I have an exchange data stream coming in that sets symbol to a random letter of the alphabet every like 10 ms in an infinite while loop that is calling func(pair, time). Symbol is the trading pair for simplification here. I have used A and Z to show the range.
Using the method below, I have to create a lot of if-statements when I want to count i for each letter. IE, I have to create iA, iB, iC, .. iZ. In reality, there is about 20 lines code to execute instead of the i-iteration shown here. This is very messy.
I am a beginner in coding and stuck with finding a more elegant and perhaps computationally faster way to do this.
def func(symbol, cur_time):
if future_timeA > cur_timeA and symbol = A:
iA += iA
return -1
if future_timeA < cur_timeA and symbol = A:
future_timeA = cur_timeA + 1
valueA = iA
return valueA
if future_timeZ > cur_timeZ and symbol = Z:
iZ += iZ
return -1
if future_timeZ < cur_timeZ and symbol = Z:
future_timeZ = cur_timeZ + 1
value = iZ
return valueZ
Since you need to check for each of the 26 letters, at least this much code would be there. This is available in Python 3.10.
match symbol:
case "A":
do whatever for A
case "B":
do whatever for B
....
case "Z":
do whatever for Z
Instead of symbol, use a number between 0 and 25 to represent all possible letters. This way you can create an array with 26 inputs to calculate the amount of all letters, and put this in a loop for cleaner code.
Hope that helps you further!
Python3.10 introduced a match case statement for that kind of problem: See this answer.
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I am researching wireless security and trying to write a python script to generate passwords, not random, but a dictionary of hex numbers. The letters need to be capital, and it has to go from 12 characters to 20 characters. I went from 11 f's to 20 f's, this seems like it would meet the requirements. I then tried to place them in a text file. After I made the file, I chmod'ed it to 777 and then clicked run. It has been a few minutes, but I cannot tell if it is working or not. I am running it in kali right now, on a 64 bit core i3 with 8gb of ram. I'm not sure how long it would be expected to take, but this is my code, let me know if it looks right please:
# generate 10 to 32 character password list using hex numbers, 0-9 A-F
def gen_pwd(x):
x = range(17592186044415 -295147905179352830000)
def toHex(dec):
x = (dec % 16)
digits = "0123456789ABCDEF"
rest = dec / 16
if (rest == 0):
return digits[x]
return toHex(rest) + digits[x]
for x in range(x):
print toHex(x)
f = open(/root/Home/sdnlnk_pwd.txt)
print f
value = x
string = str(value)
f.write(string)
gen_pwd
how bout just
password = hex(random.randint(1000000,100000000))[2:]
or
pw_len = 12
my_alphabet = "1234567890ABCDEF"
password = "".join(random.choice(my_alphabet) for _ in range(pw_len))
or what maybe closer to what you are trying to do
struct.pack("Q",12365468987654).encode("hex").upper()
basically you are overcomplicating a very simple task
to do exactly what you are asking you can simplify it
import itertools, struct
def int_to_chars(d):
'''
step 1: break into bytes
'''
while d > 0: # while we have not consumed
yield struct.pack("B",d&0xFF) # decode char
d>>=8 # shift right one byte
yield "" # a terminator just in case its empty
def to_password(d):
# this will convert an arbitrarily large number to a password
return "".join(int_to_chars(d)).encode("hex").upper()
# you could probably just get away with `return hex(d)[2:]`
def all_the_passwords(minimum,maximum):
#: since our numbers are so big we need to resort to some trickery
all_pw = itertools.takewhile(lambda x:x<maximum,
itertools.count(minimum))
for pw in all_pw:
yield to_password(pw)
all_passwords = all_the_passwords( 0xfffffffffff ,0xffffffffffffffffffff)
#this next bit is gonna take a while ... go get some coffee or something
for pw in all_passwords:
print pw
#you will be waiting for it to finish for a very long time ... but it will get there
You can use time.time() to get the execution time. and if you are using python 2 use xrange() instead range because xrange return an iterator :
import time
def gen_pwd(x):
def toHex(dec):
x = (dec % 16)
digits = "0123456789ABCDEF"
rest = dec / 16
if (rest == 0):
return digits[x]
return toHex(rest) + digits[x]
for x in range(x):
print toHex(x)
f = open("/root/Home/sdnlnk_pwd.txt")
print f
value = x
string = str(value)
f.write(string)
start= time.time()
gen_pwd()
last=time.time()-start
print last
Note : you need () to call your function and "" in your open() function. also i think your first range is an extra command , as its wrong , you need to remove it.
