Learning Python via a little word frequency program

[Andrew Savige]

I'm learning Python by reading David Beazley's "Python Essential Reference"
book and writing a few toy programs. To get a feel for hashes and sorting,
I set myself this little problem today (not homework, BTW):

Given a string containing a space-separated list of names:

names = "freddy fred bill jock kevin andrew kevin kevin jock"

produce a frequency table of names, sorted descending by frequency.
then ascending by name. For the above data, the output should be:

kevin : 3
jock : 2
andrew : 1
bill : 1
fred : 1
freddy : 1

Here's my first attempt:

names = "freddy fred bill jock kevin andrew kevin kevin jock"
freq = {}
for name in names.split():
freq[name] = 1 + freq.get(name, 0)
deco = zip([-x for x in freq.values()], freq.keys())
deco.sort()
for v, k in deco:
print "%-10s: %d" % (k, -v)

I'm interested to learn how more experienced Python folks would solve
this little problem. Though I've read about the DSU Python sorting idiom,
I'm not sure I've strictly applied it above ... and the -x hack above to
achieve a descending sort feels a bit odd to me, though I couldn't think
of a better way to do it.

I also have a few specific questions. Instead of:

for name in names.split():
freq[name] = 1 + freq.get(name, 0)

I might try:

for name in names.split():
try:
freq[name] += 1
except KeyError:
freq[name] = 1

Which is preferred?

Ditto for:

deco = zip([-x for x in freq.values()], freq.keys())

versus:

deco = zip(map(operator.neg, freq.values()), freq.keys())

Finally, I might replace:

for v, k in deco:
print "%-10s: %d" % (k, -v)

with:

print "\n".join("%-10s: %d" % (k, -v) for v, k in deco)

Any feedback on good Python style, performance tips, good books
to read, etc. is appreciated.

Thanks,
/-\


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[Ant]

I'm interested to learn how more experienced Python folks would solve

this little problem.

I think I'd do the following:

from collections import defaultdict

names = "freddy fred bill jock kevin andrew kevin kevin jock"
freq = defaultdict(lambda: 0)

for name in names.split():
freq[name] += 1

pairs = [(v, k) for k, v in freq.iteritems()]

for v, k in reversed(sorted(pairs)):
print "%-10s: %d" % (k, v)


defaultdict makes the frequency accumulation neater.

reversed(sorted(pairs)) avoids the little -v hack and makes it more
obvious what you are doing. Of course this could also be achieved by
doing pairs.sort() and pairs.reverse() before iterating over the pairs
list.

Cheers,

 

[Peter Otten]

I'm learning Python by reading David Beazley's "Python Essential
Reference" book and writing a few toy programs. To get a feel for hashes
and sorting, I set myself this little problem today (not homework, BTW):

Given a string containing a space-separated list of names:

names = "freddy fred bill jock kevin andrew kevin kevin jock"

produce a frequency table of names, sorted descending by frequency.
then ascending by name. For the above data, the output should be:

kevin : 3
jock : 2
andrew : 1
bill : 1
fred : 1
freddy : 1

Here's my first attempt:

names = "freddy fred bill jock kevin andrew kevin kevin jock" freq = {}
for name in names.split():
freq[name] = 1 + freq.get(name, 0)
deco = zip([-x for x in freq.values()], freq.keys()) deco.sort() for v,
k in deco:
print "%-10s: %d" % (k, -v)

I'm interested to learn how more experienced Python folks would solve
this little problem. Though I've read about the DSU Python sorting
idiom, I'm not sure I've strictly applied it above ... and the -x hack
above to achieve a descending sort feels a bit odd to me, though I
couldn't think of a better way to do it.

You can specify a reverse sort with

deco.sort(reverse=True)

Newer versions of Python have the whole idiom built in:
.... print "%-10s: %d" % item
....
kevin : 3
jock : 2
bill : 1
andrew : 1
fred : 1
freddy : 1

You can pass an arbitrary function as key. itemgetter(1) is equivalent to

def key(item): return item[1]

I also have a few specific questions. Instead of:

for name in names.split():
freq[name] = 1 + freq.get(name, 0)

I might try:

for name in names.split():
try:
freq[name] += 1
except KeyError:
freq[name] = 1

Which is preferred?

