的Python OUTFILE不寫的一切

Question

我有代碼，讓我（至少對我來說）執行一個相當複雜的任務：

import csv 
import os.path 
#open files + readlines 
with open("C:/Users/Ivan Wong/Desktop/Placement/Lists of targets/Mouse/UCSC to Ensembl.csv", "r") as f: 
    reader = csv.reader(f, delimiter = ',') 
    #find files with the name in 1st row 
    for row in reader: 
     graph_filename = os.path.join("C:/Python27/Scripts/My scripts/Top targets",row[0]+"_nt_counts.txt.png") 
     if os.path.exists(graph_filename): 
      y = row[0]+'_nt_counts.txt' 
      r = open('C:/Users/Ivan Wong/Desktop/Placement/fp_mesc_nochx/'+y, 'r') 
      k = r.readlines() 
      r.close 
      del k[:1] 
      k = map(lambda s: s.strip(), k) 
      interger = map(int, k) 
      import itertools 
      #adding the numbers for every 3 rows 
      def grouper(n, iterable, fillvalue=None): 
       "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx" 
       args = [iter(iterable)] * n 
       return itertools.izip_longest(*args, fillvalue=fillvalue) 
      result = map(sum, grouper(3, interger, 0))  
      e = row[0] 
      print e 
      cDNA = open('C:/Users/Ivan Wong/Desktop/Placement/Downloaded seq/Mouse/MOUSE_mRNAs.txt', 'r') 
      seq = cDNA.readlines() 
      # get all lines that have a gene name 
      lineNum = 0; 
      lineGenes = [] 
      for line in seq: 
       lineNum = lineNum +1 
       if '>' in line: 
        lineGenes.append(str(lineNum)) 
       if '>'+e in line: 
        lineBegin = lineNum 

      cDNA.close 

      # which gene is this 
      index1 = lineGenes.index(str(lineBegin)) 
      lineEnd = lineGenes[index1+1]   
# linebegin and lineEnd now give you, where to look for your sequence, all that 
# you have to do is to read the lines between lineBegin and lineEnd in the file 
# and make it into a single string.    
      lineEnd = lineGenes[index1+1] 
      Lastline = int(lineEnd) -1 

# in your code you have already made a list with all the lines (q), first delete 
# \n and other symbols, then combine all lines into a big string of nucleotides (like this)  
      qq = seq[lineBegin:Lastline] 
      qq = map(lambda s: s.strip(), qq) 
      string = '' 
      for i in range(len(qq)): 
       string = string + qq[i] 
# now you want to get a list of triplets, again you can use the for loop: 
# first get the length of the string 
      lenString = len(string); 
# this is your list codons 
      listCodon = [] 
      for i in range(0,lenString/3): 
       listCodon.append(string[0+i*3:3+i*3]) 
      proper_result = '\n'.join('%s, %s' % (nr, codon) for nr, codon in zip(result, listCodon)) 
      with open(e+'.csv','wb') as outfile: 
       outfile.writelines(proper_result) 

這些代碼讀取從.csv文件，從文件的文件夾識別具有相同的名稱，如果存在，那麼它繼續處理一些數據，並將其寫入到.csv 和他們在一起，我outfiles現在看起來是這樣

它看起來完全沒問題，但有一個問題，我知道從我的數據（我以不同的方式檢查過）第二列應該比我得到的長。我認爲這是因爲代碼正在寫入文件時，兩個結果（數字）和listCodon（字母）都存在，因此我失去了一些東西。我該如何解決它？

我試圖打印listCodon文件寫入之前，發現了所有的三胞胎仍然存在，所以我猜問題是內這裏：

proper_result = '\n'.join('%s, %s' % (nr, codon) for nr, codon in zip(result, listCodon)) 

Answer 1

zip將停止儘快任何其iterables停止（因爲否則將不知道該怎麼填補空白用！）：

返回的列表的長度被截斷，以最短的說法序列的長度。

如果要將較短的迭代次數填充到最長的長度，請使用izip_longest（它將可選參數用作填充值）。