Disclaimer
I'd like to comment on the OP question but I need to show some code and also the output that said code produces, so that I eventually decided to present my comment in the format of an answer.
OTOH, I hope that this comment persuades the OP that her/his undertaking, while conceptually simple (see my previous answer, 6 lines of Python code), is not feasible with available resources (I mean, available on Planet Earth).
Code
import locale
locale.setlocale(locale.LC_ALL, 'en_US.UTF-8')
pg = lambda n: locale.format("%d", n, grouping=True)
def count_bytes(low, hi):
count = low+1
for i in range(low+1,hi+1):
nn = 15*16**(i-1)
nc = i+1
count = count + nn*nc
return count
n_b = count_bytes(10,20)
n_d = n_b/4/10**12
dollars = 139.99*n_d
print "Total number of bytes to write on disk:", pg(n_b)
print """
Considering the use of
WD Green WD40EZRX 4TB IntelliPower 64MB Cache SATA 6.0Gb/s 3.5\" Internal Hard Drives,
that you can shop at $139.99 each
(see <http://www.newegg.com/Product/Product.aspx?Item=N82E16822236604>,
retrieved on December 29th, 2014)."""
print "\nNumber of 4TB hard disk drives necessary:", pg(n_d)
print "\nCost of said hard disks: $" + pg(dollars)
Output
Total number of bytes to write on disk: 25,306,847,157,254,216,063,385,611
Considering the use of
WD Green WD40EZRX 4TB IntelliPower 64MB Cache SATA 6.0Gb/s 3.5" Internal Hard Drives,
that you can shop at $139.99 each
(see <http://www.newegg.com/Product/Product.aspx?Item=N82E16822236604>,
retrieved on December 29th, 2014).
Number of 4TB hard disk drives necessary: 6,326,711,789,313
Cost of said hard disks: $885,676,383,385,926
My comment on what the OP wants to do
Quite a bit of disk storage (and money) is needed to accomplish your undertaking.
Perspective
Projected US Federal debt at the end of fiscal year 2014 is $18.23 trillion, my estimated cost, not considering racks, power supplies and energy bills, is $886 trillion.
Recommended reading
Combinatorial_Explosion#SussexUniversity,
There is hope
If you are still convinced to pursue your research project on wireless security in the direction you've described, it is possible that you can get a substantial volume discount on the drives'purchase.
characters=["a","b","c"]
for x,y in zip(range(5),characters):
print (hex(x)+y)
Output:
>>>
0x0a
0x1b
0x2c
>>>
You see, its actually doing that with a short way. It is not possible if you use a range like that, keep it small and try to add another things to your result.
Also for file process, here is a better way:
with open("filepath/name","a+") as f:
f.write("whateveryouwanttowrite")
I was working with password generators, well better if you define a dict with complicated characters and compile them like:
passw={"h":"_*2ac","e":"=.kq","y":"%.hq1"}
x=input("Wanna make some passwords? Enter a sentence or word: ")
for i in x:
print (passw[i],end="")
with open("passwords.txt","a+") as f:
f.write(passw[i])
Output:
>>>
Wanna make some passwords? Enter a sentence or word: hey
_*2ac=.kq%.hq1
>>>
So, just define a dict with keys=alphabet and values=complicated characters, and you can make very strong passwords with simple words-sentences.I just wrote it for an example, of course you can add them to dict later, you dont have to write. But basic way is for that is better I think.
Preamble
I don't want to comment on what you want to do.
Code MkI
Your code can be trimmed (quite a bit) to the following
with open("myfile", "w") as f:
for x in xrange(0xff,0xff*2+1): f.write("%X\n"%x)
Comments on my code
Please note that
You can write hex numbers in source code as, ehm, hex numbers and you can mix hex and decimal notation as well
The to_hex function is redundant as python has (surprise!) a number of different ways to format your output as you please (here I've used so called string interpolation).
Of course you have to change the filename in the open statement and
adjust the extremes of the interval generated by xrange (it seems
you're using python 2.x) to your content.
