Hello, group: I've just begun some introductory tutorials in Python.
Taking off from the "word play" exercise at
<http://www.greenteapre ss.com/thinkpython/html/book010.html#to c96>
I've written a mini-program to tabulate the number of characters in each
word in a file. Once the data have been collected in a list, the output
is produced by a while loop that steps through it by incrementing an
index "i", saying
print '%2u %6u %4.2f' % \
(i, wordcounts[i], 100.0 * wordcounts[i] / wordcounts[0])
My problem is with the last entry in each line, which isn't getting
padded:
1 0 0.00
2 85 0.07
3 908 0.80
4 3686 3.24
5 8258 7.26
6 14374 12.63
7 21727 19.09
8 26447 23.24
9 16658 14.64
10 9199 8.08
11 5296 4.65
12 3166 2.78
13 1960 1.72
14 1023 0.90
15 557 0.49
16 261 0.23
17 132 0.12
18 48 0.04
19 16 0.01
20 5 0.00
21 3 0.00
I've tried varying the number before the decimal in the formatting
string; "F", "g", and "G" conversions instead of "f"; and a couple of
other permutations (including replacing the arithmetical expression in
the tuple with a variable, defined on the previous line), but I can't
seem to get the decimal points to line up. I'm sure I'm missing
something obvious, but I'd appreciate a tip -- thanks in advance!
FWIW I'm running
Python 2.3.5 (#1, Oct 5 2005, 11:07:27)
[GCC 3.3 20030304 (Apple Computer, Inc. build 1809)] on darwin
from the Terminal on Mac OS X v10.4.11.
P.S. Is there a preferable technique for forcing floating-point division
of two integers to that used above, multiplying by "100.0" first? What
about if I just wanted a ratio: is "float(n / m)" better than "1.0 * n /
m"?
--
Odysseus
9  1525 
Odysseus wrote:
Hello, group: I've just begun some introductory tutorials in Python.
Taking off from the "word play" exercise at
<http://www.greenteapre ss.com/thinkpython/html/book010.html#to c96>
I've written a mini-program to tabulate the number of characters in each
word in a file. Once the data have been collected in a list, the output
is produced by a while loop that steps through it by incrementing an
index "i", saying
print '%2u %6u %4.2f' % \
(i, wordcounts[i], 100.0 * wordcounts[i] / wordcounts[0])
Using 4.2 is the problem. The first digit (your 4) give the total
number of characters to use for the number. Your numbers require at
least 5 characters, two digits, one decimal point, and two more
digits. So try 5.2, or 6.2 or 7.2 or higher if your numbers can grow
into the hundreds or thousands or higher. If you want to try one of the
floating point formats, then your first number must be large enough to
account for digits (before and after) the decimal point, the 'E', and
any digits in the exponent, as well as signs for both the number and the
exponent.
Gary Herron
My problem is with the last entry in each line, which isn't getting
padded:
1 0 0.00
2 85 0.07
3 908 0.80
4 3686 3.24
5 8258 7.26
6 14374 12.63
7 21727 19.09
8 26447 23.24
9 16658 14.64
10 9199 8.08
11 5296 4.65
12 3166 2.78
13 1960 1.72
14 1023 0.90
15 557 0.49
16 261 0.23
17 132 0.12
18 48 0.04
19 16 0.01
20 5 0.00
21 3 0.00
I've tried varying the number before the decimal in the formatting
string; "F", "g", and "G" conversions instead of "f"; and a couple of
other permutations (including replacing the arithmetical expression in
the tuple with a variable, defined on the previous line), but I can't
seem to get the decimal points to line up. I'm sure I'm missing
something obvious, but I'd appreciate a tip -- thanks in advance!
FWIW I'm running
Python 2.3.5 (#1, Oct 5 2005, 11:07:27)
[GCC 3.3 20030304 (Apple Computer, Inc. build 1809)] on darwin
from the Terminal on Mac OS X v10.4.11.
P.S. Is there a preferable technique for forcing floating-point division
of two integers to that used above, multiplying by "100.0" first? What
about if I just wanted a ratio: is "float(n / m)" better than "1.0 * n /
m"?
On Jan 13, 3:15 pm, Odysseus <odysseus1479.. .@yahoo-dot.cawrote:
[snip]
>
P.S. Is there a preferable technique for forcing floating-point division
of two integers to that used above, multiplying by "100.0" first? What
about if I just wanted a ratio: is "float(n / m)" better than "1.0 * n /
m"?
