Re: file I/O and arithmetic calculation
Mark Lawrence
> Dear all,
>
> I would appreciate if someone could write a simple python code for the
> purpose below:
>
> I have five text files each of 10 columns by 10 rows as follows:
>
>
> |file_one= 'C:/test/1.txt'
> file_two= 'C:/test/2.txt'
> . . .
> file_five= 'C:/test/5.txt'|
>
> I want to calculate the mean of first row (10 elements) for each file (5
> files), if mean of first column (10 elements) of each file (5 files) is 50.
>
> Thank you in advance.
>
> Keira
>
Sorry but we don't do homework.
--
If you're using GoogleCrap� please read this
http://wiki.python.org/moin/GoogleGroupsPython.
Mark Lawrence
Carlos Nepomuceno
### 1strow_average.py ###
#Assuming you have CSV (comma separated values) files such as:
#1.txt = '0,1,2,3,4,5,6,7,8,9\n' \
# '10,11,12,13,14,15,16,17,18,19\n' \
# '20,21,22,23,24,25,26,27,28,29\n' ...
#
# Usage: contents[file][row][column]
# contents[0] : file '1.txt'
# contents[1][2] : 3rd row of file '2.txt'
# contents[3][4][5] : 6th column of 5th row of file '4.txt'
# len(contents) : quantity of files
# len(contents[4]) : quantity of lines in file '5.txt'
# len(contents[4][0]: quantity of values in the 1st line of file '5.txt'
filenames = ['1.txt', '2.txt', '3.txt', '4.txt', '5.txt']
contents = [[[int(z) for z in y.split(',')] for y in open(x).read().split()] for x in filenames]
s1c = [sum([r[0] for r in f]) for f in contents]
a1r = [sum(f[0])/float(len(f[0])) for f in contents]
print '\n'.join([x for x in ['File "{}" has 1st row average = {:.2f}'.format(n,a1r[i]) if s1c[i]==50 else '' for i,n in enumerate(filenames)] if x])
________________________________
> From: wilk...@gmail.com
> Date: Thu, 23 May 2013 01:13:19 +0900
> Subject: file I/O and arithmetic calculation
> To: pytho...@python.org
>
>
> Dear all,
>
> I would appreciate if someone could write a simple python code for the
> purpose below:
>
> I have five text files each of 10 columns by 10 rows as follows:
>
>
>
> file_one = 'C:/test/1.txt'
> file_two = 'C:/test/2.txt'
> . . .
> file_five = 'C:/test/5.txt'
>
> I want to calculate the mean of first row (10 elements) for each file
> (5 files), if mean of first column (10 elements) of each file (5
> files) is 50.
>
> Thank you in advance.
>
> Keira
>
>
Oscar Benjamin
>
> filenames = ['1.txt', '2.txt', '3.txt', '4.txt', '5.txt']
> contents = [[[int(z) for z in y.split(',')] for y in open(x).read().split()] for x in filenames]
> s1c = [sum([r[0] for r in f]) for f in contents]
> a1r = [sum(f[0])/float(len(f[0])) for f in contents]
> print '\n'.join([x for x in ['File "{}" has 1st row average = {:.2f}'.format(n,a1r[i]) if s1c[i]==50 else '' for i,n in enumerate(filenames)] if x])
this and I certainly wouldn't use it when explaining something to a
beginner.
Rather than repeated list comprehensions you should consider using a
single loop e.g.:
for filename in filenames:
# process each file
This will make the code a lot simpler.
Oscar
Carlos Nepomuceno
> From: oscar.j....@gmail.com
[...]
>
> Do you find this code easy to read? I wouldn't write something like
> this and I certainly wouldn't use it when explaining something to a
> beginner.
>
> Rather than repeated list comprehensions you should consider using a
> single loop e.g.:
>
> for filename in filenames:
> # process each file
>
> This will make the code a lot simpler.
>
>
> Oscar
Indeed, but for that you can use Pascal.
List comprehensions it's what Python does best!
The code is pretty obvious to me, I mean there's no obfuscation at all.
Carlos Nepomuceno
BTW, it should read
# contents[3][4][5] : 6th value of 5th row of file '4.txt'
Denis McMahon
> I would appreciate if someone could write a simple python code for the
> purpose below:
import re
def v(s):
l=len(s)
t=0.
for i in range(l):
t=t+(abs(ord(s[i]))*1.)
return t/(l*1.)
for n in range(5):
m="c:/test/"+str(n+1)+".txt"
f=open(m,"r")
d=[]
t=0.
for l in range(10):
d=d+[re.findall(r"[0-9.eE+-]+",f.readline())]
t=t+v(d[l][0])
f.close()
c=t/10.
if c==50.:
t=0.
for u in range(10):
t=t+v(d[0][u])
r=t/10.
print "%s C1: %f R1: %f"%(m,c,r)
--
Denis McMahon, denismf...@gmail.com
Carlos Nepomuceno
> From: denismf...@gmail.com
[...]
> http://mail.python.org/mailman/listinfo/python-list
Can you send it again without tabs?
Oscar Benjamin
>
> The code is pretty obvious to me, I mean there's no obfuscation at all.
Carlos Nepomuceno
> From: oscar.j....@gmail.com
> Date: Thu, 23 May 2013 01:34:37 +0100
> Subject: Re: file I/O and arithmetic calculation
> To: carlosne...@outlook.com
> CC: pytho...@python.org
>
> On 23 May 2013 00:49, Carlos Nepomuceno <carlosne...@outlook.com> wrote:
>>
>> The code is pretty obvious to me, I mean there's no obfuscation at all.
