We Keep Coding

Can you for loop completely through a range, but starting from the nth element?

I would like to know if there exists a base solution to do something like this:
for n in range(length=8, start_position= 3, direction= forward)
The problem I'm encountering is I would like the loop to continue past the final index, and pick up again at idx =0, then idx=1, etc. and stop at idx= 3, the start_position.
To give context, I seek all possible complete solutions to the n-queen problem.

Based on your latest edit, you need a "normal" range and the modulo operator:
for i in range(START, START + LEN):
do_something_with(i % LEN)

from itertools import chain
for n in chain(range(3,8), range(3)):
...
The chain() returns an iterator with 3, 4, ..., 7, 0, 1, 2

Another option for solving this is to use modular arithmetic. You could do something like this, for example:
for i in range(8)
idx = (i + 3) % 8
# use idx
This easily can be generalized to work with different lengths and offsets.

def loop_around_range(length, start_position, direction='forward'):
looped_range = [k % length for k in range(start_position, start_position+length)]
if direction == 'forward':
return looped_range
else:
return looped_range[::-1]

You could implement this for an arbitrary iterable by using itertools.cycle.
from itertools import cycle
def circular_iterator(iterable, skip=0, length=None, reverse=False):
"""Produces a full cycle of #iterable#, skipping the first #skip# elements
then tacking them on to the end.
if #iterable# does not implement #__len__#, you must provide #length#
"""
if reverse:
iterable = reversed(iterable)
cyc_iter = cycle(iterable)
for _ in range(skip):
next(cyc_iter, None)
if length:
total_length = length
else:
total_length = len(iterable)
for _ in range(total_length):
yield next(cyc_iter, None)
>>> lst = [x for x in range(1, 9)]
# [1, 2, 3, 4, 5, 6, 7, 8]
>>> list(circular_iterator(lst, skip=3))
[4, 5, 6, 7, 8, 1, 2, 3]

how to find a continuous string using python

Given a string (e.g., jaghiuuabc ), i want to find a string with subsequent letter in alphabet
here is my code
import string
alpha = list(string.ascii_lowercase)
s = 'jaghiuuabc'
a = []
for i in range(len(alpha)-1):
for j in range(len(s)-1)
if s[j] in alpha[i]:
a.append(s[j])
print(a)

