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I have a cvs file containing the following format of numbers 4,16221E+13 I don't really know what it means.
How can I replace them? is there a special function or script to do it?
Thank you,
Hani.
It means 4.16221 x 10^13, i.e. 41622100000000.
You can use:
float('4,16221E+13'.replace(',', '.'))
>>> 41622100000000.0
Python needs . as decimal point.
If you take the text, replace the , with an . then you can get the number it's likely supposed to be, eg:
>>> f = '4,16221E+13'
>>> float(f.replace(',', '.'))
41622100000000.0
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I am trying to create a blogging site kind of like medium.com. The only problem I am facing is the headings in the blog. I want to do what stack overflow does with bold text.
**text**
But I can't seem to figure out how to make this. Sorry if this question is not detailed enough and Thanks for giving this question your time.
you can find string between two subStrings using python regular expression.
import re
s = '**text**'
result = re.search('\*\*(.*)\*\*', s)
print(result.group(1)) #output :==> text
you really should learn regular expression 😀.
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I have such a string below and I want to extract the url just after the imgurl:. How could you do in Python in handy way?
{ns:"images",k:"5049",mid:"551FC833EDC139718135AA91A46D6B09FE89E85C",surl:"http://www.thewritingnut.com/blog-challenge/az-day-25-yellow-symbolisms/",imgurl:"http://www.thewritingnut.com/wp-content/uploads/2011/04/yellow-rose-800.jpg",oh:"199",tft:"0",oi:"http://www.thewritingnut.com/wp-content/uploads/2011/04/yellow-rose-800.jpg"}
Here it is, using a regular expression:
data = '{ns:"images",k:"5049",mid:"551FC833EDC139718135AA91A46D6B09FE89E85C",surl:"http://www.thewritingnut.com/blog-challenge/az-day-25-yellow-symbolisms/",imgurl:"http://www.thewritingnut.com/wp-content/uploads/2011/04/yellow-rose-800.jpg",oh:"199",tft:"0",oi:"http://www.thewritingnut.com/wp-content/uploads/2011/04/yellow-rose-800.jpg"}'
re.search('imgurl:"([^"]+)', data).group(1)
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I have a string in python, which is a URL:
http://weibo.com\/p\/aa\/weibo?from=bb&wvr=5.1&mod=weibomore#cc
I want to a real URL that I can paste it in my chrome:
http://weibo.com/p/aa/weibo?from=bb&wvr=5.1&mod=weibomore#cc
Please tell me how to chang it IN PYTHON~~~
It appears that you simply need to remove the backward slashes from your original string.
You can do that using replace as such..
url = string.replace("\\", "")
<your url string>.replace("\\","") will do the trick, but as has been pointed out, it would be better to fix the source of this.
The solution will also remove any backquotes ("\"). Even legitimate ones - so it's not a general solution. Backslashes are allowed in urls*, though its very unusual and causes problems
*Then again perhaps not: http://www.ietf.org/rfc/rfc2396.txt - it looks like this is a confusing question.
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I have the following sentence:
Graft Concepts has partnered with\u00a0several sites\u00a0to offer customization options for backplates, so customers will be able to design their own with\u00a0
I'd like to get rid of all instances of "\u00a0", whether or not they are connected to other words, e.g. "with\u00a0several"
How would I do this using regex with python? I've tried experimenting with re.compile() and re.findall(), but I couldn't get this working?
Thanks.
You can just use .replace:
s = s.replace('\\u00a0', ' ')
This works under Python 2:
import re
st='''Graft Concepts has partnered with\u00a0several sites\u00a0to offer customization options for backplates, so customers will be able to design their own with\u00a0'''
st=re.sub(ur'\\u00a0','',st,re.UNICODE)
And this under Python 3:
st=re.sub(u'\\u00a0','',st,re.UNICODE)
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I found this code on github: https://github.com/Abrand88/Reddit-Tor-proxy-voting-bot/blob/master/reddit_proxy_voting_bot.py
I know that its a bot and potentially malicious but I thinks its cool nonetheless. From an educational perspective how do I get it working?: When I run the program it outputs "socket:" twice and then gives the error "KeyError: '127.0.0.1'"
Any thoughts?
it's using a dictionary and can't find the key for your local host.
this is probably causing the error breaking:
ip_hash={}; # the dictionary is made
and
ip_hash[ip] = 1; # the dictionary is incorrectly accessed
this part alone isn't a good idea:
if ip in ip_hash:
print " repeat " + ip;
else:
ip_hash[ip] = 1;