I want to take user input for message. Then I generate a random key using random package in python.
But how to shift each letter in message using key's ascii value to produce output as string only?
Example :
message = hi
random key generated = bi
encrypted message = "something in alphabets only like xh or mo."
use the objectype dictionary to map each character to another so that you create a new alphabet, and the loop through the dictionary and replace them using the dictionary
stringa = input()
swapa = {"A":"Q", "B":"A","C":"L"...}
for i in swapa:
stringa = stringa.replace(i,swapa[i])
print(stringa)
you could also take it a step fourther and encryot and decrypt using a keyword
ite = int(input())
for itar in range(ite):
keyw = list(input())
# removes dublicates and keep order
rem = set()
for i in keyw:
if keyw.count(i) > 1:
rem.add(i)
for i in rem:
keyw= keyw[::-1]
keyw.remove(i)
keyw= keyw[::-1]
keyw = "".join(keyw)
# sets up alfabet
linaalfa = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
linaalfa += " "*(len(keyw)-len(linaalfa)%len(keyw)) # adds spaces on the last line
linaalf = "".join([i for i in linaalfa if i not in keyw]) # removes dublicates
linaalf1 = [keyw]+[linaalf[x:x+len(keyw)] for x in range(0,len(linaalf),len(keyw))]
# alfa order find
order = dict()
for w,i in enumerate(sorted(keyw)):
order[keyw.index(i)] = w
# order in alfa ordder
orderalfa = ["".join([i[q] for i in linaalf1]) for q in range(len(keyw))] # now read by column
temp = [""]*len(orderalfa)
for w,i in enumerate(orderalfa):
temp[order[w]] = i
orderalfa = temp
orderalfa = "".join([x.strip() for x in orderalfa])
# relate to base alfabet
translate = dict()
encrypt = dict()
for w,i in enumerate(orderalfa):
translate[i] = linaalfa[w]
encrypt[linaalfa[w]] = i
# trans late message
lina = input().split()
outa = list()
for i in lina:
temp = ""
for q in i:
temp += translate[q]
outa.append(temp)
print(" ".join(outa))
here it takes the input
n # number of quaries
keyword
string that need to be operated
the last bit however is set to decrypt the message if you want it to encrupt it you need to replace the line
temp += translate[q]
to
temp += encrypt[q]
This takes a keyword and
removes dublicate letters in said keyword
Set the keyword in front of the an alfabet(a normal one)
and splits the alfabet up in peaces the same length as the keyword
places them above each other
orderes them in column order based on if the keyword's letter were written in alfabetical order
takes each coloumn and creats a new alphabet based on the each coloum put continualy after each other
this new alphabet now works as the new alphabet
for example if the beginning of the new alphabet were "HGJ" than A would be H, and B would be G and J would be C.
This is stil only a monoalphabetical encryption though.
Related
I have a string with characters repeated. My Job is to find starting Index and ending index of each unique characters in that string. Below is my code.
import re
x = "aaabbbbcc"
xs = set(x)
for item in xs:
mo = re.search(item,x)
flag = item
m = mo.start()
n = mo.end()
print(flag,m,n)
Output :
a 0 1
b 3 4
c 7 8
Here the end index of the characters are not correct. I understand why it's happening but how can I pass the character to be matched dynamically to the regex search function. For instance if I hardcode the character in the search function it provides the desired output
x = 'aabbbbccc'
xs = set(x)
mo = re.search("[b]+",x)
flag = item
m = mo.start()
n = mo.end()
print(flag,m,n)
output:
b 2 5
The above function is providing correct result but here I can't pass the characters to be matched dynamically.
It will be really a help if someone can let me know how to achieve this any hint will also do. Thanks in advance
String literal formatting to the rescue:
import re
x = "aaabbbbcc"
xs = set(x)
for item in xs:
# for patterns better use raw strings - and format the letter into it
mo = re.search(fr"{item}+",x) # fr and rf work both :) its a raw formatted literal
flag = item
m = mo.start()
n = mo.end()
print(flag,m,n) # fix upper limit by n-1
Output:
a 0 3 # you do see that the upper limit is off by 1?
b 3 7 # see above for fix
c 7 9
Your pattern does not need the [] around the letter - you are matching just one anyhow.