There's a nice example in the Python 2.6 itertools docs that shows how to find consecutive sequences. To quote:
Find runs of consecutive numbers using groupby. The key to the
solution is differencing with a range so that consecutive numbers all
appear in same group.
For some strange reason, that example is not in the later versions of the docs. That code works for sequences of numbers, the code below shows how to adapt it to work on letters.
from itertools import groupby
s = 'jaghiuuabc'
def keyfunc(t):
''' Subtract the character's index in the string
from its Unicode codepoint number.
'''
i, c = t
return ord(c) - i
a = []
for k, g in groupby(enumerate(s), key=keyfunc):
# Extract the chars from the (index, char) tuples in the group
seq = [t[1] for t in g]
if len(seq) > 1:
a.append(''.join(seq))
print(a)
output
['ghi', 'abc']
How it works
The heart of this code is
groupby(enumerate(s), key=keyfunc)
enumerate(s) generates tuples containing the index number and character for each character in s. For example:
s = 'ABCEF'
for t in enumerate(s):
print(t)
output
(0, 'A')
(1, 'B')
(2, 'C')
(3, 'E')
(4, 'F')
groupby takes items from a sequence or iterator and gathers adjacent equal items together into groups. By default, it simply compares the values of the items to see if they're equal. But you can also give it a key function. When you do that, it passes each item to the key function and uses the result returned by that key function for its equality test.
Here's a simple example. First, we define a function div_by_10 that divides a number by 10, using integer division. This basically gets rid of the last digit in the number.
def div_by_10(n):
return n // 10
a = [2, 5, 10, 13, 17, 21, 22, 29, 33, 35]
b = [div_by_10(u) for u in a]
print(a)
print(b)
output
[2, 5, 10, 13, 17, 21, 22, 29, 33, 35]
[0, 0, 1, 1, 1, 2, 2, 2, 3, 3]
So if we use div_by_10 as the key function to groupby it will ignore the last digit in each number and thus it will group adjacent numbers together if they only differ in the last digit.
from itertools import groupby
def div_by_10(n):
return n // 10
a = [2, 5, 10, 13, 17, 21, 22, 29, 33, 35]
print(a)
for key, group in groupby(a, key=div_by_10):
print(key, list(group))
output
[2, 5, 10, 13, 17, 21, 22, 29, 33, 35]
0 [2, 5]
1 [10, 13, 17]
2 [21, 22, 29]
3 [33, 35]
My keyfunc receives a (index_number, character) tuple and subtracts that index_number from the character's code number and returns the result. Let's see what that does with my earlier example of 'ABCEF':
def keyfunc(t):
i, c = t
return ord(c) - i
for t in enumerate('ABCEF'):
print(t, keyfunc(t))
output
(0, 'A') 65
(1, 'B') 65
(2, 'C') 65
(3, 'E') 66
(4, 'F') 66
The code number for 'A' is 65, the code number for 'B' is 66, the code number for 'C' is 67, etc. So when we subtract the index from the code number for each of 'A', 'B', and 'C' we get 65. But we skipped over 'D' so when we do the subtractions for 'E' and 'F' we get 66. And that's how groupby can put 'A', 'B', & 'C' in one group and 'E' & 'F' in the next group.
This can be tricky stuff. Don't expect to understand it all completely straight away. But if you do some experiments yourself I'm sure it will gradually sink in. ;)
Just for fun, here's the unreadable multiply-nested list comprehension version of that code. ;)
print([z for _, g in groupby(enumerate(s),lambda t:ord(t[1])-t[0])for z in[''.join([*zip(*g)][1])]if len(z)>1])
Here's another version which was inspired by Amit Tripathi's answer. This one doesn't use any imports because it does the grouping manually. prev contains the codepoint number of the previous character. We initialize prev to -2 so that the first time the if i != prev + 1 test is performed it's guaranteed to be true because the smallest possible value of ord(ch) is zero, so a new empty list will be added to groups.
s = 'jaghiuuabcxyzq'
prev, groups = -2, []
for ch in s:
i = ord(ch)
if i != prev + 1:
groups.append([])
groups[-1].append(ch)
prev = i
print(groups)
a = [''.join(u) for u in groups if len(u) > 1]
print(a)
output
[['j'], ['a'], ['g', 'h', 'i'], ['u'], ['u'], ['a', 'b', 'c'], ['x', 'y', 'z'], ['q']]
['ghi', 'abc', 'xyz']

This can be done easily with pure Python
Python 3(should work with Python 2 also) implementation. A simple 8 liner
s = 'jaghiuuabc'
prev, counter, dct = None, 0, dict()
for i in s:
if prev is not None:
if not chr(ord(prev) + 1) == i:
counter += 1
prev = i
dct.setdefault(counter, []).append(prev)
[''.join(dct[d]) for d in dct if len(dct[d]) > 1]
Out[51]: ['ghi', 'abc']
ord converts char to equivalent ASCII number
chr converts a number to equivalent ASCII char
setdefault set default value as list if a key doesn't exists

What about some recursion without any external module ?
a='jaghiuuabc'
import string
alpha = list(string.ascii_lowercase)
def trech(string_1,chr_list,new_string):
final_list=[]
if not string_1:
return 0
else:
for chunk in range(0,len(string_1),chr_list):
for sub_chunk in range(2,len(string_1)+1):
if string_1[chunk:chunk + sub_chunk] in ["".join(alpha[i:i + sub_chunk]) for i in range(0, len(alpha), 1)]:
final_list.append(string_1[chunk:chunk + sub_chunk])
if final_list:
print(final_list)
return trech(string_1[1:],chr_list-1,new_string)
print(trech(a,len(a),alpha))
output:
['gh', 'ghi']
['hi']
['ab', 'abc']
['bc']
0

What does the random.sample() method in Python do?

I want to know the use of random.sample() method and what does it give? When should it be used and some example usage.

