Suppose I have a list of 17 objects, out of these 17 objects some have a certain property say P1. I separate them and say they are n in number where n < 17. Out of these n objects some have another property say P2. I separate them and say they are m in number where m < n. Out of these m objects some have another property say P3. I separate them and say they are k in number where k < m. I want to print these k objects only.
I was thinking of a long way that is I separate n, m and k objects all from 17 objects according to their respective property and then look for common index, the index that appear in all of three calculations.
Either I need to derive this common index or I do what I have written in the first paragraph that is to filter through and through according to the three properties.
Example:
list_1 = [17, 23, 15, 37, 43, 52, 57, 93, 55, 85, 11, 13, 7, 22, 24]
list_odd = [17, 23, 15, 37, 43, 57, 93, 55, 85, 11, 13, 7] #P1 is a number is odd
list_odd_div3 = [15, 57, 93] #P2 is a number divisible by 3
list_odd_div5 = [15, 55, 85] #P3 is a number divisible by 5
required_list = [15] #A number having P1, P2 and P3
Related
I am having trouble finding the number of distinct elements in a 2D-array using for loops. I know how to do it if its a 1D-array but can't seem to figure out how to do it for 2D-array.
I tried searching for it, but can't seem to quite understand how some of the example works.
I recommend you to use std::array and use the find() method for finding specific element in the array.
int array[5][4] = {{ 34, 56, 79, 12},
{ 25, 37, 41, 18 },
{ 59, 29, 38, 47 },
{ 55, 11, 88, 34 },
{ 45, 19, 34, 66 } };
And use
find(array[0], array[n-1]+m, x)
//array is your 2D array, n is the first dimension, m is the second and x is your value
I have just started my journey with Wolfram Mathematica and I want to implement a simple genetic algorithm. The construction of the data is given and I have to start with such rows/columns.
Here is what I have:
chromosome := RandomSample[CharacterRange["A", "G"], 7]
chromosomeList = Table[chromosome, 7] // MatrixForm
This gives me a matrix, where every row represents a chromosome:
yPos = Flatten[Position[chromosomeList, #], 1] & /# {"A", "B", "C",
"D", "E", "F", "G"};
yPos = yPos[[All, 3 ;; 21 ;; 3]] // Transpose
Now every column represents a letter (From A to G) and every row it's index in every chromosome:
Here is a given efficiency matrix, where very row represents different letter (From A to G) and every column gives the value that should be applied on the particular position:
efficiencyMatrix = {
{34, 31, 20, 27, 24, 24, 18},
{14, 14, 22, 34, 26, 19, 22},
{22, 16, 21, 27, 35, 25, 30},
{17, 21, 24, 16, 31, 22, 20},
{17, 29, 22, 31, 18, 19, 26},
{26, 29, 37, 34, 37, 20, 21},
{30, 28, 37, 28, 29, 23, 19}}
What I want to do is to create a matrix with values that correspond to the letter and it's position. I have done it like that:
values = Transpose[{ efficiencyMatrix[[1, yPos[[1]]]],
efficiencyMatrix[[2, yPos[[2]]]],
efficiencyMatrix[[3, yPos[[3]]]],
efficiencyMatrix[[4, yPos[[4]]]],
efficiencyMatrix[[5, yPos[[5]]]],
efficiencyMatrix[[6, yPos[[6]]]],
efficiencyMatrix[[7, yPos[[7]]]]}]
How can I write it in more elegant way?
You can apply a list of functions to some variable using the function Through, which is helpful when applying Position multiple times. Because Position[patt][expr] == Position[expr, patt], we can do
Through[ (Position /# CharacterRange["A","C"])[{"B", "C", "A"}] ]
to get {3, 1, 2}.
Position can also operate on lists, so we can simplify finding ypos by doing
Transpose#Map[Last, Through[(Position /# characters)[chromosomeList]], {2}]
where characters is the relevant output of CharacterRange.
We can also simplify dealing with ranges of integers by mapping over the Range function, so in total we end up with
characters = CharacterRange["A","G"]
efficiencies = ...
chromosomes = ...
ypos = Transpose#Map[Last, Through[(Position /# characters)[chromosomes]], {2}];
efficiencies[[#, ypos[[#]]]]& /# Range[Length[characters]] //Transpose ]
i'm trying to iterate over an arraylist saving in every loop the highest/lowest difference of the consecutive values.
e1=([ 0 , 0, 0, 0, 15, 28, 28, 28, 27, 27, 35, 44, 43, 43, 42, 39])
Hodiffmax = 0
Hodiffmin = 0
for k in e1:
diff1= e1[k+1] - e1[k]
if diff1 > Hodiffmax:
Hodiffmax=diff1
if diff1 < Hodiffmin:
Hodiffmin=diff1
The problem is i get an "index out of bound" error. How can i iterate through an arraylist with [k+1]? I tried a bunch of things now but i dont get smarter. I appreciate any help!
