Counting

[Andy]

Hi, the file below will print all the keywords in a file and also the
line # of the keyword. What I couldn't figure out is to count those
keywords per line. For example - "Line #1 has 3 keywords"

Can I do like -

total[j] = total[j] + numwords(k)
"Line number %d has %d keywords" % (j, total[j])

Seems sort of "illegal" in Python?



-------------------------------------------------
import keyword, sys, string, fileinput
def numwords(s):
list = string.split(s)
return len(list)

# Get the file name either from the command-line or the user
if len(sys.argv) != 2:
name = raw_input("Ente r the file name: ")
else:
name = sys.argv[1]

inp = open(name,"r")
linelist = inp.readlines()
total, words,lines = 0, 0, 0

for i in range(len(linel ist)):
line = linelist[i]
tempwords = line.split()
for k in tempwords:
if keyword.iskeywo rd(k):
total = total + numwords(k)
j = i + 1
print" The word * %s * belongs in line number: %d" % (k,
j)

print "Total keywords in this file are: %d" %(total)

[James Stroud]

Re: Counting

Andy wrote:

Hi, the file below will print all the keywords in a file and also the
line # of the keyword. What I couldn't figure out is to count those
keywords per line. For example - "Line #1 has 3 keywords"
>
Can I do like -
>
total[j] = total[j] + numwords(k)
"Line number %d has %d keywords" % (j, total[j])
>
Seems sort of "illegal" in Python?
>
>
>
-------------------------------------------------
import keyword, sys, string, fileinput
def numwords(s):
list = string.split(s)
return len(list)
>
# Get the file name either from the command-line or the user
if len(sys.argv) != 2:
name = raw_input("Ente r the file name: ")
else:
name = sys.argv[1]
>
inp = open(name,"r")
linelist = inp.readlines()
total, words,lines = 0, 0, 0
>
for i in range(len(linel ist)):
line = linelist[i]
tempwords = line.split()
for k in tempwords:
if keyword.iskeywo rd(k):
total = total + numwords(k)
j = i + 1
print" The word * %s * belongs in line number: %d" % (k,
j)
>
print "Total keywords in this file are: %d" %(total)
>

You probably want something that goes a little like this:

for i,line in enumerate(linel ist):
for k in line.split():
if keyword.iskeywo rd(k):
total += line.count(k)
print "The word '%s' belongs in line num: %d" % (k, i+1)

print "Total keyords are: %d" % total

James

[James Stroud]

Re: Counting

James Stroud wrote:

Andy wrote:

>Hi, the file below will print all the keywords in a file and also the
>line # of the keyword. What I couldn't figure out is to count those
>keywords per line. For example - "Line #1 has 3 keywords"
>>
>Can I do like -
>>
>total[j] = total[j] + numwords(k)
>"Line number %d has %d keywords" % (j, total[j])
>>
>Seems sort of "illegal" in Python?
>>
>>
>>
>-------------------------------------------------
>import keyword, sys, string, fileinput
>def numwords(s):
> list = string.split(s)
> return len(list)
>>
># Get the file name either from the command-line or the user
>if len(sys.argv) != 2:
> name = raw_input("Ente r the file name: ")
>else:
> name = sys.argv[1]
>>
>inp = open(name,"r")
>linelist = inp.readlines()
>total, words,lines = 0, 0, 0
>>
>for i in range(len(linel ist)):
> line = linelist[i]
> tempwords = line.split()
> for k in tempwords:
> if keyword.iskeywo rd(k):
> total = total + numwords(k)
> j = i + 1
> print" The word * %s * belongs in line number: %d" % (k,
>j)
>>
>print "Total keywords in this file are: %d" %(total)
>>

>
You probably want something that goes a little like this:
>
for i,line in enumerate(linel ist):
for k in line.split():
if keyword.iskeywo rd(k):
total += line.count(k)
print "The word '%s' belongs in line num: %d" % (k, i+1)
>
print "Total keyords are: %d" % total
>
James

Oops, that over-counts, I forgot to put a continue in. Also, keeping a
cache of the split line will probably be faster.

for i,line in enumerate(linel ist):
line = line.split()
for k in line:
if keyword.iskeywo rd(k):
total += line.count(k)
print "The word '%s' belongs in line num: %d" % (k, i+1)
continue

print "Total keyords are: %d" % total


James

[James Stroud]

Re: Counting

James Stroud wrote:

James Stroud wrote:

>Andy wrote:

