FAQ
Here is a function which takes any list and creates a freq table,
which can be printed unsorted, sorted by cases or items. It's supposed
to mirror the proc freq in SAS.

Dirk

def freq(seq,order='unsorted',prin=True):
#order can be unsorted, cases, items

freq={}
for s in seq:
if s in freq:
freq[s]+=1
else:
freq[s]=1
if prin==True:
print 'Items=',len(seq),'Cases=',len(freq)
print '------------------------'
if order=='unsorted':
for k in freq.keys():
print k,freq[k],float(freq[k])/len(seq)
elif order=='cases':
#http://blog.client9.com/2007/11/sorting-python-dict-by-
value.html
freq2=sorted(freq.iteritems(), key=lambda (k,v):
(v,k),reverse=True)
for f in freq2:
print f[0],f[1],float(f[1])/len(seq)
elif order=='items':
for k in sorted(freq.iterkeys()):
print k,freq[k],float(freq[k])/len(seq)
print '------------------------'
return freq

#test

import random

rand=[]
for i in range(10000):
rand.append(str(int(100*random.random())))

fr=freq(rand)
fr2=freq(rand,order='items')
fr2=freq(rand,order='cases')

## Search Discussions

•  at Aug 22, 2010 at 8:16 am ⇧

On Sun, Aug 22, 2010 at 1:31 PM, Dirk Nachbar wrote:

Here is a function which takes any list and creates a freq table,
which can be printed unsorted, sorted by cases or items. It's supposed
to mirror the proc freq in SAS.

Dirk

def freq(seq,order='unsorted',prin=True):
#order can be unsorted, cases, items

freq={}
for s in seq:
if s in freq:
freq[s]+=1
else:
freq[s]=1
The above code can be replaced with this:
freq = {}
for s in seqn:
freq[s] = freq.get(s,0) + 1

if prin==True:
print 'Items=',len(seq),'Cases=',len(freq)
print '------------------------'
if order=='unsorted':
for k in freq.keys():
print k,freq[k],float(freq[k])/len(seq)
elif order=='cases':
#http://blog.client9.com/2007/11/sorting-python-dict-by-
value.html
freq2=sorted(freq.iteritems(), key=lambda (k,v):
(v,k),reverse=True)
for f in freq2:
print f[0],f[1],float(f[1])/len(seq)
elif order=='items':
for k in sorted(freq.iterkeys()):
print k,freq[k],float(freq[k])/len(seq)
print '------------------------'
return freq

#test

import random

rand=[]
for i in range(10000):
rand.append(str(int(100*random.random())))

fr=freq(rand)
fr2=freq(rand,order='items')
fr2=freq(rand,order='cases')
--
I feel the code you wrote is bloated a bit. You shall definately give
another try to improvise it.

--
~l0nwlf
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•  at Aug 22, 2010 at 8:34 am ⇧

On Sun, Aug 22, 2010 at 1:16 AM, Shashwat Anand wrote:
On Sun, Aug 22, 2010 at 1:31 PM, Dirk Nachbar wrote:
Here is a function which takes any list and creates a freq table,
which can be printed unsorted, sorted by cases or items. It's supposed
to mirror the proc freq in SAS.

Dirk
<snip>
? ?freq={}
? ?for s in seq:
? ? ? ?if s in freq:
? ? ? ? ? ?freq[s]+=1
? ? ? ?else:
? ? ? ? ? ?freq[s]=1
The above code can be replaced with this:
?freq = {}
?for s in seq:
?? ? ? ? ?freq[s] = freq.get(s,0) + 1
Which can be further replaced by:

from collections import Counter
freq = Counter(seq)

Using collections.defaultdict is another possibility if one doesn't
have Python 2.7.

Cheers,
Chris
--
It really bothers me that Counter isn't a proper Bag.
http://blog.rebertia.com
•  at Aug 22, 2010 at 12:29 pm ⇧

Dirk Nachbar wrote:

Here is a function which takes any list and creates a freq table,
which can be printed unsorted, sorted by cases or items. It's supposed
to mirror the proc freq in SAS.

Dirk

def freq(seq,order='unsorted',prin=True):
#order can be unsorted, cases, items

freq={}
for s in seq:
if s in freq:
freq[s]+=1
else:
freq[s]=1
if prin==True:
print 'Items=',len(seq),'Cases=',len(freq)
print '------------------------'
if order=='unsorted':
for k in freq.keys():
print k,freq[k],float(freq[k])/len(seq)
elif order=='cases':
#http://blog.client9.com/2007/11/sorting-python-dict-by-
value.html
freq2=sorted(freq.iteritems(), key=lambda (k,v):
(v,k),reverse=True)
Sorting in two steps gives a slightly better result when there are items
with equal keys. Compare
freq = {"a": 2, "b": 1, "c": 1, "d": 2}
sorted(freq.iteritems(), key=lambda (k, v): (v, k), reverse=True)
[('d', 2), ('a', 2), ('c', 1), ('b', 1)]

with
freq2 = sorted(freq.iteritems(), key=lambda (k, v): k)
freq2.sort(key=lambda (k, v): v, reverse=True)
freq2
[('a', 2), ('d', 2), ('b', 1), ('c', 1)]

Here the keys within groups of equal frequency are in normal instead of
reversed order.

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