letter frequency counter / your thoughts..

[umpsumps]

Hello,

Here is my code for a letter frequency counter. It seems bloated to
me and any suggestions of what would be a better way (keep in my mind
I'm a beginner) would be greatly appreciated..

def valsort(x):
res = []
for key, value in x.items():
res.append((val ue, key))
return res

def mostfreq(strng) :
dic = {}
for letter in strng:
if letter not in dic:
dic.setdefault( letter, 1)
else:
dic[letter] += 1
newd = dic.items()
getvals = valsort(newd)
getvals.sort()
length = len(getvals)
return getvals[length - 3 : length]

thanks much!!

[Paul Melis]

um******@gmail. com wrote:

Here is my code for a letter frequency counter. It seems bloated to
me and any suggestions of what would be a better way (keep in my mind
I'm a beginner) would be greatly appreciated..

Not bad for a beginner I think :)

def valsort(x):
res = []
for key, value in x.items():
res.append((val ue, key))
return res

def mostfreq(strng) :
dic = {}
for letter in strng:
if letter not in dic:
dic.setdefault( letter, 1)
else:
dic[letter] += 1
newd = dic.items()
getvals = valsort(newd)
getvals.sort()
length = len(getvals)
return getvals[length - 3 : length]

thanks much!!

Slightly shorter:

def mostfreq(strng) :
dic = {}
for letter in strng:
if letter not in dic:
dic[letter] = 0
dic[letter] += 1
# Swap letter, count here as we want to sort on count first
getvals = [(pair[1],pair[0]) for pair in dic.iteritems()]
getvals.sort()
return getvals[-3:]

I'm not sure if you wanted the function mostfreq to return the 3 most
frequent letters of the first 3 letters? It seems to do the latter. The
code above uses the former, i.e. letters with highest frequency.

Paul

[castironpi]

On May 7, 12:30*pm, Paul Melis <p...@floorba ll-flamingos.nlwro te:

umpsu...@gmail. com wrote:

Here is my code for a letter frequency counter. *It seems bloated to
me and any suggestions of what would be a better way (keep in my mind
I'm a beginner) would be greatly appreciated..

Not bad for a beginner I think :)

def valsort(x):
* *res = []
* *for key, value in x.items():
* * * * * *res.append((va lue, key))
* *return res

def mostfreq(strng) :
* *dic = {}
* *for letter in strng:
* * * * * *if letter not in dic:
* * * * * * * * * *dic.setdefault (letter, 1)
* * * * * *else:
* * * * * * * * * *dic[letter] += 1
* *newd = dic.items()
* *getvals = valsort(newd)
* *getvals.sort()
* *length = len(getvals)
* *return getvals[length - 3 : length]

thanks much!!

Slightly shorter:

def mostfreq(strng) :
* * *dic = {}
* * *for letter in strng:
* * * * *if letter not in dic:
* * * * * * *dic[letter] = 0
* * * * *dic[letter] += 1
* * *# Swap letter, count here as we want to sort on count first
* * *getvals = [(pair[1],pair[0]) for pair in dic.iteritems()]
* * *getvals.sort()
* * *return getvals[-3:]

I'm not sure if *you wanted the function mostfreq to return the 3 most
frequent letters of the first 3 letters? It seems to do the latter. The
code above uses the former, i.e. letters with highest frequency.

Paul- Hide quoted text -

- Show quoted text -

I think I'd try to get a deque on disk. Constant indexing. Store
disk addresses in b-trees. How long does 'less than' take? Is a
sector small, and what's inside?

[Arnaud Delobelle]

um******@gmail. com writes:

Hello,

Here is my code for a letter frequency counter. It seems bloated to
me and any suggestions of what would be a better way (keep in my mind
I'm a beginner) would be greatly appreciated..

def valsort(x):
res = []
for key, value in x.items():
res.append((val ue, key))
return res

def mostfreq(strng) :
dic = {}
for letter in strng:
if letter not in dic:
dic.setdefault( letter, 1)
else:
dic[letter] += 1
newd = dic.items()
getvals = valsort(newd)
getvals.sort()
length = len(getvals)
return getvals[length - 3 : length]

thanks much!!

I won't comment on the algorithm, but I think you should try to find
better names for your variables. In the snippet above you have x,
res, dic, newd, length, getvals which don't give much of a clue as to
what they are used for.

e.g.

