Primes (Python):
This program is to produce prime numbers less than and equal to n.
Showing posts with label Python. Show all posts
Showing posts with label Python. Show all posts
Sunday, November 11, 2012
Coordinates of numpy array from index and shape (Python)
Coordinates of numpy array from index and shape (Python):
returns the coordinates of a numpy array given the index and the shape. A first_index_et function is given as example code
returns the coordinates of a numpy array given the index and the shape. A first_index_et function is given as example code
Human readable file/memory sizes v2 (Python)
Human readable file/memory sizes v2 (Python):
In writing a application to display the file sizes of set of files, I wanted to provide a human readable size rather then just displaying a byte count (which can get rather big).
I developed this useful short recipe that extends the format specifier mini Language to add new presentation type s- which will intelligently convert the value to be displayed into a known human readable size format - i.e. b, Kb,Mb, Gb, B, KB etc. It honours the rest of the format specification language (http://docs.python.org/2/library/string.html#format-string-syntax)
It uses a factor of 1024 for IEC and common formats, and factor of 1000 for SI units.
In writing a application to display the file sizes of set of files, I wanted to provide a human readable size rather then just displaying a byte count (which can get rather big).
I developed this useful short recipe that extends the format specifier mini Language to add new presentation type s- which will intelligently convert the value to be displayed into a known human readable size format - i.e. b, Kb,Mb, Gb, B, KB etc. It honours the rest of the format specification language (http://docs.python.org/2/library/string.html#format-string-syntax)
It uses a factor of 1024 for IEC and common formats, and factor of 1000 for SI units.
Terry Jones: A simple way to calculate the day of the week for any day of a given year
Terry Jones: A simple way to calculate the day of the week for any day of a given year:
Image: Jeremy Church
The other day I read a tweet about how someone was impressed that a friend had been able to tell them the day of the week given an arbitrary date.
There are a bunch of general methods to do this listed on the Wikipedia page for Determination of the day of the week. Typically, there are several steps involved, and you need to memorize some small tables of numbers.
I used to practice that mental calculation (and many others) when I was about 16. Although all the steps are basic arithmetic, it’s not easy to do the calculation in your head in a couple of seconds. Given that most of these questions that you’re likely to face in day-to-day life will be about the current year, it seemed like it might be a poor tradeoff to learn the complicated method to calculate the day of the week for any date if there was a simpler way to do it for a specific year.
The method I came up with after that observation is really simple. It just requires you to memorize a single 12-digit sequence for the current year. The 12 digits correspond to the first Monday for each month.
For example, the sequence for 2012 is 265-274-263-153. Suppose you’ve memorized the sequence and you need to know the day of the week for March 29th. You can trivially calculate that it’s a Thursday. You take the 3rd digit of the sequence (because March is the 3rd month), which is 5. That tells you that the 5th of March was a Monday. Then you just go backward or forward as many weeks and days as you need. The 5th was a Monday, so the 12th, 19th, and 26th were too, which means the 29th was a Thursday.
It’s nice because the amount you need to memorize is small, and you can memorize less digits if you only want to cover a shorter period. The calculation is very simple and always the same in every case, and you never have to think about leap years. At the start of each year you just memorize a single sequence, which is quickly reinforced once you use it a few times.
Here’s Python code to print the sequence for any year.
import datetime, sys
try:
year = int(sys.argv[1])
except IndexError:
year = datetime.datetime.today().year
firstDayToFirstMonday = ['1st', '7th', '6th', '5th', '4th', '3rd', '2nd']
months = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun',
'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
summary = ''
for month in range(12):
firstOfMonth = datetime.datetime(year, month + 1, 1).weekday()
firstMonday = firstDayToFirstMonday[firstOfMonth]
print months[month], firstMonday
summary += firstMonday[0]
print 'Summary:', '-'.join(summary[x:x+3] for x in range(0, 12, 3))
The output for 2012 looks like
Here are the sequences through 2032:
Image: Jeremy Church
There are a bunch of general methods to do this listed on the Wikipedia page for Determination of the day of the week. Typically, there are several steps involved, and you need to memorize some small tables of numbers.
I used to practice that mental calculation (and many others) when I was about 16. Although all the steps are basic arithmetic, it’s not easy to do the calculation in your head in a couple of seconds. Given that most of these questions that you’re likely to face in day-to-day life will be about the current year, it seemed like it might be a poor tradeoff to learn the complicated method to calculate the day of the week for any date if there was a simpler way to do it for a specific year.