Code MkII
Joran Beasley remarked that (at least in Python 2.7) xrange internally uses a C long and as such it cannot step up to the task of representing
0XFFFFFFFFFFFFFFFFFFFF. This alternative code may be a possibility:
f = open("myfile", "w")
cursor = 0XFFFFFFFFFF
end = 0XFFFFFFFFFFFFFFFFFFFF
while cursor <= end:
f.write("%X\n"%cursor)
cursor += 1
all of this is well and good, however, none of it accomplishes my purpose. if python cannot handle such large numbers, i will have to use something else. as i stated, i do not want to generate random anything, i need a list of sequential hex characters which are anywhere from 12 characters to 20 characters long. it is to make a dictionary of passwords which are nothing more than a hex number that should be about 16 characters long.
does anyone have any suggestions on what i can use for this purpose? i think some type of c language should do the trick, but i know less about c or c++ than python. sounds like this will take a while, but that's ok, it is just a research project.
i have come up with another possibility, counting in hex starting from 11 f's and going until i reach 20 f's. this would produce about 4.3 billion numbes, which should fit in a 79 million page word document. sounds like it is a little large, but if i go from 14 f's to 18 f's, that should be manageable. here is the code i am proposing now:
x = 0xffffffffffffff
def gen_pwd(x):
while x <= 0xffffffffffffffffff:
return x
string = str(x)
f = open("root/Home/sdnlnk_pwd.txt")
print f.upper(string, 'a')
f.write(string)
x = x + 0x1
gen_pwd()
I have this code which computes the Longest Common Subsequence between random strings to see how accurately one can reconstruct an unknown region of the input. To get good statistics I need to iterate it many times but my current python implementation is far too slow. Even using pypy it currently takes 21 seconds to run once and I would ideally like to run it 100s of times.
#!/usr/bin/python
import random
import itertools
#test to see how many different unknowns are compatible with a set of LCS answers.
def lcs(x, y):
n = len(x)
m = len(y)
# table is the dynamic programming table
table = [list(itertools.repeat(0, n+1)) for _ in xrange(m+1)]
for i in range(n+1): # i=0,1,...,n
for j in range(m+1): # j=0,1,...,m
if i == 0 or j == 0:
table[i][j] = 0
elif x[i-1] == y[j-1]:
table[i][j] = table[i-1][j-1] + 1
else:
table[i][j] = max(table[i-1][j], table[i][j-1])
# Now, table[n, m] is the length of LCS of x and y.
return table[n][m]
def lcses(pattern, text):
return [lcs(pattern, text[i:i+2*l]) for i in xrange(0,l)]
l = 15
#Create the pattern
pattern = [random.choice('01') for i in xrange(2*l)]
#create text start and end and unknown.
start = [random.choice('01') for i in xrange(l)]
end = [random.choice('01') for i in xrange(l)]
unknown = [random.choice('01') for i in xrange(l)]
lcslist= lcses(pattern, start+unknown+end)
count = 0
for test in itertools.product('01',repeat = l):
test=list(test)
testlist = lcses(pattern, start+test+end)
if (testlist == lcslist):
count += 1
print count
I tried converting it to numpy but I must have done it badly as it actually ran more slowly. Can this code be sped up a lot somehow?
Update. Following a comment below, it would be better if lcses used a recurrence directly which gave the LCS between pattern and all sublists of text of the same length. Is it possible to modify the classic dynamic programming LCS algorithm somehow to do this?
The recurrence table table is being recomputed 15 times on every call to lcses() when it is only dependent upon m and n where m has a maximum value of 2*l and n is at most 3*l.
If your program only computed table once, it would be dynamic programming which it is not currently. A Python idiom for this would be
table = None
def use_lcs_table(m, n, l):
global table
if table is None:
table = lcs(2*l, 3*l)
return table[m][n]
Except using an class instance would be cleaner and more extensible than a global table declaration. But this gives you an idea of why its taking so much time.
Added in reply to comment:
Dynamic Programming is an optimization that requires a trade-off of extra space for less time. In your example you appear to be doing a table pre-computation in lcs() but you build the whole list on every single call and then throw it away. I don't claim to understand the algorithm you are trying to implement, but the way you have it coded, it either:
Has no recurrence relation, thus no grounds for DP optimization, or
Has a recurrence relation, the implementation of which you bungled.