EDIT (that works neither):
for k in e1:
for w in k:
diff1= e1[w+1] - e1[w]
if diff1 > Hodiffmax:
Hodiffmax=diff1
if diff1 < Hodiffmin:
Hodiffmin=diff1
Error: for w in k - TypeError: 'numpy.int32' object is not iterable
With [y - x for x, y in zip(e1, e1[1:])] you can get consecutive differences without worrying for the indexes:
>>> e1 = [ 0 , 0, 0, 0, 15, 28, 28, 28, 27, 27, 35, 44, 43, 43, 42, 39]
>>> l = [y - x for x, y in zip(e1, e1[1:])]
>>> Hodiffmax, Hodiffmin = max(l), min(l)
>>> Hodiffmax, Hodiffmin
15, -3
Use the grouper recipe:
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)
from itertools import izip_longest # required by grouper
i = [0, 0, 0, 0, 15, 28, 28, 28, 27, 27, 35, 44, 43, 43, 42, 39]
lowest = None
highest = None
for z,q in grouper(i, 2):
v = z-q
if v < lowest:
lowest = v
if v > highest:
highest = v
print(lowest)
print(highest)
The problem here is that you are iterating over the elements of the list. By doing
for k in e1:
k will get the values of the elements on e1. k=0, k=0, k=0, k=0, k=15, k=28 and so on. What you want instead is to iterate over the range of the list.
for k in range(len(e1)):
k will get the values of indexes on e1. k=0, k=1, k=2, k=3, k=4, k=5 and so on. I think you were looking for something like this:
e1 = [0, 0, 0, 0, 15, 28, 28, 28, 27, 27, 35, 44, 43, 43, 42, 39]
for k in range(len(e1)):
print k
if k > 0:
diff1 = e1[k] - e1[k-1]
if diff1 > Hodiffmax:
Hodiffmax=diff1
if diff1 < Hodiffmin:
Hodiffmin=diff1
print 'Hodiffmax ' + str(Hodiffmax)
print 'Hodiffmin ' + str(Hodiffmin)
I'm trying to implement simplified DES for learning purposes in python, but I am having trouble figuring out how to do the permutations based on a "schedule." Essentially, I have a tuple with the appropriate permutations, and I need to bit shift to the correct location.
For example, using a key:
K = 00010011 00110100 01010111 01111001 10011011 10111100 11011111 11110001
Would move the 57st bit to the first bit spot, 49th bit to the second bit spot, etc...
K+ = 1111000 0110011 0010101 0101111 0101010 1011001 1001111 0001111
Current code:
def keyGen(key):
PC1table = (57, 49, 41, 33, 25, 17, 9,
1, 58, 50, 42, 34, 26, 18,
10, 2, 59, 51, 43, 35, 27,
19, 11, 3, 60, 52, 44, 36,
63, 55, 47, 39, 31, 23, 15,
7, 62, 54, 46, 38, 30, 22,
14, 6, 61, 53, 45, 37, 29,
21, 13, 5, 28, 20, 12, 4)
keyBinary = bin(int(key, 16))[2:].zfill(64)
print keyBinary
permute(PC1table, keyBinary)
def permute(permutation, permuteInput):
elements = list(enumerate(permutation))
for bit in permuteInput:
***magic bitshifting goes here***
keyGen("133457799BBCDFF1")
The logic I thought would work was to enumerate the tuple of permutations, and for each bit of my old key, look in the enumeration to find the index corresponding the the bit, and bit shift the appropriate number of times, but I just can't figure out how to go about doing this. It may be that I am approaching the problem from the wrong angle, but any guidance would be greatly appreciated!
Ok, I ended up figuring a way to make this work, although this probably isn't the most efficient way...
prior to calling the function, turn the binary number into a list:
keyBinary = bin(int(key, 16))[2:].zfill(64)
keyBinary = [int(i) for i in keyBinary]
Kplus = permute(PC1table, keyBinary)
def permute(mapping, permuteInput):
permuteOutput = []
for i in range(len(mapping)):
permuteOutput.append(permuteInput[mapping[i % 56] - 1])
return permuteOutput
if anyone has a better way of tackling this, I'd love to see your solutions!
I've written a similar question which was closed I would like to ask not the code but an efficiency tip. I haven't coded but if I can't find any good hint in here I'll go and code straightforward. My question:
Suppose you have a function listNums that take a as lower bound and b as upper bound.
For example a=120 and b=400
I want to print numbers between these numbers with one rule. 120's permutations are 102,210,201 etc. Since I've got 120 I would like to skip printing 201 or 210.
Reason: The upper limit can go up to 1020 and reducing the permutations would help the running time.
Again just asking for efficiency tips.
I am not sure how you are handling 0s (eg: after outputting 1 do you skip 10, 100 etc since technically 1=01=001..).
The trick is to select a number such that all its digits are in increasing order (from left to right).
You can do it recursively. AT every recursion add a digit and make sure it is equal to or higher than the one you recently added.
EDIT: If the generated number is less than the lower limit then permute it in such a way that it is greater than or equal to the lower limit. If A1A2A3..Ak is your number and it is lower than limit), then incrementally check if any of A2A1A3...Ak, A3A1A2...Ak, ... , AkA1A2...Ak-1 are within limit. If need arises, repeat this step to with keeping Ak as first digit and finding a combination of A1A2..Ak-1.
Eg: Assume we are selecting 3 digits and lower limit is 99. If the combination is 012, then the lowest permutation that is higher than 99 is 102.
When the lower bound is 0, an answer is given by the set of numbers with non-decreasing digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 23, 24, 25, 26, 27, 28, 29, 33, 34, 35, 36, 37, 38, 39, 44, 45, 46, 47, 48, 49, 55, 56, 57, 58, 59, 66, 67, 68, 69, 77, 78, 79, 88, 89, 99, 111, 112...) that fall in the requested range.
This sequence is easily formed by incrementing an integer, and when there is a carry, replicate the digit instead of carrying. Exemple: 73 is followed by 73+1 = 74 (no carry); 79 is followed by 79+1 = 80 (carry), so 88 instead; 22356999 is followed by 22356999+1 = 22357000, hence 22357777.
# Python code
A= 0 # CAUTION: this version only works for A == 0 !
B= 1000
N= A
while N < B:
# Detect zeroes at the end
S= str(N)
P= S.find('0')
if P > 0:
# Replicate the last nonzero digit
S= S[:P] + ((len(S) - P) * S[P-1])
N= eval(S)
# Next candidate
print N
N+= 1
Dealing with a nonzero lower bound is a lot more tricky.