According to documentation:
random.sample(population, k)
Return a k length list of unique elements
chosen from the population sequence. Used for random sampling without
replacement.
Basically, it picks k unique random elements, a sample, from a sequence:
>>> import random
>>> c = list(range(0, 15))
>>> c
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
>>> random.sample(c, 5)
[9, 2, 3, 14, 11]
random.sample works also directly from a range:
>>> c = range(0, 15)
>>> c
range(0, 15)
>>> random.sample(c, 5)
[12, 3, 6, 14, 10]
In addition to sequences, random.sample works with sets too:
>>> c = {1, 2, 4}
>>> random.sample(c, 2)
[4, 1]
However, random.sample doesn't work with arbitrary iterators:
>>> c = [1, 3]
>>> random.sample(iter(c), 5)
TypeError: Population must be a sequence or set. For dicts, use list(d).

random.sample() also works on text
example:
> text = open("textfile.txt").read()
> random.sample(text, 5)
> ['f', 's', 'y', 'v', '\n']
\n is also seen as a character so that can also be returned
you could use random.sample() to return random words from a text file if you first use the split method
example:
> words = text.split()
> random.sample(words, 5)
> ['the', 'and', 'a', 'her', 'of']

random.sample(population, k)
It is used for randomly sampling a sample of length 'k' from a population. returns a 'k' length list of unique elements chosen from the population sequence or set
it returns a new list and leaves the original population unchanged and the resulting list is in selection order so that all sub-slices will also be valid random samples
I am putting up an example in which I am splitting a dataset randomly. It is basically a function in which you pass x_train(population) as an argument and return indices of 60% of the data as D_test.
import random
def randomly_select_70_percent_of_data_from_1_to_length(x_train):
return random.sample(range(0, len(x_train)), int(0.6*len(x_train)))

from random import *
lst1 = sample(range(0, 1000), 100)
lst2 = sample(range(0, 1000), 100)
print(lst1)
print(lst2)
print(set(lst1).intersection(set(lst2)))

group list of ints by continuous sequence

I have a list of integers...
[1,2,3,4,5,8,9,10,11,200,201,202]
I would like to group them into a list of lists where each sublist contains integers whose sequence has not been broken. Like this...
[[1,5],[8,11],[200,202]]
I have a rather clunky work around...
lSequenceOfNum = [1,2,3,4,5,8,9,10,11,200,201,202]
lGrouped = []
start = 0
for x in range(0,len(lSequenceOfNum)):
if x != len(lSequenceOfNum)-1:
if(lSequenceOfNum[x+1] - lSequenceOfNum[x]) > 1:
lGrouped.append([lSequenceOfNum[start],lSequenceOfNum[x]])
start = x+1
else:
lGrouped.append([lSequenceOfNum[start],lSequenceOfNum[x]])
print lGrouped
It is the best I could do. Is there a more "pythonic" way to do this? Thanks..

Assuming the list will always be in ascending order:
from itertools import groupby, count
numberlist = [1,2,3,4,5,8,9,10,11,200,201,202]
def as_range(g):
l = list(g)
return l[0], l[-1]
print [as_range(g) for _, g in groupby(numberlist, key=lambda n, c=count(): n-next(c))]

I realised I had overcomplicated this a little, far easier to just count manually than use a slightly convoluted generator:
def ranges(seq):
start, end = seq[0], seq[0]
count = start
for item in seq:
if not count == item:
yield start, end
start, end = item, item
count = item
end = item
count += 1
yield start, end
print(list(ranges([1,2,3,4,5,8,9,10,11,200,201,202])))
Producing:
[(1, 5), (8, 11), (200, 202)]
This method is pretty fast:
This method (and the old one, they perform almost exactly the same):
python -m timeit -s "from test import ranges" "ranges([1,2,3,4,5,8,9,10,11,200,201,202])"
1000000 loops, best of 3: 0.47 usec per loop
Jeff Mercado's Method:
python -m timeit -s "from test import as_range; from itertools import groupby, count" "[as_range(g) for _, g in groupby([1,2,3,4,5,8,9,10,11,200,201,202], key=lambda n, c=count(): n-next(c))]"
100000 loops, best of 3: 11.1 usec per loop
That's over 20x faster - although, naturally, unless speed matters this isn't a real concern.
My old solution using generators:
import itertools
def resetable_counter(start):
while True:
for i in itertools.count(start):
reset = yield i
if reset:
start = reset
break
def ranges(seq):
start, end = seq[0], seq[0]
counter = resetable_counter(start)
for count, item in zip(counter, seq): #In 2.x: itertools.izip(counter, seq)
if not count == item:
yield start, end
start, end = item, item
counter.send(item)
end = item
yield start, end
print(list(ranges([1,2,3,4,5,8,9,10,11,200,201,202])))
Producing:
[(1, 5), (8, 11), (200, 202)]