>>Hi, the file below will print all the keywords in a file and also the
>>line # of the keyword. What I couldn't figure out is to count those
>>keywords per line. For example - "Line #1 has 3 keywords"
>>>
>>Can I do like -
>>>
>>total[j] = total[j] + numwords(k)
>>"Line number %d has %d keywords" % (j, total[j])
>>>
>>Seems sort of "illegal" in Python?
>>>
>>>
>>>
>>-------------------------------------------------
>>import keyword, sys, string, fileinput
>>def numwords(s):
>> list = string.split(s)
>> return len(list)
>>>
>># Get the file name either from the command-line or the user
>>if len(sys.argv) != 2:
>> name = raw_input("Ente r the file name: ")
>>else:
>> name = sys.argv[1]
>>>
>>inp = open(name,"r")
>>linelist = inp.readlines()
>>total, words,lines = 0, 0, 0
>>>
>>for i in range(len(linel ist)):
>> line = linelist[i]
>> tempwords = line.split()
>> for k in tempwords:
>> if keyword.iskeywo rd(k):
>> total = total + numwords(k)
>> j = i + 1
>> print" The word * %s * belongs in line number: %d" % (k,
>>j)
>>>
>>print "Total keywords in this file are: %d" %(total)
>>>

>>
>You probably want something that goes a little like this:
>>
>for i,line in enumerate(linel ist):
> for k in line.split():
> if keyword.iskeywo rd(k):
> total += line.count(k)
> print "The word '%s' belongs in line num: %d" % (k, i+1)
>>
>print "Total keyords are: %d" % total
>>
>James

>
Oops, that over-counts, I forgot to put a continue in. Also, keeping a
cache of the split line will probably be faster.
>
for i,line in enumerate(linel ist):
line = line.split()
for k in line:
if keyword.iskeywo rd(k):
total += line.count(k)
print "The word '%s' belongs in line num: %d" % (k, i+1)
continue
>
print "Total keyords are: %d" % total
>
>
James

I should really wait until I've had some coffee. Not continue, but break!


for i,line in enumerate(linel ist):
line = line.split()
for k in line:
if keyword.iskeywo rd(k):
total += line.count(k)
print "The word '%s' belongs in line num: %d" % (k, i+1)
break

print "Total keyords are: %d" % total

[rockmode@gmail.com]

Re: Counting

That's a short, abridged version of my code :) But, what I want is to
count total# of keywords per line and print 'em. Rather than
printing :

The word 'and' belongs in line num: 1
The word 'del' belongs in line num: 1
The word 'from' belongs in line num: 1

I want to print " Line #1 has 3 keywords"

;)

You probably want something that goes a little like this:
>
for i,line in enumerate(linel ist):
for k in line.split():
if keyword.iskeywo rd(k):
total += line.count(k)
print "The word '%s' belongs in line num: %d" % (k, i+1)
>
print "Total keyords are: %d" % total
>
James

[castironpi@gmail.com]

Re: Counting

On Apr 29, 2:11 pm, Andy <andy.rockf...@ gmail.comwrote:

Hi, the file below will print all the keywords in a file and also the
line # of the keyword. What I couldn't figure out is to count those
keywords per line. For example - "Line #1 has 3 keywords"
>
Can I do like -
>
total[j] = total[j] + numwords(k)
"Line number %d has %d keywords" % (j, total[j])
>
Seems sort of "illegal" in Python?
>
-------------------------------------------------
import keyword, sys, string, fileinput
def numwords(s):
list = string.split(s)
return len(list)
>
# Get the file name either from the command-line or the user
if len(sys.argv) != 2:
name = raw_input("Ente r the file name: ")
else:
name = sys.argv[1]
>
inp = open(name,"r")
linelist = inp.readlines()
total, words,lines = 0, 0, 0
>
for i in range(len(linel ist)):
line = linelist[i]
tempwords = line.split()
for k in tempwords:
if keyword.iskeywo rd(k):
total = total + numwords(k)
j = i + 1
print" The word * %s * belongs in line number: %d" % (k,
j)
>
print "Total keywords in this file are: %d" %(total)

tempwords = line.split()
for k in tempwords:

linec = 0

if keyword.iskeywo rd(k):
total = total + numwords(k)
j = i + 1

linec += 1

print" The word * %s * belongs in line number: %d" % (k,
j)

print "%i characters in line" % linec

And less readably,

tempwords = line.split()
for k in tempwords:

linec = j

if keyword.iskeywo rd(k):
total = total + numwords(k)
j = i + 1
print" The word * %s * belongs in line number: %d" % (k,
j)

print "%i characters in line" % ( j - linec )