I have a problem in Python I simply can't wrap my head around, even though it's fairly simple (I think).
I'm trying to make "string series". I don't really know what it's called, but it goes like this:
I want a function that makes strings that run in series, so that every time the functions get called it "counts" up once.
I have a list with "a-z0-9._-" (a to z, 0 to 9, dot, underscore, dash). And the first string I should receive from my method is aaaa, next time I call it, it should return aaab, next time aaac etc. until I reach ----
Also the length of the string is fixed for the script, but should be fairly easy to change.
(Before you look at my code, I would like to apologize if my code doesn't adhere to conventions; I started coding Python some days ago so I'm still a noob).
What I've got:
Generating my list of available characters
chars = []
for i in range(26):
chars.append(str(chr(i + 97)))
for i in range(10):
chars.append(str(i))
chars.append('.')
chars.append('_')
chars.append('-')
Getting the next string in the sequence
iterationCount = 0
nameLen = 3
charCounter = 1
def getString():
global charCounter, iterationCount
name = ''
for i in range(nameLen):
name += chars[((charCounter + (iterationCount % (nameLen - i) )) % len(chars))]
charCounter += 1
iterationCount += 1
return name
And it's the getString() function that needs to be fixed, specifically the way name gets build.
I have this feeling that it's possible by using the right "modulu hack" in the index, but I can't make it work as intended!
What you try to do can be done very easily using generators and itertools.product:
import itertools
def getString(length=4, characters='abcdefghijklmnopqrstuvwxyz0123456789._-'):
for s in itertools.product(characters, repeat=length):
yield ''.join(s)
for s in getString():
print(s)
aaaa
aaab
aaac
aaad
aaae
aaaf
...
allow me to preface this by saying that i am learning python on my own as part of my own curiosity, and i was recommended a free online computer science course that is publicly available, so i apologize if i am using terms incorrectly.
i have seen questions regarding this particular problem on here before - but i have a separate question from them and did not want to hijack those threads. the question:
"a substring is any consecutive sequence of characters inside another string. The same substring may occur several times inside the same string: for example "assesses" has the substring "sses" 2 times, and "trans-Panamanian banana" has the substring "an" 6 times. Write a program that takes two lines of input, we call the first needle and the second haystack. Print the number of times that needle occurs as a substring of haystack."
my solution (which works) is:
first = str(input())
second = str(input())
count = 0
location = 0
while location < len(second):
if location == 0:
location = str.find(second,first,0)
if location < 0:
break
count = count + 1
location = str.find(second,first,location +1)
if location < 0:
break
count = count + 1
print(count)
if you notice, i have on two separate occasions made the if statement that if location is less than 0, to break. is there some way to make this a 'global' condition so i do not have repetitive code? i imagine efficiency becomes paramount with increasing program sophistication so i am trying to develop good practice now.
how would python gurus optimize this code or am i just being too nitpicky?
I think Matthew and darshan have the best solution. I will just post a variation which is based on your solution:
first = str(input())
second = str(input())
def count_needle(first, second):
location = str.find(second,first)
if location == -1:
return 0 # none whatsoever
else:
count = 1
while location < len(second):
location = str.find(second,first,location +1)
if location < 0:
break
count = count + 1
return count
print(count_needle(first, second))
Idea:
use function to structure the code when appropriate
initialise the variable location before entering the while loop save you from checking location < 0 multiple times
Check out regular expressions, python's re module (http://docs.python.org/library/re.html). For example,
import re
first = str(input())
second = str(input())
regex = first[:-1] + '(?=' + first[-1] + ')'
print(len(re.findall(regex, second)))
As mentioned by Matthew Adams the best way to do it is using python'd re module Python re module.
For your case the solution would look something like this:
import re
def find_needle_in_heystack(needle, heystack):
return len(re.findall(needle, heystack))
Since you are learning python, best way would be to use 'DRY' [Don't Repeat Yourself] mantra. There are lots of python utilities that you can use for many similar situation.
