点分隔的元素，请使用：

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

然后您可以进行比较并提取所需的列（通过使用 open(file_name, 'r') 或您的 CSV 解析器）

迈克尔·G。

于 2012-08-13T21:41:06.677 回答