Without regex1:
x = "aaabbbbcc"
last_ch = x[0]
start_idx = 0
# process the remainder
for idx,ch in enumerate(x[1:],1):
if last_ch == ch:
continue
else:
print(last_ch,start_idx, idx-1)
last_ch = ch
start_idx = idx
print(ch,start_idx,idx)
output:
a 0 2 # not off by 1
b 3 6
c 7 8
1RegEx: And now you have 2 problems...
Looking at the output, I'm guessing that another option would be,
import re
x = "aaabbbbcc"
xs = re.findall(r"((.)\2*)", x)
start = 0
output = ''
for item in xs:
end = start + len(item[0])
output += (f"{item[1]} {start} {end}\n")
start = end
print(output)
Output
a 0 3
b 3 7
c 7 9
I think it'll be in the Order of N, you can likely benchmark it though, if you like.
import re, time
timer_on = time.time()
for i in range(10000000):
x = "aabbbbccc"
xs = re.findall(r"((.)\2*)", x)
start = 0
output = ''
for item in xs:
end = start + len(item[0])
output += (f"{item[1]} {start} {end}\n")
start = end
timer_off = time.time()
timer_total = timer_off - timer_on
print(timer_total)
I need to create a function that replaces a letter with the letter 13 letters after it in the alphabet (without using encode). I'm relatively new to Python so it has taken me a while to figure out a way to do this without using Encode.
Here's what I have so far. When I use this to type in a normal word like "hello" it works but if I pass through a sentence with special characters I can't figure out how to JUST include letters of the alphabet and skip numbers, spaces or special characters completely.
def rot13(b):
b = b.lower()
a = [chr(i) for i in range(ord('a'),ord('z')+1)]
c = []
d = []
x = a[0:13]
for i in b:
c.append(a.index(i))
for i in c:
if i <= 13:
d.append(a[i::13][1])
elif i > 13:
y = len(a[i:])
z = len(x)- y
d.append(a[z::13][0])
e = ''.join(d)
return e
EDIT
I tried using .isalpha() but this doesn't seem to be working for me - characters are duplicating for some reason when I use it. Is the following format correct:
def rot13(b):
b1 = b.lower()
a = [chr(i) for i in range(ord('a'),ord('z')+1)]
c = []
d = []
x = a[0:13]
for i in b1:
if i.isalpha():
c.append(a.index(i))
for i in c:
if i <= 12:
d.append(a[i::13][1])
elif i > 12:
y = len(a[i:])
z = len(x)- y
d.append(a[z::13][0])
else:
d.append(i)
if message[0].istitle() == True:
d[0] = d[0].upper()
e = ''.join(d)
return e
Following on from comments. OP was advised to use isalpha, and wondering why that's causing duplication (see OP's edit)
This isn't tied to the use of isalpha, it's to do with the second for loop
for i in c:
isn't necessary, and is causing the duplication. You should remove that. Instead you can do the same by just using index = a.index(i). You were already doing this, but for some reason appending to a list instead and causing confusion
Use the index variable any time you would have used i inside the for i in c loop. On a side note, in nested for loops try not to reuse the same variables. It just causes confusion...but that's a matter for code review
Assuming you do all that right it should work.
I'm using a nested list to hold data in a Cartesian coordinate type system.
The data is a list of categories which could be 0,1,2,3,4,5,255 (just 7 categories).
The data is held in a list formatted thus:
stack = [[0,1,0,0],
[2,1,0,0],
[1,1,1,3]]
Each list represents a row and each element of a row represents a data point.
I'm keen to hang on to this format because I am using it to generate images and thus far it has been extremely easy to use.
However, I have run into problems running the following code:
for j in range(len(stack)):
stack[j].append(255)
stack[j].insert(0, 255)
This is intended to iterate through each row adding a single element 255 to the start and end of each row. Unfortunately it adds 12 instances of 255 to both the start and end!