You can do this efficiently in three steps
given
list1=[1,2,3,4,5,8,9,10,11,200,201,202]
Calculate the discontinuity
[1,2,3,4,5,8,9,10,11 ,200,201,202]
- [1,2,3,4,5,8,9 ,10 ,11 ,200,201,202]
----------------------------------------
[1,1,1,1,3,1,1 ,1 ,189,1 ,1]
(index) 1 2 3 4 5 6 7 8 9 10 11
* *
rng = [i+1 for i,e in enumerate((x-y for x,y in zip(list1[1:],list1))) if e!=1]
>>> rng
[5, 9]
Add the boundaries
rng = [0] + rng + [len(list1)]
>>> rng
[0, 5, 9,12]
now calculate the actual continuity ranges
[(list1[i],list1[j-1]) for i,j in zip(list2,list2[1:])]
[(1, 5), (8, 11), (200, 202)]
LB [0, 5, 9, 12]
UB [0, 5, 9, 12]
-----------------------
indexes (LB,UB-1) (0,4) (5,8) (9,11)

The question is quite old, but I thought I'll share my solution anyway
Assuming import numpy as np
a = [1,2,3,4,5,8,9,10,11,200,201,202]
np.split(a, array(np.add(np.where(diff(a)>1),1)).tolist()[0])

pseudo code (with off-by-one errors to fix):
jumps = new array;
for idx from 0 to len(array)
if array[idx] != array[idx+1] then jumps.push(idx);
I think this is actually a case where it makes sense to work with the indices (as in C, before java/python/perl/etc. improved upon this) instead of the objects in the array.

Here's a version that should be easy to read:
def close_range(el, it):
while True:
el1 = next(it, None)
if el1 != el + 1:
return el, el1
el = el1
def compress_ranges(seq):
iterator = iter(seq)
left = next(iterator, None)
while left is not None:
right, left1 = close_range(left, iterator)
yield (left, right)
left = left1
list(compress_ranges([1, 2, 3, 4, 5, 8, 9, 10, 11, 200, 201, 202]))

Similar questions:
Python - find incremental numbered sequences with a list comprehension
Pythonic way to convert a list of integers into a string of comma-separated ranges
input = [1, 2, 3, 4, 8, 10, 11, 12, 17]
i, ii, result = iter(input), iter(input[1:]), [[input[0]]]
for x, y in zip(i,ii):
if y-x != 1:
result.append([y])
else:
result[-1].append(y)
>>> result
[[1, 2, 3, 4], [8], [10, 11, 12], [17]]
>>> print ", ".join("-".join(map(str,(g[0],g[-1])[:len(g)])) for g in result)
1-4, 8, 10-12, 17
>>> [(g[0],g[-1])[:len(g)] for g in result]
[(1, 4), (8,), (10, 12), (17,)]

Split an integer into digits to compute an ISBN checksum [duplicate]

This question already has answers here:
How to split an integer into a list of digits?
(10 answers)
Closed last month.
I'm writing a program which calculates the check digit of an ISBN number. I have to read the user's input (nine digits of an ISBN) into an integer variable, and then multiply the last digit by 2, the second last digit by 3 and so on. How can I "split" the integer into its constituent digits to do this? As this is a basic homework exercise I am not supposed to use a list.