[James Stroud]

Re: Counting

rockmode@gmail. com wrote:

That's a short, abridged version of my code :) But, what I want is to
count total# of keywords per line and print 'em. Rather than
printing :
>
The word 'and' belongs in line num: 1
The word 'del' belongs in line num: 1
The word 'from' belongs in line num: 1
>
I want to print " Line #1 has 3 keywords"
>
;)
>

I think it would be obvious how to write this:


for i,line in enumerate(linel ist):
line = line.split()
for k in line:
if keyword.iskeywo rd(k):
c = line.count(k)
total += line.count(k)
print "Line #%d has %d keywords." % (i+1, c)
break

print "Total keyords are: %d" % total

[Andy]

Re: Counting

I pretty doubt about this - "c = line.count(k)" I might wanna recheck
on that.
-------------------------------------
I think it would be obvious how to write this:

for i,line in enumerate(linel ist):
line = line.split()
for k in line:
if keyword.iskeywo rd(k):
c = line.count(k)
total += line.count(k)
print "Line #%d has %d keywords." % (i+1, c)
break

print "Total keyords are: %d" % total

[Andy]

Re: Counting

James -

I pretty doubt about this - "c = line.count(k)" You might wanna
recheck on that.
------------------------------------------------
I think it would be obvious how to write this:

for i,line in enumerate(linel ist):
line = line.split()
for k in line:
if keyword.iskeywo rd(k):
c = line.count(k)
total += line.count(k)
print "Line #%d has %d keywords." % (i+1, c)
break

print "Total keyords are: %d" % total

>
I think it would be obvious how to write this:
>
for i,line in enumerate(linel ist):
line = line.split()
for k in line:
if keyword.iskeywo rd(k):
c = line.count(k)
total += line.count(k)
print "Line #%d has %d keywords." % (i+1, c)
break
>
print "Total keyords are: %d" % total

[John Machin]

Re: Counting

On 30/04/2007 7:17 AM, James Stroud wrote:

rockmode@gmail. com wrote:

>That's a short, abridged version of my code :) But, what I want is to
>count total# of keywords per line and print 'em. Rather than
>printing :
>>
>The word 'and' belongs in line num: 1
>The word 'del' belongs in line num: 1
>The word 'from' belongs in line num: 1
>>
>I want to print " Line #1 has 3 keywords"
>>
>;)
>>

>
>
I think it would be obvious how to write this:
>
>
for i,line in enumerate(linel ist):
line = line.split()
for k in line:
if keyword.iskeywo rd(k):
c = line.count(k)
total += line.count(k)
print "Line #%d has %d keywords." % (i+1, c)
break
>
print "Total keyords are: %d" % total

I would have thought so too. But the above is ... let's just say it's
not quite right. If there are 3 different keywords (as in the OP's
example), the above code prints 3 times for the same line.

Here's a straight-forward natural way to do it:
total = 0
for i, line in enumerate(linel ist):
c = 0
line = line.split()
for k in line:
if keyword.iskeywo rd(k):
c += 1
# Alternatively, replace above 5 lines by
# c = sum(keyword.isk eyword(k) for k in line.split())
# or the equivalent using map(), depending on taste etc :-)
total += c
print "Line #%d has %d keywords." % (i+1, c)
print "Total number of keywords is", total

======

Perhaps someone should point out to the OP that using str.split as a
tokeniser is somewhat deficient:
1. comments and string literals could make the counts somewhat unreliable:
"# if not use mung(), will break while frobotzing later in code"
2. "else:"
3. "if not(0 <= n < maxn):"

HTH,
John

[Steve Holden]

Re: Counting

James Stroud wrote:

James Stroud wrote:

>James Stroud wrote:

[finally ...]

>
I should really wait until I've had some coffee. Not continue, but break!
>
>
for i,line in enumerate(linel ist):
line = line.split()
for k in line:
if keyword.iskeywo rd(k):
total += line.count(k)
print "The word '%s' belongs in line num: %d" % (k, i+1)
break
>
print "Total keyords are: %d" % total

James is actually trying to convince you of the merits of test-driven
development by cleverly showing you the disadvantages of not using it.

Clever approach to advocacy, James. Did you get your coffee yet?

regards
Steve
--
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[Andy]

Re: Counting

..o0