I have no strong opinion about that. Generally speaking try...except is
faster when you have many hits, i. e. the except suite is rarely invoked.
Starting with Python 2.5 you can alternatively use

from collections import defaultdict
freq = defaultdict(int)
for name in names.split():
freq[name] += 1

Ditto for:

deco = zip([-x for x in freq.values()], freq.keys())

versus:

deco = zip(map(operator.neg, freq.values()), freq.keys())

I think the list comprehension is slightly more readable.

Finally, I might replace:

for v, k in deco:
print "%-10s: %d" % (k, -v)

with:

print "\n".join("%-10s: %d" % (k, -v) for v, k in deco)

Again, I find the explicit for loop more readable, but sometimes use the
genexp, too.

Peter

 

[Bruno Desthuilliers]

Andrew Savige a écrit :

I'm learning Python by reading David Beazley's "Python Essential Reference"
book and writing a few toy programs. To get a feel for hashes and sorting,
I set myself this little problem today (not homework, BTW):

Given a string containing a space-separated list of names:

names = "freddy fred bill jock kevin andrew kevin kevin jock"

produce a frequency table of names, sorted descending by frequency.
then ascending by name. For the above data, the output should be:

kevin : 3
jock : 2
andrew : 1
bill : 1
fred : 1
freddy : 1

Here's my first attempt:

names = "freddy fred bill jock kevin andrew kevin kevin jock"
freq = {}
for name in names.split():
freq[name] = 1 + freq.get(name, 0)
deco = zip([-x for x in freq.values()], freq.keys())
deco.sort()
for v, k in deco:
print "%-10s: %d" % (k, -v)

I'm interested to learn how more experienced Python folks would solve
this little problem.

For a one-shot Q&D script:

names = "freddy fred bill jock kevin andrew kevin kevin jock"
freqs = [(- names.count(name), name) for name in set(names.split())]
print "\n".join("%-10s : %s" % (n, -f) for f, n in sorted(freqs))


Now I might choose a very different solution for a more serious
application, depending on detailed specs and intended use of the
"frequency table".

Though I've read about the DSU Python sorting idiom,
I'm not sure I've strictly applied it above ...

Perhaps not "strictly" since you don't really "undecorate", but that's
another application of the same principle : provided the appropriate
data structure, sort() (or sorted()) will do the right thing.

and the -x hack above to
achieve a descending sort feels a bit odd to me, though I couldn't think
of a better way to do it.

The "other" way would be to pass a custom comparison callback to sort,
which would be both slower and more complicated. Your solution is IMHO
the right thing to do here.

I also have a few specific questions. Instead of:

for name in names.split():
freq[name] = 1 + freq.get(name, 0)

I might try:

for name in names.split():
try:
freq[name] += 1
except KeyError:
freq[name] = 1

or a couple other solutions, including a defaultdict (python >= 2.5).

Which is preferred?

It's a FAQ - or it should be one. Globally: the second one tends to be
faster when there's no exception (ie the key already exists), but slower
when exceptions happen. So it mostly depends on what you expect your
dataset to be.

Now note that you don't necessarily need a dict here !-)

Ditto for:

deco = zip([-x for x in freq.values()], freq.keys())

versus:

deco = zip(map(operator.neg, freq.values()), freq.keys())

As far as I'm concerned, I'd favor the first solution here. Reads better
IMHO

Finally, I might replace:

for v, k in deco:
print "%-10s: %d" % (k, -v)

with:

print "\n".join("%-10s: %d" % (k, -v) for v, k in deco)

That's what I'd do here too - but it depends on context (ie: for huge
datasets and/or complex formating, i'd use a for loop).

 

[Bruno Desthuilliers]

Ant a écrit :

I'm interested to learn how more experienced Python folks would solve
this little problem.

I think I'd do the following:

from collections import defaultdict

names = "freddy fred bill jock kevin andrew kevin kevin jock"
freq = defaultdict(lambda: 0)

for name in names.split():
freq[name] += 1

pairs = [(v, k) for k, v in freq.iteritems()]

for v, k in reversed(sorted(pairs)):
print "%-10s: %d" % (k, v)


defaultdict makes the frequency accumulation neater.

reversed(sorted(pairs)) avoids the little -v hack and makes it more
obvious what you are doing.