For a quick overview of few very important python modules you can go through this class:
Google Python Class
which should only take you a day.
even your aproach could be imo simplified (which uses the fact, that find returns -1, while you aks it to search from non existent offset):
>>> x = 'xoxoxo'
>>> start = x.find('o')
>>> indexes = []
>>> while start > -1:
... indexes.append(start)
... start = x.find('o',start+1)
>>> indexes
[1, 3, 5]
needle = "ss"
haystack = "ssi lass 2 vecess estan ss."
print 'needle occurs %d times in haystack.' % haystack.count(needle)
Here you go :
first = str(input())
second = str(input())
x=len(first)
counter=0
for i in range(0,len(second)):
if first==second[i:(x+i)]:
counter=counter+1
print(counter)
Answer
needle=input()
haystack=input()
counter=0
for i in range(0,len(haystack)):
if(haystack[i:len(needle)+i]!=needle):
continue
counter=counter+1
print(counter)
I am making a program for my own purposes (a naming program) that completely generates a random name. The problem is I cannot assign a number to a letter, so as a being 1 and z being 26, or a being 0 and z being 25. It gives me a SyntaxError. I need to assign this because the random integer (1,26) triggers a letter (if the random integer is 1, select A) and prints the name.
EDIT:
I have implemented your advice, and it works, I am grateful for this, but I wish to have my program create readable names, or more procedural. Here is an example of a name after I tweaked my program: ddjau. Now that doesn't look like a name, so I want it my program to work as if it were creating REAL names, like Samuel or other common names. Thanks!
EDIT (2):
Thanks, Adam, but I need a sort of 'seed' for the user to enter for the start of the name is. (Seed = A, Name = Adam. Seed = G, Name = George.) Should I do this by searching the file line by line, at the very beginning? If so, how do I do this?
Short Answer
Look into Python dictionaries to allow the 1 = 'a' type assignments. Below I have working example that would generate a random name based on gender and a 'litter'.
Disclaimer
I do not fully understand (via the code) what you're trying to accomplish with char/ord and a random letter. Also note having absolutely no idea of your design goals or requirements, I have made the example more complex than it may need to be for instructional purposes.
Additional Resources
* Python Docs for dictionary
* Using Python dictionary relationship to search both ways
In response to the last edit
If you are looking to build random 'real' names, I think your best bet will be to use a large list of names and just pick a random one. If I were you I'd look into something linking to the census results: males and females. Note that male_names.txt and female_names.txt are a copy of the list found at the census website. As a disclaimer, I'm sure there is a more efficient way to load / read the file. Just use this example as a proof on concept.
Update
Here's a quick and dirty way to seed the random values. Again I am not sure that this is the most pythonic way or most efficient way, but it works.
Example
import random
import time
def get_random_name(gender, seed):
if(gender == 'male'):
file = 'male_names.txt'
elif(gender == 'female'):
file = 'female_names.txt'
fid = open(file,'r')
names = []
total_names = 0
for line in fid:
if(line.lower().startswith(seed)):
names.append(line)
total_names = total_names + 1
random_index = random.randint(0,total_names)
return names[random_index]
if (__name__ == "__main__"):
print 'Welcome to Name Database 2.2\n'
print '1. Boy'
print '2. Girl'
bog = raw_input('\nGender: ')
print 'What should the name start with?'
print 'A, Ab, Abc, B, Ba, Br, etc...'
print ''
l = raw_input('Leter(s): ').lower()
new_name = ''
if bog == '1': # Boy
print get_random_name('male',l)
elif bog == '2':
print get_random_name('female',l)
Output
Welcome to Name Database 2.2
1. Boy
2. Girl
Gender: 2
What should the name start with?
A, Ab, Abc, B, Ba, Br, etc...
Leter(s): br
BRITTA
chr (see here) and ord (see here) are the two functions you're looking for (though you already seem to know about the latter). Follow those links for a more detailed explanation.
The first gives you a one-character string based on the integer, the second does the reverse operaion (technically, it handles Unicode as well, which chr doesn't, though you have unichr for that if you need it).
You can base your code on the following:
ch = "E"
print ord (ch) - ord ("A") + 1 # should give 5 for the fifth letter
val = 7
print chr (val + ord ("A") - 1) # should give G, the seventh letter
I'm not entirely sure what you're trying to do, but you can convert a number into a letter with the chr() function. chr() takes an ASCII code, so if you want to use the range [0, 25] instead you can adapt it like so:
chr(25 + ord('a')) # 'z'