Just create a string out of it.
myinteger = 212345
number_string = str(myinteger)
That's enough. Now you can iterate over it:
for ch in number_string:
print ch # will print each digit in order
Or you can slice it:
print number_string[:2] # first two digits
print number_string[-3:] # last three digits
print number_string[3] # forth digit
Or better, don't convert the user's input to an integer (the user types a string)
isbn = raw_input()
for pos, ch in enumerate(reversed(isbn)):
print "%d * %d is %d" % pos + 2, int(ch), int(ch) * (pos + 2)
For more information read a tutorial.

while number:
digit = number % 10
# do whatever with digit
# remove last digit from number (as integer)
number //= 10
On each iteration of the loop, it removes the last digit from number, assigning it to digit.
It's in reverse, starts from the last digit, finishes with the first

list_of_ints = [int(i) for i in str(ISBN)]
Will give you a ordered list of ints. Of course, given duck typing, you might as well work with str(ISBN).
Edit: As mentioned in the comments, this list isn't sorted in the sense of being ascending or descending, but it does have a defined order (sets, dictionaries, etc in python in theory don't, although in practice the order tends to be fairly reliable). If you want to sort it:
list_of_ints.sort()
is your friend. Note that sort() sorts in place (as in, actually changes the order of the existing list) and doesn't return a new list.

On Older versions of Python...
map(int,str(123))
On New Version 3k
list(map(int,str(123)))

(number/10**x)%10
You can use this in a loop, where number is the full number, x is each iteration of the loop (0,1,2,3,...,n) with n being the stop point. x = 0 gives the ones place, x = 1 gives the tens, x = 2 gives the hundreds, and so on. Keep in mind that this will give the value of the digits from right to left, so this might not be the for an ISBN but it will still isolate each digit.

Convert it to string and map over it with the int() function.
map(int, str(1231231231))

Recursion version:
def int_digits(n):
return [n] if n<10 else int_digits(n/10)+[n%10]

Converting to str is definitely slower then dividing by 10.
map is sligthly slower than list comprehension:
convert to string with map 2.13599181175
convert to string with list comprehension 1.92812991142
modulo, division, recursive 0.948769807816
modulo, division 0.699964046478
These times were returned by the following code on my laptop:
foo = """\
def foo(limit):
return sorted(set(map(sum, map(lambda x: map(int, list(str(x))), map(lambda x: x * 9, range(limit))))))
foo(%i)
"""
bar = """\
def bar(limit):
return sorted(set([sum([int(i) for i in str(n)]) for n in [k *9 for k in range(limit)]]))
bar(%i)
"""
rac = """\
def digits(n):
return [n] if n<10 else digits(n / 10)+[n %% 10]
def rabbit(limit):
return sorted(set([sum(digits(n)) for n in [k *9 for k in range(limit)]]))
rabbit(%i)
"""
rab = """\
def sum_digits(number):
result = 0
while number:
digit = number %% 10
result += digit
number /= 10
return result
def rabbit(limit):
return sorted(set([sum_digits(n) for n in [k *9 for k in range(limit)]]))
rabbit(%i)
"""
import timeit
print "convert to string with map", timeit.timeit(foo % 100, number=10000)
print "convert to string with list comprehension", timeit.timeit(bar % 100, number=10000)
print "modulo, division, recursive", timeit.timeit(rac % 100, number=10000)
print "modulo, division", timeit.timeit(rab % 100, number=10000)