This makes no sense to me. Presumably I am missing something very trivial but I can't see what it might be. As far as I can tell it is related to the loop: if I write stack[0].append(255) outside of the loop it behaves normally.
The code is obviously part of a much larger script. The script runs multiple For loops, a couple of which are range(12) but which should have closed by the time this loop is called.
So - am I missing something trivial or is it more nefarious than that?
Edit: full code
step_size = 12, the code above is the part that inserts "right and left borders"
def classify(target_file, output_file):
import numpy
import cifar10_eval # want to hijack functions from the evaluation script
target_folder = "Binaries/" # finds target file in "Binaries"
destination_folder = "Binaries/Maps/" # destination for output file
# open the meta file to retrieve x,y dimensions
file = open(target_folder + target_file + "_meta" + ".txt", "r")
new_x = int(file.readline())
new_y = int(file.readline())
orig_x = int(file.readline())
orig_y = int(file.readline())
segment_dimension = int(file.readline())
step_size = int(file.readline())
file.close()
# run cifar10_eval and create predictions vector (formatted as a list)
predictions = cifar10_eval.map_interface(new_x * new_y)
del predictions[(new_x * new_y):] # get rid of excess predictions (that are an artefact of the fixed batch size)
print("# of predictions: " + str(len(predictions)))
# check that we are mapping the whole picture! (evaluation functions don't necessarily use the full data set)
if len(predictions) != new_x * new_y:
print("Error: number of predictions from cifar10_eval does not match metadata for this file")
return
# copy predictions to a nested list to make extraction of x/y data easy
# also eliminates need to keep metadata - x/y dimensions are stored via the shape of the output vector
stack = []
for j in range(new_y):
stack.append([])
for i in range(new_x):
stack[j].append(predictions[j*new_x + i])
predictions = None # clear the variable to free up memory
# iterate through map list and explode each category to cover more pixels
# assigns a step_size x step_size area to each classification input to achieve correspondance with original image
new_stack = []
for j in range(len(stack)):
row = stack[j]
new_row = []
for i in range(len(row)):
for a in range(step_size):
new_row.append(row[i])
for b in range(step_size):
new_stack.append(new_row)
stack = new_stack
new_stack = None
new_row = None # clear the variables to free up memory
# add a border to the image to indicate that some information has been lost
# border also ensures that map has 1-1 correspondance with original image which makes processing easier
# calculate border dimensions
top_and_left_thickness = int((segment_dimension - step_size) / 2)
right_thickness = int(top_and_left_thickness + (orig_x - (top_and_left_thickness * 2 + step_size * new_x)))
bottom_thickness = int(top_and_left_thickness + (orig_y - (top_and_left_thickness * 2 + step_size * new_y)))
print(top_and_left_thickness)
print(right_thickness)
print(bottom_thickness)
print(len(stack[0]))
# add the right then left borders
for j in range(len(stack)):
for b in range(right_thickness):
stack[j].append(255)
for b in range(top_and_left_thickness):
stack[j].insert(0, 255)
print(stack[0])
print(len(stack[0]))
# add the top and bottom borders
row = []
for i in range(len(stack[0])):
row.append(255) # create a blank row
for b in range(top_and_left_thickness):
stack.insert(0, row) # append the blank row to the top x many times
for b in range(bottom_thickness):
stack.append(row) # append the blank row to the bottom of the map
# we have our final output
# repackage this as a numpy array and save for later use
output = numpy.asarray(stack,numpy.uint8)
numpy.save(destination_folder + output_file + ".npy", output)
print("Category mapping complete, map saved as numpy pickle: " + output_file + ".npy")
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my question is the following: I have a file which contains around 70 strings, all of them have 6 characters (either a,c,g or t for every position -> these are short DNA-sequences).
For example:
accggt agctta gggatc gactta ccttgg
What I need are the strings which are completely unique. Which have on every position a different character (base) compared with the other strings.
In this case I would get two matches (I define them as lists but this is only an idea for the output format):
[accggt , gggatc]
[gggatc , ccttgg]
The elements of list one are on every position different and so are also the elements of list 2.