* dic = {}
We know it's a dict, but a dict of what?

* newd = dic.items()
Sounds like 'new dictionary', but obviously isn'tas it is a list
of key,value pairs.

* length = len(getvals)
Again, we know it's a length, but the length of what?

HTH

--
Arnaud

[Ian Kelly]

On Wed, May 7, 2008 at 11:30 AM, Paul Melis <pa**@floorba ll-flamingos.nlwro te:

dic = {}
for letter in strng:
if letter not in dic:
dic[letter] = 0
dic[letter] += 1

As a further refinement, you could use the defaultdict class from the
collections module:

dic = defaultdict(int )
for letter in strng:
dic[letter] += 1

[castironpi]

On May 7, 1:31*pm, "Ian Kelly" <ian.g.ke...@gm ail.comwrote:

On Wed, May 7, 2008 at 11:30 AM, Paul Melis <p...@floorba ll-flamingos.nlwro te:

* * dic = {}
* * for letter in strng:
* * * * if letter not in dic:
* * * * * * dic[letter] = 0
* * * * dic[letter] += 1

As a further refinement, you could use the defaultdict class from the
collections module:

* *dic = defaultdict(int )
* * * *for letter in strng:
* * * * * *dic[letter] += 1

Sounds like novel flow of control.

[umpsumps]

That's a great suggestion Arnaud. I'll keep that in mind next time I
post code. Thanks ;)
On May 7, 12:27 pm, Arnaud Delobelle <arno...@google mail.comwrote:

umpsu...@gmail. com writes:

Hello,

Here is my code for a letter frequency counter. It seems bloated to
me and any suggestions of what would be a better way (keep in my mind
I'm a beginner) would be greatly appreciated..

def valsort(x):
res = []
for key, value in x.items():
res.append((val ue, key))
return res

def mostfreq(strng) :
dic = {}
for letter in strng:
if letter not in dic:
dic.setdefault( letter, 1)
else:
dic[letter] += 1
newd = dic.items()
getvals = valsort(newd)
getvals.sort()
length = len(getvals)
return getvals[length - 3 : length]

thanks much!!

I won't comment on the algorithm, but I think you should try to find
better names for your variables. In the snippet above you have x,
res, dic, newd, length, getvals which don't give much of a clue as to
what they are used for.

e.g.

* dic = {}
We know it's a dict, but a dict of what?

* newd = dic.items()
Sounds like 'new dictionary', but obviously isn'tas it is a list
of key,value pairs.

* length = len(getvals)
Again, we know it's a length, but the length of what?

HTH

--
Arnaud

[John Machin]

On May 8, 6:00 am, umpsu...@gmail. com wrote:

That's a great suggestion Arnaud. I'll keep that in mind next time I
post code. Thanks ;)

It's a suggestion for YOUR benefit, not ours. Consider keeping it in
mind next time you WRITE code, whether you intend publishing it or
not.

[bruno.desthuilliers]

On 7 mai, 23:50, Paul Rubin <http://phr...@NOSPAM.i nvalidwrote:
(snip)

Someone else suggested the heapq module, which is a good approach
though it might be considered a little bit high-tech. If you
want to use sorting (conceptually simpler), you could use the
sorted function instead of the in-place sorting function:

# return the second element of a 2-tuple. Note how we
# use tuple unpacking: this is really a function of one argument
# (the tuple) but we specify the arg as (a,b) so the tuple
# is automatically unpacked on entry to the function.
# this is a limited form of the "pattern matching" found in
# languages like ML.
def snd((a,b)): return b

operator.itemge tter does this already

return sorted(dic.iter items, key=snd, reverse=True)[-3:]

you want to actually call iteritems here !-)

return sorted(dic.iter items(), key=operator.it emgetter(1),
reverse=True)

Thanks for reminding me the 'key' argument to sorted anyway - I too
often tend to forget it.

[bruno.desthuilliers]

On 7 mai, 23:51, "bruno.desthuil [e-mail] "
<bruno.desthuil [e-mail] wrote:
(snip)

Small improvement thanks to Paul Rubin:

from collections import defaultdict
from operator import itemgetter

def get_letters_fre quency(source):
letters_count = defaultdict(int )
for letter in source:
letters_count[letter] += 1
return sorted(
letters_count.i teritems(),
key=itemgetter( 1),
reverse=True
)