我有一个包含这样的数据的文本文件
AA 331
line1 ...
line2 ...
% information here
AA 332
line1 ...
line2 ...
line3 ...
%information here
AA 1021
line1 ...
line2 ...
% information here
AA 1022
line1 ...
% information here
AA 1023
line1 ...
line2 ...
% information here
我想仅针对位于行"AA 331"
和行"AA 1021"
之后的最小整数之后的“信息”执行操作,而不是在行"AA 332"
,"AA 1022"
之后执行操作"AA 1023"
。
P.s这只是大文件的示例数据
下面的代码我尝试解析文本文件并获取列表“list1”中“AA”之后的整数,并在第二个函数中将它们分组以获得“list2”中的最小值。这将返回[331,1021,...]之类的整数。所以我想要提取“AA 331”之后的行并执行动作,但我不知道如何继续。
from itertools import groupby
def getlineindex(textfile):
with open(textfile) as infile:
list1 = []
for line in infile :
if line.startswith("AA"):
intid = line[3:]
list1.append(intid)
return list1
def minimalinteger(list1):
list2 = []
for k,v in groupby(list1,key=lambda x: x//10):
minimalint = min(v)
list2.append(minimalint)
return list2
list2包含“AA”之后的最小整数[331,1021,..]
答案 0 :(得分:2)
您可以使用以下内容:
import re
matcher = re.compile("AA ([\d]+)")
already_was = []
good_block = False
with open(filename) as f:
for line in f:
m = matcher.match(line)
if m:
v = int(m.groups(0)) / 10
else:
v = None
if m and v not in already_was:
good_block = True
already_was.append(m)
if m and v in already_was:
good_block = False
if not m and good_block:
do_action()
仅当组中的第一个值为最小值时,这些代码才有效。
答案 1 :(得分:1)
好的,这是我的解决方案。在高级别,我逐行,看着AA线知道我何时找到了数据块的开始/结束,并观察我所谓的运行编号,以了解我们是否应该处理下一个块。然后,我有一个子程序来处理任何给定的块,基本上读取所有相关的行并在需要时处理它们。该子程序用于监视 next AA线,以便知道它何时完成。
import re
runIdRegex = re.compile(r'AA (\d+)')
def processFile(fileHandle):
lastNumber = None # Last run number, necessary so we know if there's been a gap or if we're in a new block of ten.
line = fileHandle.next()
while line is not None: # None is being used as a special value indicating we've hit the end of the file.
processData = False
match = runIdRegex.match(line)
if match:
runNumber = int(match.group(1))
if lastNumber == None:
# Startup/first iteration
processData = True
elif runNumber - lastNumber == 1:
# Continuation, see if the tenths are the same.
lastNumberTens = lastNumber / 10
runNumberTens = runNumber / 10
if lastNumberTens != runNumberTens:
processData = True
else:
processData = True
# Always remember where we were.
lastNumber = runNumber
# And grab and process data.
line = dataBlock(fileHandle, process=processData)
else:
try:
line = fileHandle.next()
except StopIteration:
line = None
def dataBlock(fileHandle, process=False):
runData = []
try:
line = fileHandle.next()
match = runIdRegex.match(line)
while not match:
runData.append(line)
line = fileHandle.next()
match = runIdRegex.match(line)
except StopIteration:
# Hit end of file
line = None
if process:
# Data processing call here
# processData(runData)
pass
# Return line so we don't lose it!
return line
给你一些笔记。首先,我同意Jimilian的意见,你应该使用正则表达式匹配AA线。
其次,我们在处理数据时所讨论的逻辑是在processFile中。特别是这些行:
processData = False
match = runIdRegex.match(line)
if match:
runNumber = int(match.group(1))
if lastNumber == None:
# Startup/first iteration
processData = True
elif runNumber - lastNumber == 1:
# Continuation, see if the tenths are the same.
lastNumberTens = lastNumber / 10
runNumberTens = runNumber / 10
if lastNumberTens != runNumberTens:
processData = True
else:
processData = True
我认为我们不想处理数据,然后确定我们何时这样做。从逻辑上讲,您可以执行与此相反的操作,并假设您要处理数据,然后确定何时不处理数据。接下来,我们需要存储 last 运行的值,以便了解我们是否需要处理此运行的数据。 (并注意第一次运行边缘情况)我们知道我们想要在序列被破坏时处理数据(两次运行之间的差异大于1),这由else语句处理。我们也知道我们想要在序列递增数十位的数字时处理数据,这是由我的整数除以10来处理的。
第三,注意dataBlock的返回数据。如果你不这样做,你将失去导致dataBlock停止迭代的AA线,并且processFile需要该线以便知道是否应该处理下一个数据块。
最后,我选择使用fileHandle.next()和异常处理来确定何时到达文件末尾。但不要以为这是唯一的方法。 :)
如果您有任何问题,请在评论中告诉我。