Is there a build-in function which can do it? I also thought of regular expression but I'm not that familar with this approach.
Thanks in advance!
Edit:
Ok, it seems it is not that easy to describe. So lets go into more detail:
Let's take the five strings mentioned above:
I would start to compare the first string with all the other strings and then continue with string 2 comparing with all other strings and so on.
The first character of the first string is an a.
The first character of the second string is also an a.
This means I would discard the second string.
The first character of the third string is an g.
Fine.
The second character of the first string is an c.
The second character of the third string is an g.
Fine.
The third character of the first string is an c.
The third character of the third string is an g.
Fine.
The fourth ... and so on.
And if all characters of a string are different from the characters of another string (on every position like described above) I would keep those two strings and would search for the next strings which are different on every position compared to the strings I already found. Because I only have four letters there should be only four possibilities fo different strings.
I should end up with, probably a list, which contains the groups of strings which are different in every position.
I hope this helps.
You can use the following algorithm: iterate through all possible word combinations in your string and check each pair for equality with if [x == y for (x, y) in zip(word, nextWord)].count(True) == 0:.
Here is a snippet:
s = "accggt agctta gggatc gactta ccttgg"
chks = s.split(" ");
for word in chks:
for nextWord in chks:
if word != nextWord:
if [x == y for (x, y) in zip(word, nextWord)].count(True) == 0:
print([word, nextWord])
Result of the IDEONE demo:
['accggt', 'gggatc']
['gggatc', 'accggt']
['gggatc', 'ccttgg']
['ccttgg', 'gggatc']
UPDATE
You can deduplicate the list with a custom function. Here is an updated snippet:
def dedup(lst):
seen = set()
result = []
for item in lst:
fs = frozenset(item)
if fs not in seen:
result.append(item)
seen.add(fs)
return result
res = []
s = "accggt agctta gggatc gactta ccttgg"
chks = s.split(" ");
for word in chks:
for nextWord in chks:
if word != nextWord:
if [x == y for (x, y) in zip(word, nextWord)].count(True) == 0:
res.append([word, nextWord])
print(dedup(res))
Result: [['accggt', 'gggatc'], ['gggatc', 'ccttgg']].
To check the words by 3, you need to create all possible permutations of the string into 3-word combinations and use something like:
from itertools import permutations
def dedup(lst):
seen = set()
result = []
for item in lst:
fs = frozenset(item)
if fs not in seen:
result.append(item)
seen.add(fs)
return result
res = []
s = "accggt agctta gggatc gactta ccttgg"
chks = s.split(" ");
perms = [p for p in permutations(chks, 3)]
for perm in perms:
if [(x == y or y == z or x == z) for (x, y, z) in zip(*perm)].count(True) == 0:
res.append(perm)
print(dedup(res))
To find the DNA strings which are completely different on every character you have to check every string against any other string if any character of the given string is the same character on the same position in the comparing string.
Here is an example code for that:
# read all dna strings into a list of strings
dna = ['accggt', 'agctta', 'gggatc', 'gactta', 'ccttgg', '123456']
def compare_two_dna(dna1, dna2):
i = 0
l = len(dna1)
while(i < l):
if dna1[i] == dna2[i]:
return True
i += 1
return False
def is_dna_unique(d, dna_strings):
return len(filter(lambda x: compare_two_dna(d, x), dna_strings)) == 1
# filter all items which only occure once in the list
unique_dna = filter(lambda d: is_dna_unique(d, dna), dna)
print(unique_dna)
The result here is: 123456
var dnaList = "accggt agctta gggatc gactta ccttgg".split( " " );
function getUniqueDnas( dna_list ){
var result = [];
for( var d1 in dna_list ){
var isRepeat = false;
var dna1 = dna_list[ d1 ];
for( var d2 in dna_list ){
var dna2 = dna_list[ d2 ];
if( dna1 == dna2 ){
isRepeat = true;
break;
}
}
if( !isRepeat )
result.push( dna1 );
}
return result;
}
var uniqueDnaList = getUniqueDnas( dnaList );
I am trying to convert the below Matlab code into C++ using codegen. However it fails at build and I get the error:
"??? Unless 'rows' is specified, the first input must be a vector. If the vector is variable-size, the either the first dimension or the second must have a fixed length of 1. The input [] is not supported. Use a 1-by-0 or 0-by-1 input (e.g., zeros(1,0) or zeros(0,1)) to represent the empty set."
