我对Python很陌生，只有通过大量网页找到的知识。
话虽这么说，我正在尝试搜索一个文件（~10k行）来查找我写的一套过滤器类似的标准，然后我希望它打印符合条件的行和一行前面的X行数。
我创建了以下脚本来打开所述文件，逐行迭代，并将符合过滤条件的行打印到输出文件，但是我对如何将其合并到当前脚本感到困惑。
import os
output_file = 'Output.txt'
filename = 'BigFile.txt'
numLines = 0
numWords = 0
numChrs = 0
numMes = 0
f1 = open(output_file, 'w')
print 'Output File has been Opened'
with open(filename, 'r') as file:
for line in file:
wordsList = line.split()
numLines += 1
numWords += len(wordsList)
numChrs += len(line)
if "X" in line and "Y" not in line and "Z" in line:
numMes += 1
print >>f1, line
print 'Object found and Catalogued in Output.txt'
print "Lines: %i\nWords: %i\nCharacters: %i" % (numLines, numWords, numChrs)
print >>f1, "Lines: %i\nWords: %i\nCharacters: %i" % (numLines, numWords, numChrs)
print "There are a total of %i thing in this file" % (numMes)
print >>f1, "There are a total of %i things in this file" % (numMes)
f1.close()
print 'Output Files have been Closed'
我的第一个猜测是使用line.enumeration，但我不认为我可以说明lines - 5之类的内容，以便在lines之前打印5行：
lines = f1.enumeration()
if "blah blah" in line and "so so" not in line:
print >>f1, lines
print >>f1, [lines - 5]
最好的部分还没有到来，因为我必须获取Output.txt文件并与另一个文件进行比较以输出两个文件中的匹配条件......但是一次一步，对吧？
- 也可以自由添加“正确”的模糊内容。技术......我确信这个剧本可以写得更好，所以请教我一些我做错的事。
提前感谢您的帮助！
更新： 已成功实施此修复程序，感谢以下帮助：
import os
output_file = 'Output.txt'
filename = 'BigFile.txt'
numLines = 0
numWords = 0
numChrs = 0
numMulMes = 0
last5 = []
f1 = open(output_file, 'w')
print 'Output Files have been Opened'
with open(filename, 'r') as file:
for line in file:
wordsList = line.split()
numLines += 1
numWords += len(wordsList)
numChrs += len(line)
last5[:] = last5[-5:]+[line]
if "X" in line and "Y" not in line and "Z" not in line:
del last5[1:5] ###the missing piece of the puzzle!
numMulMes += 1
print >>f1, last5
print 'Object found and Catalogued in Output.txt'
print "Lines: %i\nWords: %i\nCharacters: %i" % (numLines, numWords, numChrs)
print >>f1, "Lines: %i\nWords: %i\nCharacters: %i" % (numLines, numWords, numChrs)
print "There are a total of %i messages in this file" % (numMulMes)
print >>f1, "There are a total of %i messages in this file" % (numMulMes)
f1.close()
f3.close()
print 'Output Files have been Closed'
我一直试图通过另一个单独的脚本修改输出文件，并且在最长的时间内我正在与str操作和错误操作进行对抗。刚决定回到原来的剧本，随心所欲地将它扔进那里，并且中提琴。
感谢你推动我朝着正确的方向前进，很容易从那里弄明白！
答案 0 :(得分：4)
你自己解决了大部分问题（计算单词，行，亚麻等） - 您可以在浏览文件时记住最后n行。
示例：
t = """"zero line
one line
two line
three line
four line
five line
six line
seven line
eight line
"""
last5 = [] # memory cell
for l in t.split("\n"): # similar to your for line in file:
last5[:] = last5[-4:]+[l] # keep last 4 and add current line, inplace list mod
if "six" in l:
print last5
您还可以查看deque并指定最大长度（您需要导入它）
from collections import deque
last5 = deque(maxlen=5)
for l in t.split("\n"):
last5.append(l) # will automatically only keep 5 (maxlen)
if "six" in l:
print last5
输出：
# list version
['two line', 'three line', 'four line ', 'five line ', 'six line']
