Search string in a list

Question

data = ['HTTP/1.1 200 OK', 'CACHE-CONTROL: max-age=1810', 'DATE: 2014年5月14日 12:15:19 GMT', 'EXT:', 'LOCATION: http://192.168.94.57:9000/DeviceDescription.xml', 'SERVER: Windows NT/5.0, UPnP/1.0, pvConnect UPnP SDK/1.0', 'ST: uuid:7076436f-6e65-1063-8074-78542e239ff5', 'USN: uuid:7076436f-6e65-1063-8074-78542e239ff5', 'Content-Length: 0', '', '']

From the above list, I have to extract the .xml link.

My code:

for element in data:
 if 'LOCATION' in element:
 xmllink = element.split(': ').[1]

It's taking too much time. How can I make this faster?

Actually I am doing SSDP discovery for finding devices in a network. After sending the M-SEARCH command, devices send a datagram packet which I have taken in a data variable. From this I have to extract the file link of that device for processing it.

When I use indexing to extract, it was done quickly.

• Marc-Andre: I cannot understand how a split and a small array like that can take too much time. Have you use some sort of profiling to make sure that this part is the problem?
• Patrick: @Marc-Andre actually i am doing ssdp dicovery for devices in network, after sending M-SEARCH command devices respond with a datagram packet which i have taken in "data" and its taking so much time to process this, earlier i have used direct indexing to find "LOCATION" and it was done quickly
• Marc-Andre: those information are important for a review! You should edit the question.

Answer 1

You want to test for element.startswith('LOCATION: '). You are doing 'LOCATION' in element, which is not only slower since it has to check at every position of every element, it might also lead to false matches.

Also, element and data are poor names. I suggest header and headers, respectively.

My suggestion:

LOC = 'LOCATION: '
xmllinks = [header[len(LOC):] for header in headers if header.startswith(LOC)]
if xmllinks:
 xmllink = xmllinks[0]

• BeetDemGuise: Is it safe to assume there will be a single space after 'LOCATION:' based on his original implementation? Searching for simply 'LOCATION:' then using strip() to remove any additional white space would probably be more secure.

Answer 2

I am not sure what is causing problem. This piece of code should not take up a lot of time. But here are some of the suggestions to speed up your code:

1. Instead of list, create a set. The lookup time in set is constant.
2. But the main problem with set is uses up more memory in comparison to list. So another option is to keep a sorted list (if possible) and use the bisect module to search the list.

Now some style suggestions:

for element in data:
 if 'LOCATION' in element:
 xmllink = element.split(': ').[1]

rewrite it as

for element in data:
 if 'LOCATION' in element and ':' in element:
 xmllink = element.split(':')[1].strip()

This ensures that if a string like 'LOCATION something' is in list then it will not raise any errors.

The .strip() is a better way to remove trailing whitespaces

• Simon Forsberg: Using a set only works if you want to check the whole element, set.contains("LOCATION: http://www.example.com/test.xml") would be fast, but what is needed here is something like set.startsWith("LOCATION: "), which doesn't exist.

Answer 3

With such a small input it should be lightning fast. Anyway, shouldn't you break after finding the element? I'd write:

xmllink = next(s.split(":", 1)[1].strip() for s in data if s.startswith("LOCATION:")

• Pranav Raj: how does this code ensures speedup ?
• tokland: because it breaks after finding the match. Not sure how the OP is having speed problems here, though
• Marc-Andre: @tokland Read the comments under the question, he's giving a bit more information about the speed problem.
• Simon Forsberg: While I agree with the break suggestion, I think using regex is totally overkill.
• tokland: some refactors.