>
> I honestly can't tell if you're joking.
I'm not! lol
Carlos Nepomuceno
### 1strow_average.py ###
#Assuming you have CSV (comma separated values) files such as:
#1.txt = '0,1,2,3,4,5,6,7,8,9\n' \
# '10,11,12,13,14,15,16,17,18,19\n' \
# '20,21,22,23,24,25,26,27,28,29\n' ...
#
# Usage: contents[file][row][column]
# contents[0] : file '1.txt'
# contents[1][2] : 3rd row of file '2.txt'
# contents[3][4][5] : value on the 6th column of 5th row of file '4.txt'
# len(contents) : quantity of files
# len(contents[4]) : quantity of lines in file '5.txt'
# len(contents[4][0]: quantity of values in the 1st line of file '5.txt'
filenames = ['1.txt', '2.txt', '3.txt', '4.txt', '5.txt']
contents = [[[int(z) for z in y.split(',')] for y in open(x).read().split()] for x in filenames]
s1c = [sum([r[0] for r in f]) for f in contents]
a1r = [sum(f[0])/float(len(f[0])) for f in contents]
print '\n'.join(['File "{}" has 1st row average = {:.2f}'.format(n,a1r[i]) for i,n in enumerate(filenames) if s1c[i]==50])
Oscar Benjamin
> The last line of my noob piece can be improved. So this is it:
> filenames = ['1.txt', '2.txt', '3.txt', '4.txt', '5.txt']
> contents = [[[int(z) for z in y.split(',')] for y in open(x).read().split()] for x in filenames]
> s1c = [sum([r[0] for r in f]) for f in contents]
> a1r = [sum(f[0])/float(len(f[0])) for f in contents]
> print '\n'.join(['File "{}" has 1st row average = {:.2f}'.format(n,a1r[i]) for i,n in enumerate(filenames) if s1c[i]==50])
like this:
list2 = [func1(x) for x in list1]
list3 = [func2(y) for y in list2]
list4 = [func3(y) for y in list2]
In this case it is usually better to write a single loop
for x in list1:
y = func1(x)
v = func2(y)
w = func3(y)
With that your code becomes:
filenames = ['1.txt', '2.txt', '3.txt', '4.txt', '5.txt']
s1c = sum([r[0] for r in contents])
a1r = sum(f[0])/float(len(contents[0]))
if s1c == 50:
print('File "{}" has 1st row average = {:.2f}'.format(filename,a1r))
However you shouldn't really be doing open(x).read().split() part. You
should use the with statement to open the files:
with open(filename, 'rb') as inputfile:
contents = [map(int, line.split()) for line in inputfile]
Of course if you don't have so many list comprehensions in your code
then your lines will be shorter and you won't feel so much pressure to
use such short variable names. It's also better to define a mean
function as it makes it clearer to read:
# Needed by the mean() function in Python 2.x
from __future__ import division
def mean(numbers):
return sum(numbers) / len(numbers)
filenames = ['1.txt', '2.txt', '3.txt', '4.txt', '5.txt']
with open(filename, 'rb') as inputfile:
matrix = [map(int, line.split()) for line in inputfile]
column1 = [row[0] for row in matrix]
row1 = matrix[0]
if mean(column1) == 50:
print('File "{}" has 1st row average =
{:.2f}'.format(filename, mean(row1)))
It's all a little easier if you use numpy:
import numpy as np
filenames = ['1.txt', '2.txt', '3.txt', '4.txt', '5.txt']
matrix = np.loadtxt(filename, dtype=int)
column1 = matrix[:, 0]
row1 = matrix[0, :]
if sum(column1) == 50 * len(column1):
print('File "{}" has 1st row average =
{:.2f}'.format(filename, np.mean(row1)))
Then again in practise I wouldn't be testing for equality of the mean.
Oscar
Chris Angelico
<carlosne...@outlook.com> wrote:
> ----------------------------------------
>> From: oscar.j....@gmail.com
>> Date: Thu, 23 May 2013 01:34:37 +0100
>> Subject: Re: file I/O and arithmetic calculation
>> To: carlosne...@outlook.com
>> CC: pytho...@python.org
>>
>>>
>>> The code is pretty obvious to me, I mean there's no obfuscation at all.
>>
>> I honestly can't tell if you're joking.
>
> I'm not! lol
Leonard: <obviously sarcastic comment>
Sheldon: "Was that sarcasm?"
Leonard: "Noooooo..."
Sheldon: "Was THAT sarcasm?"
Leonard: -- no answer needed.
Normally don't do this much list comprehension in an example. It may
be fine for the code, but it doesn't help explain stuff. :)
ChrisA
Denis McMahon
> Dear all who involved with responding to my question - Thank you so much
> for your nice code which really helped me.
Hold on a sec? Someone posted code that gave the correct answer to a
homework question?
--
Denis McMahon, denismf...@gmail.com
Carlos Nepomuceno
> From: denismf...@gmail.com
[...]
>> Dear all who involved with responding to my question - Thank you so much
>> for your nice code which really helped me.
>
> Hold on a sec? Someone posted code that gave the correct answer to a
> homework question?
>
> --
> Denis McMahon, denismf...@gmail.com
Not sure what she've asked, but mine certainly do what I think she've asked.