After own diligent searches I found several solutions, where each has advantages and disadvantages. Use the most suitable for your task.
All examples tested with the CPython 3.5 on the operation system GNU/Linux Debian 8.
Using a recursion
Code
def get_digits_from_left_to_right(number, lst=None):
"""Return digits of an integer excluding the sign."""
if lst is None:
lst = list()
number = abs(number)
if number < 10:
lst.append(number)
return tuple(lst)
get_digits_from_left_to_right(number // 10, lst)
lst.append(number % 10)
return tuple(lst)
Demo
In [121]: get_digits_from_left_to_right(-64517643246567536423)
Out[121]: (6, 4, 5, 1, 7, 6, 4, 3, 2, 4, 6, 5, 6, 7, 5, 3, 6, 4, 2, 3)
In [122]: get_digits_from_left_to_right(0)
Out[122]: (0,)
In [123]: get_digits_from_left_to_right(123012312312321312312312)
Out[123]: (1, 2, 3, 0, 1, 2, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 1, 2)
Using the function divmod
Code
def get_digits_from_right_to_left(number):
"""Return digits of an integer excluding the sign."""
number = abs(number)
if number < 10:
return (number, )
lst = list()
while number:
number, digit = divmod(number, 10)
lst.insert(0, digit)
return tuple(lst)
Demo
In [125]: get_digits_from_right_to_left(-3245214012321021213)
Out[125]: (3, 2, 4, 5, 2, 1, 4, 0, 1, 2, 3, 2, 1, 0, 2, 1, 2, 1, 3)
In [126]: get_digits_from_right_to_left(0)
Out[126]: (0,)
In [127]: get_digits_from_right_to_left(9999999999999999)
Out[127]: (9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9)
Using a construction tuple(map(int, str(abs(number)))
In [109]: tuple(map(int, str(abs(-123123123))))
Out[109]: (1, 2, 3, 1, 2, 3, 1, 2, 3)
In [110]: tuple(map(int, str(abs(1412421321312))))
Out[110]: (1, 4, 1, 2, 4, 2, 1, 3, 2, 1, 3, 1, 2)
In [111]: tuple(map(int, str(abs(0))))
Out[111]: (0,)
Using the function re.findall
In [112]: tuple(map(int, re.findall(r'\d', str(1321321312))))
Out[112]: (1, 3, 2, 1, 3, 2, 1, 3, 1, 2)
In [113]: tuple(map(int, re.findall(r'\d', str(-1321321312))))
Out[113]: (1, 3, 2, 1, 3, 2, 1, 3, 1, 2)
In [114]: tuple(map(int, re.findall(r'\d', str(0))))
Out[114]: (0,)
Using the module decimal
In [117]: decimal.Decimal(0).as_tuple().digits
Out[117]: (0,)
In [118]: decimal.Decimal(3441120391321).as_tuple().digits
Out[118]: (3, 4, 4, 1, 1, 2, 0, 3, 9, 1, 3, 2, 1)
In [119]: decimal.Decimal(-3441120391321).as_tuple().digits
Out[119]: (3, 4, 4, 1, 1, 2, 0, 3, 9, 1, 3, 2, 1)

Use the body of this loop to do whatever you want to with the digits
for digit in map(int, str(my_number)):

I have made this program and here is the bit of code that actually calculates the check digit in my program
#Get the 10 digit number
number=input("Please enter ISBN number: ")
#Explained below
no11 = (((int(number[0])*11) + (int(number[1])*10) + (int(number[2])*9) + (int(number[3])*8)
+ (int(number[4])*7) + (int(number[5])*6) + (int(number[6])*5) + (int(number[7])*4) +
(int(number[8])*3) + (int(number[9])*2))/11)
#Round to 1 dp
no11 = round(no11, 1)
#explained below
no11 = str(no11).split(".")
#get the remainder and check digit
remainder = no11[1]
no11 = (11 - int(remainder))
#Calculate 11 digit ISBN
print("Correct ISBN number is " + number + str(no11))
Its a long line of code, but it splits the number up, multiplies the digits by the appropriate amount, adds them together and divides them by 11, in one line of code. The .split() function just creates a list (being split at the decimal) so you can take the 2nd item in the list and take that from 11 to find the check digit. This could also be made even more efficient by changing these two lines:
remainder = no11[1]
no11 = (11 - int(remainder))
To this:
no11 = (11 - int(no11[1]))
Hope this helps :)

Similar to this answer but more a more "pythonic" way to iterate over the digis would be:
while number:
# "pop" the rightmost digit
number, digit = divmod(number, 10)

How about a one-liner list of digits...
ldigits = lambda n, l=[]: not n and l or l.insert(0,n%10) or ldigits(n/10,l)

Answer: 165
Method: brute-force! Here is a tiny bit of Python (version 2.7) code to count'em all.
from math import sqrt, floor
is_ps = lambda x: floor(sqrt(x)) ** 2 == x
count = 0
for n in range(1002, 10000, 3):
if n % 11 and is_ps(sum(map(int, str(n)))):
count += 1
print "#%i: %s" % (count, n)

Just assuming you want to get the i-th least significant digit from an integer number x, you can try:
(abs(x)%(10**i))/(10**(i-1))
I hope it helps.