Odysseus
You obviously haven't tried float(n / m), or you wouldn't be asking.
Go ahead and try it.
"Preferable " depends on whether you want legibility or speed.
Most legible and slowest first:
1. float(n) / float(m)
2. n / float(m)
3. 1.0 * n / m
# Rationale so far: function calls are slow
4. If you have a lot of this to do, and you really care about the
speed and m (the denominator) is constant throughout, do fm = float(m)
once, and then in your loop do n / fm for each n -- and make sure you
run properly constructed benchmarks ...
Recommendation: go with (2) until you find you've got a program with
a real speed problem (and then it probably won't be caused by this
choice).
HTH,
John
On Jan 12, 10:15 pm, Odysseus <odysseus1479.. .@yahoo-dot.cawrote:
P.S. Is there a preferable technique for forcing floating-point division
of two integers to that used above, multiplying by "100.0" first?
Put this at the beginning of your program:
from __future__ import division
This forces all divisions to yield floating points values:
>>1/3
0
>>from __future__ import division
1/3
0.3333333333333 3331
>>>
HTH,
--
Roberto Bonvallet
In article
<a2************ *************** *******@d4g2000 prg.googlegroup s.com>,
John Machin <sj******@lexic on.netwrote:
<snip>
>
You obviously haven't tried float(n / m), or you wouldn't be asking.
True, it was a very silly idea.
Most legible and slowest first:
1. float(n) / float(m)
2. n / float(m)
3. 1.0 * n / m
Recommendation: go with (2) until you find you've got a program with
a real speed problem (and then it probably won't be caused by this
choice).
I had actually used n / float(m) at one point; somehow the above struck
me as an improvement while I was writing the message. Thanks for the
reality check.
--
Odysseus
In article
<e6************ *************** *******@v46g200 0hsv.googlegrou ps.com>,
Roberto Bonvallet <rb******@gmail .comwrote:
Put this at the beginning of your program:
from __future__ import division
This forces all divisions to yield floating points values:
Thanks for the tip. May I assume the div operator will still behave as
usual?
--
Odysseus
On Jan 13, 8:43 pm, Odysseus <odysseus1479.. .@yahoo-dot.cawrote:
In article
<e6f9c0c3-d248-46ae-af0f-fbb8ab279...@v4 6g2000hsv.googl egroups.com>,
Roberto Bonvallet <rbonv...@gmail .comwrote:
Put this at the beginning of your program:
from __future__ import division
This forces all divisions to yield floating points values:
Thanks for the tip. May I assume the div operator will still behave as
usual?
div operator? The integer division operator is //
>>from __future__ import division
1 / 3
0.3333333333333 3331
>>1 // 3
0
>>22 / 7
3.1428571428571 428
>>22 // 7
3
>>22 div 7
File "<stdin>", line 1
22 div 7
^
SyntaxError: invalid syntax
>>>
In article <ma************ *************** *********@pytho n.org>,
Gary Herron <gh*****@island training.comwro te:
Odysseus wrote:
<snip>
print '%2u %6u %4.2f' % \
(i, wordcounts[i], 100.0 * wordcounts[i] / wordcounts[0])
Using 4.2 is the problem. The first digit (your 4) give the total
number of characters to use for the number.
Thanks; I was thinking the numbers referred to digits before and after
the decimal. The largest figures have five characters in all, so they
were 'overflowing'.
--
Odysseus
Odysseus wrote:
Hello, group: I've just begun some introductory tutorials in Python.
Taking off from the "word play" exercise at
<http://www.greenteapre ss.com/thinkpython/html/book010.html#to c96>
I've written a mini-program to tabulate the number of characters in each
word in a file. Once the data have been collected in a list, the output
is produced by a while loop that steps through it by incrementing an
index "i", saying
print '%2u %6u %4.2f' % \
(i, wordcounts[i], 100.0 * wordcounts[i] / wordcounts[0])
This isn't very important, but instead of keeping track of the index
yourself, you can use enumerate():
>>mylist = ['a', 'b', 'c']
for i, item in enumerate(mylis t):
.... print i, item
....
0 a
1 b
2 c
>>>
Err, it doesn't look like you can make it start at 1 though.
<snip>
--
In article
<7d************ *************** *******@q77g200 0hsh.googlegrou ps.com>,
John Machin <sj******@lexic on.netwrote:
<snip>
>
div operator? The integer division operator is //
Yes, sorry, that's what I meant.
--
Odysseus This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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