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# large dictionary creation takes a LOT of time.

this code here:
def wordcount(lines ):
for i in range(len(lines )/8):
words = lines[i].split(" ")
if not locals().has_ke y("frequency" ):
frequency = {}
for word in words:
if frequency.has_k ey(word):
frequency[word] += 1
else:
frequency[word] = 1
return frequency
wordcount(lines )

is taking over six minutes to run on a two megabyte text file. i
realize that's a big text file, a really big one (it's actually the
full text of don quixote.). i'm trying to figure out how. is there a
better way for me to do a frequency count of all the words in the text?
it seems to me like this should scale linearly, but perhaps it isn't?
i don't know much about algorithmic complexity. if someone could give
a breakdown of this functions complexity as well i'd be much obliged.

lines is expected to be a list of lines as provided by file.readline()

Jul 19 '05 #1
12 1555
"possibilitybox " <po************ @gmail.com> writes:
this code here:
def wordcount(lines ):
for i in range(len(lines )/8):
words = lines[i].split(" ")
if not locals().has_ke y("frequency" ):
frequency = {}
for word in words:
if frequency.has_k ey(word):
frequency[word] += 1
else:
frequency[word] = 1
return frequency
wordcount(lines )

is taking over six minutes to run on a two megabyte text file. i
realize that's a big text file, a really big one (it's actually the
full text of don quixote.). i'm trying to figure out how. is there a
better way for me to do a frequency count of all the words in the text?
2MB is not that large. Your method is ok and shouldn't be that slow
unless you're on a pretty slow PC. Could your machine be short of
memory and paging a lot? You could tweak the code somewhat by moving
the initialization of the frequency dict out of the loop and combining
a few other statements. Also you should use xrange instead of range,
to avoid allocating a big list in memory:

def wordcount(lines ):
frequency = {}
for i in xrange(len(line s)/8):
for word in lines[i].split():
frequency[word] = 1 + frequency.get(w ord, 0)
return frequency
wordcount(lines )
it seems to me like this should scale linearly, but perhaps it isn't?
i don't know much about algorithmic complexity. if someone could give
a breakdown of this functions complexity as well i'd be much obliged.

It should be close to linear, or at worst n log n, depending on what
happens when dicts have to be enlarged as the # of elements increases.
Why are you only processing 1/8th of the lines?
Jul 19 '05 #2
possibilitybox wrote:
this code here:
def wordcount(lines ):
for i in range(len(lines )/8):
words = lines[i].split(" ")
if not locals().has_ke y("frequency" ):
frequency = {}
for word in words:
if frequency.has_k ey(word):
frequency[word] += 1
else:
frequency[word] = 1
return frequency
wordcount(lines )

is taking over six minutes to run on a two megabyte text file. i
realize that's a big text file, a really big one (it's actually the
full text of don quixote.). i'm trying to figure out how. is there a
better way for me to do a frequency count of all the words in the text?
it seems to me like this should scale linearly, but perhaps it isn't?
i don't know much about algorithmic complexity. if someone could give
a breakdown of this functions complexity as well i'd be much obliged.

lines is expected to be a list of lines as provided by file.readline()

Here is a little cleaner version. It takes about a second to run on my PC. What hardware are you
running on?

path = 'DonQuixote.txt '

frequency = {}

for line in open(path):
for word in line.split():
if frequency.has_k ey(word):
frequency[word] += 1
else:
frequency[word] = 1

print len(frequency), 'words'
Kent
Jul 19 '05 #3
>>>>> "Kent" == Kent Johnson <ke****@tds.net > writes:

Kent> if frequency.has_k ey(word):
Kent> frequency[word] += 1
Kent> else:
Kent> frequency[word] = 1

This is a good place to use 'get' method of dict:

frequency[word] = frequency.get(w ord,0) + 1

--
Ville Vainio http://tinyurl.com/2prnb
Jul 19 '05 #4
Ville Vainio:
This is a good place to use 'get' method of dict:
frequency[word] = frequency.get(w ord,0) + 1

I think Kent Johnson is longer, but a bit faster...

Bye,
Bearophile

Jul 19 '05 #5
In article <11************ **********@z14g 2000cwz.googleg roups.com>,
"possibilitybox " <po************ @gmail.com> wrote:
this code here:
def wordcount(lines ):
for i in range(len(lines )/8):
words = lines[i].split(" ")
if not locals().has_ke y("frequency" ):
frequency = {}
for word in words:
if frequency.has_k ey(word):
frequency[word] += 1
else:
frequency[word] = 1
return frequency
wordcount(lines )

is taking over six minutes to run on a two megabyte text file. i
realize that's a big text file, a really big one (it's actually the
full text of don quixote.).
Something doesn't make sense with your timing.

428 kwords, 40 klines. This includes the text of the novel itself, plus a
little boilerplate text added by the Gutenberg Project folks.