It then points to [id,m,n] = unique(id); being the culprit. Why doesn't it build and what's the best way to fix it?
function [L,num,sz] = label(I,n) %#codegen
% Check input arguments
error(nargchk(1,2,nargin));
if nargin==1, n=8; end
assert(ndims(I)==2,'The input I must be a 2-D array')
sizI = size(I);
id = reshape(1:prod(sizI),sizI);
sz = ones(sizI);
% Indexes of the adjacent pixels
vec = #(x) x(:);
if n==4 % 4-connected neighborhood
idx1 = [vec(id(:,1:end-1)); vec(id(1:end-1,:))];
idx2 = [vec(id(:,2:end)); vec(id(2:end,:))];
elseif n==8 % 8-connected neighborhood
idx1 = [vec(id(:,1:end-1)); vec(id(1:end-1,:))];
idx2 = [vec(id(:,2:end)); vec(id(2:end,:))];
idx1 = [idx1; vec(id(1:end-1,1:end-1)); vec(id(2:end,1:end-1))];
idx2 = [idx2; vec(id(2:end,2:end)); vec(id(1:end-1,2:end))];
else
error('The second input argument must be either 4 or 8.')
end
% Create the groups and merge them (Union/Find Algorithm)
for k = 1:length(idx1)
root1 = idx1(k);
root2 = idx2(k);
while root1~=id(root1)
id(root1) = id(id(root1));
root1 = id(root1);
end
while root2~=id(root2)
id(root2) = id(id(root2));
root2 = id(root2);
end
if root1==root2, continue, end
% (The two pixels belong to the same group)
N1 = sz(root1); % size of the group belonging to root1
N2 = sz(root2); % size of the group belonging to root2
if I(root1)==I(root2) % then merge the two groups
if N1 < N2
id(root1) = root2;
sz(root2) = N1+N2;
else
id(root2) = root1;
sz(root1) = N1+N2;
end
end
end
while 1
id0 = id;
id = id(id);
if isequal(id0,id), break, end
end
sz = sz(id);
% Label matrix
isNaNI = isnan(I);
id(isNaNI) = NaN;
[id,m,n] = unique(id);
I = 1:length(id);
L = reshape(I(n),sizI);
L(isNaNI) = 0;
if nargout>1, num = nnz(~isnan(id)); end
Just an FYI, if you are using MATLAB R2013b or newer, you can replace error(nargchk(1,2,nargin)) with narginchk(1,2).
As the error message says, for codegen unique requires that the input be a vector unless 'rows' is passed.
If you look at the report (click the "Open report" link that is shown) and hover over id you will likely see that its size is neither 1-by-N nor N-by-1. The requirement for unique can be seen if you search for unique here:
http://www.mathworks.com/help/coder/ug/functions-supported-for-code-generation--alphabetical-list.html
You could do one of a few things:
Make id a vector and treat it as a vector for the computation. Instead of the declaration:
id = reshape(1:prod(sizI),sizI);
you could use:
id = 1:numel(I)
Then id would be a row vector.
You could also keep the code as is and do something like:
[idtemp,m,n] = unique(id(:));
id = reshape(idtemp,size(id));
Obviously, this will cause a copy, idtemp, to be made but it may involve fewer changes to your code.
Remove the anonymous function stored in the variable vec and make vec a subfunction:
function y = vec(x)
coder.inline('always');
y = x(:);
Without the 'rows' option, the input to the unique function is always interpreted as a vector, and the output is always a vector, anyway. So, for example, something like id = unique(id) would have the effect of id = id(:) if all the elements of the matrix id were unique. There is no harm in making the input a vector going in. So change the line
[id,m,n] = unique(id);
to
[id,m,n] = unique(id(:));