The method I came up with after that observation is really simple. It just requires you to memorize a single 12-digit sequence for the current year. The 12 digits correspond to the first Monday for each month.
For example, the sequence for 2012 is 265-274-263-153. Suppose you’ve memorized the sequence and you need to know the day of the week for March 29th. You can trivially calculate that it’s a Thursday. You take the 3rd digit of the sequence (because March is the 3rd month), which is 5. That tells you that the 5th of March was a Monday. Then you just go backward or forward as many weeks and days as you need. The 5th was a Monday, so the 12th, 19th, and 26th were too, which means the 29th was a Thursday.
It’s nice because the amount you need to memorize is small, and you can memorize less digits if you only want to cover a shorter period. The calculation is very simple and always the same in every case, and you never have to think about leap years. At the start of each year you just memorize a single sequence, which is quickly reinforced once you use it a few times.
Here’s Python code to print the sequence for any year.
#!/usr/bin/env pythonimport datetime, sys
try:
year = int(sys.argv[1])
except IndexError:
year = datetime.datetime.today().year
firstDayToFirstMonday = ['1st', '7th', '6th', '5th', '4th', '3rd', '2nd']
months = ['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun',
'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
summary = ''
for month in range(12):
firstOfMonth = datetime.datetime(year, month + 1, 1).weekday()
firstMonday = firstDayToFirstMonday[firstOfMonth]
print months[month], firstMonday
summary += firstMonday[0]
print 'Summary:', '-'.join(summary[x:x+3] for x in range(0, 12, 3))
The output for 2012 looks like
Jan 2nd Feb 6th Mar 5th Apr 2nd May 7th Jun 4th Jul 2nd Aug 6th Sep 3rd Oct 1st Nov 5th Dec 3rd Summary: 265-274-263-153The memory task is made simpler by the fact that there are only 14 different possible sequences. Or, if you consider just the last 10 digits of the sequences (i.e., starting from March), there are only 7 possible sequences. There are only 14 different sequences, so if you use this method in the long term, you’ll find the effort of remembering a sequence will pay off when it re-appears. E.g., 2013 and 2019 both have sequence 744-163-152-742. There are other nice things you can learn that can also make the memorization and switching between years easier (see the Corresponding months section on the above Wikipedia page).
Here are the sequences through 2032:
2012 265-274-263-153 2013 744-163-152-742 2014 633-752-741-631 2015 522-641-637-527 2016 417-426-415-375 2017 266-315-374-264 2018 155-274-263-153 2019 744-163-152-742 2020 632-641-637-527 2021 411-537-526-416 2022 377-426-415-375 2023 266-315-374-264 2024 154-163-152-742 2025 633-752-741-631 2026 522-641-637-527 2027 411-537-526-416 2028 376-315-374-264 2029 155-274-263-153 2030 744-163-152-742 2031 633-752-741-631 2032 521-537-526-416
Friday, October 12, 2012
Find the location and value of an extremum of a matrix
I needed to find the location and the value of the maximum or minimum of a 2-d array. Here is how I did it.
index = A.argmax() #use argmin() to get the minimum
row = index / A.shape[1]
col = index % A.shape[1]
# The following two statements will give you the value of the maximum
A.max()
A[row][col]
import numpy as np
b=np.where(A==np.max())
row = b[0][0]
col = b[1][0]
The result is an array of indices indicating the column number where the maximum or minimum occurs. So the size of this array is the same size as your rows in A (A.shape[0]))
The result is an array of indices indicating the row number where the maximum or minimum occurs. So the size of this array is the same size as your columns in A (A.shape[1]))
A Simple way
Assume that the data is stored in a 2-d array A.index = A.argmax() #use argmin() to get the minimum
row = index / A.shape[1]
col = index % A.shape[1]
# The following two statements will give you the value of the maximum
A.max()
A[row][col]
Another way
I saw another solution on Stack Overflow. Again, is A is my array:import numpy as np
b=np.where(A==np.max())
row = b[0][0]
col = b[1][0]
Get the location of the maximum/minimum in each row
A.argvmax(axis=1) or A.argvmin(axis=1)The result is an array of indices indicating the column number where the maximum or minimum occurs. So the size of this array is the same size as your rows in A (A.shape[0]))
Get the location of the maximum/minimum in each column
A.argvmax(axis=0) or A.argvmin(axis=0)The result is an array of indices indicating the row number where the maximum or minimum occurs. So the size of this array is the same size as your columns in A (A.shape[1]))
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