I ran your program against it on my PowerBook (1 GHz PowerPC, Python 2.3.4,
768 Mbytes RAM). It took about 0.4 seconds. When I got rid of the "/8" in
the range() call (so it processed the whole text), it took about 1.8
seconds (0.24 of which were just reading the file). Some other posters
reported similiar findings.

What kind of system are you running it on? The only thing I can think of
is that you've got way too little memory and your machine is just
thrashing. My Python process is about 31 Mb when it first starts up, grows
to 35 when the file is read into a list, then gets to 38 after the call to
wordcount().
i'm trying to figure out how. is there a
better way for me to do a frequency count of all the words in the text?
it seems to me like this should scale linearly, but perhaps it isn't?
i don't know much about algorithmic complexity. if someone could give
a breakdown of this functions complexity as well i'd be much obliged.
Well, I don't see anything in your code which isn't linear. There are some
places where you could do things a little more efficiently, but these would
all be replacing one linear process with a better linear process. Small
speedups, but not the drastic kind of speedups you would expect by
replacing a quadratic process with a linear one.

Here's a few things I would change:
if not locals().has_ke y("frequency" ):
frequency = {}
This is kind of silly. Just factor this out of the main loop, and do
"frequency = {}" before the first for loop. This won't speed things up
other than trivially, but it'll make your code easier to read and
understand.

Next, replace
for i in range(len(lines )):
words = lines[i].split(" ")
with something like
for line in lines:
words = line.split()
It's marginally faster, easier to read, and is actually more correct;
calling split() with no arguments makes it split on arbitrary white space,
which is probably what you really want:
"foo bar baz".split(" ") ['foo', 'bar', '', '', '', 'baz']
"foo bar baz".split()

['foo', 'bar', 'baz']

if frequency.has_k ey(word): frequency[word] += 1
else:
frequency[word] = 1

I would do:

try:
frequency[word] += 1
except KeyError:
frequency[word] = 1

which is usually a little bit faster. Somebody else mentioned using the
get() method of dictionaries here; that might be even better, but I learned
the try/except trick before get() existed, so I tend to stick to that :-)

But, as I said, all of these are just minor tweaks. Overall, your code
looks like it should run in O(n), so fixing your code is not where you
should be looking. Something external (i.e. memory thrashing or some other
exceptionally bad bit of system performance) has to be causing the
horrendously bad performance you're seeing, and that's where you should be

BTW, if you suspected you had some kind of non-linear algorithm, and didn't
trust code inspection to verify its existance, you could just run your
program a bunch of times on different sized data sets, and plot runtime vs.
input size to see what kind of curve you get.
Jul 19 '05 #6
Kent Johnson wrote:

Here is a little cleaner version. It takes about a second to run on my
PC. What hardware are you running on?

path = 'DonQuixote.txt '

frequency = {}

for line in open(path):
for word in line.split():
if frequency.has_k ey(word):
frequency[word] += 1
else:
frequency[word] = 1

print len(frequency), 'words'
Kent for line in open(path):

the line of your example raise another question: opened file will be read at once time, as method readlines() do, or it will be read line by line as method readline() do.
as far i know, it is depends on implementation of method "__iter__" of the object that "open()" returns, so another question: where i can find such an information (about how does such a functions works)?
--
Best regards,
Maksim Kasimov
mailto: ka*****@i.com.u a
Jul 19 '05 #7
Maksim Kasimov wrote:
Kent Johnson wrote:
> for line in open(path): the line of your example raise another question: opened file will be

as far i know, it is depends on implementation of method "__iter__" of
the object that "open()" returns, so another question: where i can find
such an information (about how does such a functions works)?

http://docs.python.org/lib/built-in-funcs.html
http://docs.python.org/lib/bltin-file-objects.html

Kent
Jul 19 '05 #8
In fact, as one of the Peter's (either Otten or Hansen) explained to me,

for line in open(file):

is actually both faster (being buffered) and generally better for very
large files because it doesn't read the whole file into memory, like
readlines does (if you have a memory limitation).

On Fri, 29 Apr 2005 12:00:37 -0400, Kent Johnson <ke****@tds.net > wrote:
Maksim Kasimov wrote:
Kent Johnson wrote:
> for line in open(path):

the line of your example raise another question: opened file will be

as far i know, it is depends on implementation of method "__iter__" of
the object that "open()" returns, so another question: where i can find
such an information (about how does such a functions works)?

http://docs.python.org/lib/built-in-funcs.html
http://docs.python.org/lib/bltin-file-objects.html

Kent

Jul 19 '05 #9
On Friday 29 April 2005 11:53, Ville Vainio wrote:
>> "Kent" == Kent Johnson <ke****@tds.net > writes:

Kent> if frequency.has_k ey(word):
Kent> frequency[word] += 1
Kent> else:
Kent> frequency[word] = 1

This is a good place to use 'get' method of dict:

frequency[word] = frequency.get(w ord,0) + 1

try/except might be fastest of all:

http://gumuz.looze.net/wordpress/ind...-optimisation/

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Jul 19 '05 #10

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