But fails to implement the specs (emphasis is mine):
"""
produce a frequency table of names, sorted descending by frequency.
*then ascending by name*. For the above data, the output should be:

kevin : 3
jock : 2
andrew : 1
bill : 1
fred : 1
freddy : 1
"""

With your solution, you get:

kevin : 3
jock : 2
freddy : 1
fred : 1
bill : 1
andrew : 1

 

[MRAB]

Andrew Savige a écrit :

I'm learning Python by reading David Beazley's "Python Essential Reference"
book and writing a few toy programs. To get a feel for hashes and sorting,
I set myself this little problem today (not homework, BTW):

Given a string containing a space-separated list of names:

names = "freddy fred bill jock kevin andrew kevin kevin jock"

produce a frequency table of names, sorted descending by frequency.
then ascending by name. For the above data, the output should be:

kevin : 3
jock : 2
andrew : 1
bill : 1
fred : 1
freddy : 1

Here's my first attempt:

names = "freddy fred bill jock kevin andrew kevin kevin jock"
freq = {}
for name in names.split():
freq[name] = 1 + freq.get(name, 0)
deco = zip([-x for x in freq.values()], freq.keys())
deco.sort()
for v, k in deco:
print "%-10s: %d" % (k, -v)

I'm interested to learn how more experienced Python folks would solve
this little problem.

For a one-shot Q&D script:

names = "freddy fred bill jock kevin andrew kevin kevin jock"
freqs = [(- names.count(name), name) for name in set(names.split())]
print "\n".join("%-10s : %s" % (n, -f) for f, n in sorted(freqs))

[snip]
That actually prints:

kevin : 3
fred : 2
jock : 2
andrew : 1
bill : 1
freddy : 1

It says that "fred" occurs twice because of "freddy".

names = "freddy fred bill jock kevin andrew kevin kevin jock"
name_list = names.split()
freqs = [(- name_list.count(name), name) for name in set(name_list)]
print "\n".join("%-10s : %s" % (n, -f) for f, n in sorted(freqs))

 

[Bruno Desthuilliers]

MRAB a écrit :

On Jan 9, 12:19 pm, Bruno Desthuilliers <bruno.
(e-mail address removed)> wrote: (snip)
That actually prints:

kevin : 3
fred : 2
jock : 2
andrew : 1
bill : 1
freddy : 1

It says that "fred" occurs twice because of "freddy".

oops ! My bad, didn't spot that one

Thanks for pointing this out.

 

[rent]

import collections

names = "freddy fred bill jock kevin andrew kevin kevin jock"
freq = collections.defaultdict(int)
for name in names.split():
freq[name] += 1
keys = freq.keys()
keys.sort(key = freq.get, reverse = True)
for k in keys:
print "%-10s: %d" % (k, freq[k])

 

[Paul Rubin]

keys = freq.keys()
keys.sort(key = freq.get, reverse = True)
for k in keys:
print "%-10s: %d" % (k, freq[k])

I prefer (untested):

def snd((x,y)): return y # I wish this was built-in
sorted_freq = sorted(freq.iteritems(), key=snd, reverse=True)
for k,f in sorted_freq:
print "%-10s: %d" % (k, f)

 

[Mike Meyer]

keys = freq.keys()
keys.sort(key = freq.get, reverse = True)
for k in keys:
print "%-10s: %d" % (k, freq[k])

I prefer (untested):

def snd((x,y)): return y # I wish this was built-in

What's wrong with operator.itemgetter?

sorted_freq = sorted(freq.iteritems(), key=snd, reverse=True)

(still untested)

from operator import itemgetter
sorted_freq = sorted(freq.iteritems(), key=itemgetter(2), reverse=True)

<mike

 

[Hrvoje Niksic]

keys = freq.keys()
keys.sort(key = freq.get, reverse = True)
for k in keys:
print "%-10s: %d" % (k, freq[k])

I prefer (untested):

def snd((x,y)): return y # I wish this was built-in

What's wrong with operator.itemgetter?

sorted_freq = sorted(freq.iteritems(), key=snd, reverse=True)

(still untested)

from operator import itemgetter
sorted_freq = sorted(freq.iteritems(), key=itemgetter(2), reverse=True)

It should be itemgetter(1). See how easy it is to get it wrong?


(Okay, this was too easy a shot to miss out on; I actually like
itemgetter.)