# deque version
deque(['two line', 'three line', 'four line ', 'five line ', 'six line'], maxlen=5)
答案 1 :(得分：2)
这里与@PatricArtner建议的解决方案相同，但是使用了环形缓冲区。它可能（或可能不是，我没有检查）使用大文件更快地工作。
这个想法非常简单：我们可以创建一个包含所需大小的列表（您应该保留的行数）和当前记录位置cnt的计数器。对于每个新行，我们应该将cnt增加1并使用缓冲区的大小来模数。因此cnt在列表中循环。例如，如果列表大小为5 cnt = (cnt+1)%5，则会提供0 1 2 3 4 0 1 2，依此类推。 cnt的每一步都将指向列表中最旧的数据，这些数据将被新数据替代。下面是一个实现的例子。
t = """"zero line
six line - surprize
one line
two line
three line
four line
five line
six line
seven line
eight line
"""
last5 = [None,None,None,None,None]
cnt = 0
for l in t.split("\n"):
last5[cnt]=l
if 'six' in l:
print last5[(cnt+1)%5]
print last5[(cnt+2)%5]
print last5[(cnt+3)%5]
print last5[(cnt+4)%5]
print last5[(cnt+0)%5]
print
cnt = (cnt+1)%5
输出非常简单：
None
None
None
"zero line
six line - surprize
two line
three line
four line
five line
six line
注意：如果您从文件中读取，并且文件非常大并且您需要保留的字符串很大（例如，基因序列）并且您的条件不会触发通常，要聪明，不要在记忆中保留字符串。在文件中创建最后一个字符串开始的位置列表，并在需要时重新读取它们。下面是一个如何快速实现它的例子......
from numpy import random as rnd
print "Creating the file ...."
DNA=["G","C","T","A"]
with open("bigdatafile","w") as fd:
for i in xrange(5000):
fd.write("".join([ DNA[rnd.randint(4)] for x in xrange(2000)])+"\n")
print "DONE"
print
print "SEARCHING GGGGGGGGGGG"
last5, cnt = [0,0,0,0,0], 1
with open("bigdatafile","r") as fd:
for i,l in enumerate(fd.readlines()):
last5[cnt] = last5[(cnt+4)%5]+len(l)
if "GGGGGGGGGGG" in l:
print "FIND!"
fd.seek(last5[(cnt+1)%5])
print fd.read(last5[cnt]-last5[(cnt+1)%5])
cnt = (cnt+1)%5
答案 2 :(得分：0)
我没有写入文件，而是将内容输出到字典中。处理完整个文件后，摘要数据字典将以json的形式转储到文件中。使用Artner的测试文件。
import os
import json
output_file = 'Output.txt'
filename = 'BigFile.txt'
#initiate output container
outDict = {}
for fields in ['numLines', 'numWords', 'numChrs', 'numMes']:
outDict[fields] = 0
outDict['lineNum'] = []
with open(filename, 'r') as file:
for line in file:
wordsList = line.strip().split("\s")
outDict['numLines'] += 1
outDict['numWords'] += len(wordsList)
outDict['numChrs'] += len(line)
#find items in the line
if "t" in line:
outDict['numMes'] += 1
#save line number
outDict['lineNum'].append(outDict['numLines'])
#save line content
outDict['lineList'].append(line)
#record output
with open(output_file, 'w') as f1:
f1.write(json.dumps(outDict))
##print lines of desire
#x number of lines before
x=5
with open(filename, 'r') as file:
for i, line in enumerate(file):
#iterate over line numbers for which condition is met
for j in range(0,len(outDict['lineNum'])):
#if line number is between found line num and line num minus x, print
if (outDict['lineNum'][j]-x) <= i <= outDict['lineNum'][j]:
print(line)
答案 3 :(得分：0)
由于我在comments中提到过，以下是使用grep的{{3}}功能在* nix机器上执行相同操作的方法。
首先假设您有以下文本文件test.txt：
zero line
one line
two line
three line
four line
five line
six line
seven line
eight line
如果您想在匹配前获得N行，可以使用-B选项。例如，"six"之前的5行：
$ grep -B 5 six test.txt
one line
two line
three line
four line
five line
six line
还有-A选项可用于在匹配后获得N行，-C可用于获取N行之前和之后的行。