blocks|key|975394|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|975395|975396|将它们全部读入一个列表，将每一行分成两个数字，这样你就有了一个包含所有numbers.|unordered-list-item|975397|Remove的列表|975398|列表中的第一个和最后一个项目|975399|写出你的列表，一次两个项目，中间有点。|975400|975401|下面是一个例子：|975402|a+=+"""190..255
+++++++337..2799
+++++++2801..3733
+++++++3734..5020
+++++++5234..5530
+++++++5683..6459
+++++++8238..9191
+++++++9306..9893"""
a_list+=+a.replace('..','\n').split()
b_list+=+a_list[1:-1]
b+=+''
for+i+in+range(len(a_list)/2):
++++b+%2B=+'..'.join(b_list[2*i:2*i%2B2])+%2B+'\n'|code-block|syntax|javascript|975403|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|W|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|X|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|Y|8|@]|9|@]|A|$]]|$1|H|3|I|5|E|7|Z|8|@]|9|@]|A|$]]|$1|J|3|K|5|E|7|10|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|11|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|12|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|13|8|@]|9|@]|A|$R|S]]|$1|T|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|U|$]]

Try this:

<ul>
<li>Read all of them into one list, split each line into two numbers, so you have one list of all your numbers.</li>
<li>Remove the first and last item from your list</li>
<li>Write out your list, two items at a time, with dots in between them.</li>
</ul>

Here's an example:

<pre><code>a = """190..255
 337..2799
 2801..3733
 3734..5020
 5234..5530
 5683..6459
 8238..9191
 9306..9893"""
a_list = a.replace('..','\n').split()
b_list = a_list[1:-1]
b = ''
for i in range(len(a_list)/2):
 b += '..'.join(b_list[2*i:2*i+2]) + '\n'
</code></pre>

blocks|key|1089404|text|lines+=+"""190..255
337..2799
2801..3733"""

values+=+[int(v)+for+line+in+lines.split()+for+v+in+line.split('..')]
#+values+=+[190,+255,+337,+2799,+2801,+3733]

pairs+=+zip(values[1:-1:2],+values[2:-1:2])
#+pairs+=+[(255,+337),+(2799,+2801)]

out+=+'\n'.join('%25d..%25d'+%25+pair+for+pair+in+pairs)
#+out+=+"255..337\n2799..2801"|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1089405|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>lines = """190..255
337..2799
2801..3733"""

values = [int(v) for line in lines.split() for v in line.split('..')]
# values = [190, 255, 337, 2799, 2801, 3733]

pairs = zip(values[1:-1:2], values[2:-1:2])
# pairs = [(255, 337), (2799, 2801)]

out = '\n'.join('%d..%d' % pair for pair in pairs)
# out = "255..337\n2799..2801"
</code></pre>

blocks|key|1083526|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1083527|with+open(filename,+'r')+as+f:
++++lines+=+f.readlines()

numbers+=+[]
for+row+in+lines:
++++numbers.extend(row.split('..'))

numbers+=+numbers[1:len(numbers)-1]
newLines+=+['..'.join(numbers[idx:idx%2B2])+for+idx+in+xrange(0,+len(numbers),+2]

with+open(filename,+'w')+as+f:
++++for+line+in+newLines:
++++++++f.write(line)
++++++++f.write('\n')|code-block|syntax|javascript|1083528|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this:

<pre><code>with open(filename, 'r') as f:
 lines = f.readlines()

numbers = []
for row in lines:
 numbers.extend(row.split('..'))

numbers = numbers[1:len(numbers)-1]
newLines = ['..'.join(numbers[idx:idx+2]) for idx in xrange(0, len(numbers), 2]

with open(filename, 'w') as f:
 for line in newLines:
 f.write(line)
 f.write('\n')
</code></pre>

blocks|key|1083530|text|也许这会有所帮助：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1083531|def+makeColumns(listOfNumbers):
++++n+=+int()
++++while+n+<+len(listOfNumbers):
++++++++print(listOfNumbers[n],+'..',+listOfNumbers[(n%2B1)])
++++++++n+%2B=+2

def+trim(listOfNumbers):
++++listOfNumbers.pop(0)
++++listOfNumbers.pop((len(listOfNumbers)+-+1))

listOfNumbers+=+[190,+255,+337,+2799,+2801,+3733,+3734,+5020,+5234,+5530,+5683,+6459,+8238,+9191,+9306,+9893]

makeColumns(listOfNumbers)
print()

trim(listOfNumbers)
makeColumns(listOfNumbers)|code-block|syntax|javascript|1083532|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Maybe this will help:

<pre><code>def makeColumns(listOfNumbers):
 n = int()
 while n &lt; len(listOfNumbers):
 print(listOfNumbers[n], '..', listOfNumbers[(n+1)])
 n += 2

def trim(listOfNumbers):
 listOfNumbers.pop(0)
 listOfNumbers.pop((len(listOfNumbers) - 1))

listOfNumbers = [190, 255, 337, 2799, 2801, 3733, 3734, 5020, 5234, 5530, 5683, 6459, 8238, 9191, 9306, 9893]

makeColumns(listOfNumbers)
print()

trim(listOfNumbers)
makeColumns(listOfNumbers)
</code></pre>

blocks|key|975430|text|temp+=+[]
with+open('temp.txt')+as+ofile:
++++for+x+in+ofile:
++++++++temp.append(x.rstrip("\n"))
for+x+in+range(0,+len(temp)+-+1):
++++print+temp[x].split("..")[1]+%2B".."%2B+temp[x%2B1].split("..")[0]
++++x+%2B=+1|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|975431|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>temp = []
with open('temp.txt') as ofile:
 for x in ofile:
 temp.append(x.rstrip("\n"))
for x in range(0, len(temp) - 1):
 print temp[x].split("..")[1] +".."+ temp[x+1].split("..")[0]
 x += 1
</code></pre>

blocks|key|1083550|text|我认为这也可能是有用的。我正在从文件名列表中读取数据。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1083551|data+=+open("list","r")
temp+=+[]
value+=+[]
print+data
for+line+in+data:
++++temp+=+line.split("..")
++++value.append(temp[0])
++++value.append(temp[1])

for+i+in+range(1,(len(value)-1),2):
++++print+value[i].strip()%2B".."%2Bvalue[i%2B1]

print+value|code-block|syntax|javascript|1083552|在读取数据后，我拆分并将其存储在临时list.After中，我将数据复制到主列表值中，该值包含我从第二个元素迭代到倒数第二个元素的所有data.Then，以获得感兴趣的输出。为了从值中删除'\n‘字符，使用了strip函数。|1083553|您可以稍后将这些值写入文件，而不是打印出来。|1083554|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I think this might be useful too. I am reading data from a file name list. 

<pre><code>data = open("list","r")
temp = []
value = []
print data
for line in data:
 temp = line.split("..")
 value.append(temp[0])
 value.append(temp[1])

for i in range(1,(len(value)-1),2):
 print value[i].strip()+".."+value[i+1]

print value
</code></pre>

After reading the data I split and store it in the temporary list.After that, I copy data to the main list value which have all of the data.Then I iterate from the second element to second last element to get the output of interest. strip function is used in order to remove the '\n' character from the value.

You can later write these values to a file Instead of printing out.

So I have a text file consisting of one column, each column consist two numbers

<pre><code>190..255
337..2799
2801..3733
3734..5020
5234..5530
5683..6459
8238..9191
9306..9893
</code></pre>

I would like to discard the very 1st and the very last number, in this case, 190 and 9893.
and basically moves the rest of the numbers one spot forward. like this

My desired output

<pre><code>255..337
2799..2801
3733..3734
5020..5234
5530..5683
6459..8238
9191..9306
</code></pre>

I hope that makes sense I'm not sure how to approach this

How to rearrange numbers from different lines of a text file in python?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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所以我有一个由一列组成的文本文件，每列包含两个数字190..255337..27992801..37333734..50205234..55305683..64598238..91919306..9893我想丢弃第一个也是最后一个数字，在本例中是190和9893。基本上把剩下的数字前移一位。像这样我想要的输出255.....

问如何在python中重新排列文本文件中不同行的数字？
EN

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问如何在python中重新排列文本文件中不同行的数字？EN