Related
from tokenizers import Tokenizer, models, normalizers, pre_tokenizers, decoders, trainers
tokenizer = Tokenizer(models.Unigram())
tokenizer.normalizer = normalizers.NFKC()
tokenizer.pre_tokenizer = pre_tokenizers.ByteLevel()
tokenizer.decoder = decoders.ByteLevel()
trainer = trainers.UnigramTrainer(
vocab_size=30000,
initial_alphabet=pre_tokenizers.ByteLevel.alphabet(),
special_tokens=["<PAD>", "<BOS>", "<EOS>", '<s>', '</s>', '<unk>', '<mask>'],
min_frequency = 2
)
def batch_iterator(batch_size=10, size=5000):
for i in range(100):
query = f"select note_text from db.note where id > {i * size} limit 50;"
df = pd.read_sql(sql=query, con=db)
for x in range(0, size, batch_size):
yield list(df['note_text'].loc[0:5000])[x:x + batch_size]
tokenizer.train_from_iterator(batch_iterator(), trainer=trainer, length=100*5000)
A single note may look something like this:
!~!~!~!~!~!~!~!~!~!~!~!~!~!~Discussing settlement with Amy.!~!~
The output looks as follows:
out = tokenizer.encode('There should be an inspection come Monday 1/2/2022!')
['ĠThe', 'r', 'e', 'Ġsh', 'ould', 'Ġbe', 'Ġan', 'Ġinspect', 'ion', 'Ġ', 'com', 'e', 'Ġ', 'M', 'ond', 'a', 'y', 'Ġ', '1', '/', '2', '/', '20', '2', '2', '!']
I'm looking for simple password-based obfuscation/security of strings.
I've pretty much gone over each example of > Simple way to encode a string according to a password?
And none of them work with my python 3.7.
I got the error with ord() so I updated the code, but even after, its still broken. For examle:
from itertools import cycle
def encode_zip_cycle(key, clear):
enc = [chr((ord(clear_char) + ord(key_char)) % 256)
for clear_char, key_char in zip(clear, cycle(key))]
return base64.urlsafe_b64encode("".join(enc).encode())
def decode_zip_cycle(key, enc):
enc = base64.urlsafe_b64decode(enc)
dec = [chr((256 + enc_char - ord(key_char)) % 256)
for enc_char, key_char in zip(enc, cycle(key))]
print(dec)
return "".join(dec)
text = "ATTACKATONCEfor Live 2154125-21-512^!££613-123!"
s = "1235348udgfjff"
print("Text : " + text)
print("Shift : " + str(s))
print("Cipher: ", encode_zip_cycle(s, text)) # , type(encode(s, text)))
print("Original text: ", decode_zip_cycle(s, encode_zip_cycle(s, text)))
Gives me
Text : ATTACKATONCEfor Live 2154125-21-512^!££613-123!
Shift : 1235348udgfjff
Cipher: b'csKGwod2dn95w4nCs8K1wqnCr8OMw5XCo1J_wp7CqcKZWMKVwoTCmcKXwp_CmsKXY2dgZ2RhbcKmwpbDhcKHDQnCnGJlYGZlZ1k='
['A', '\x90', 'S', '\x8d', 'T', 'B', '>', '\n', '\x15', '\\', '#', 'X', 'M', '\\', '\x84', '\x90', 'v', '\x8d', '|', '\x8f', 'T', 'N', '1', '[', '=', 'è', '\x19', '\\', 'm', '\x90', 'v', '\x8d', 'f', '$', '\x8a', ' ', '^', '\x1d', '\\', '/', '\\', '1', '\x91', 'm', '\x8f', 'e', '\x8f', 'c', '+', 'ò', 'ü', '\x00', 'þ', '÷', '\x07', '\\', 'u', '\x90', 'c', '\x8e', 'R', '\x8e', 'O', '\x98', '¥', '[', '6', 'ø', 'ÿ', 'ú', '5', '3', '4', '$']
Original text: ASTB>
\#XM\v|TN1[=è\mvf$ ^\/\1mec+òü þ÷\ucRO¥[6øÿú534$
In encode_zip_cycle you encode the "encrypted" string into utf-8 before doing the second encoding into base64. Yet, you don't revert this operation later in decode_zip_cycle.
This is the correct decode_zip_cycle function:
def decode_zip_cycle(key, enc):
enc = base64.urlsafe_b64decode(enc).decode()
dec = [chr((256 + ord(enc_char) - ord(key_char)) % 256)
for enc_char, key_char in zip(enc, cycle(key))]
print(dec)
return "".join(dec)
I want to use list comprehension, instead of itertools.product
def pr(l):
return [''.join(i) for i in [(x,y) for x in l for y in l]]
operators = ['/','*','+','-']
pr(operators)
['//', '/*', '/+', '/-', '*/', '**', '*+', '*-', '+/', '+*', '++', '+-', '-/', '-*', '-+', '--']
This works, but I want to modify my function such that it returns combinations in pairs of repeat:
def pr(l, repeat=1):
# list comprehension code here
o = ['/','*','+','-']
pr(operators, repeat=4)
pr(operators, repeat=3)
['////', '///*', '///+', '///-', '//*/', '//**', '//*+', '//*-', '//+/', '//+*', '//++', '//+-', '//-/', '//-*', '//-+', '//--', '/*//', '/*/*', '/*/+', '/*/-', '/**/', '/***', '/**+', '/**-', '/*+/', '/*+*', '/*++', '/*+-', '/*-/', '/*-*', '/*-+', '/*--', '/+//', '/+/*', '/+/+', '/+/-', '/+*/', '/+**', '/+*+', '/+*-', '/++/', '/++*', '/+++', '/++-', '/+-/', '/+-*', '/+-+', '/+--', '/-//', '/-/*', '/-/+', '/-/-', '/-*/', '/-**', '/-*+', '/-*-', '/-+/', '/-+*', '/-++', '/-+-', '/--/', '/--*', '/--+', '/---', '*///', '*//*', '*//+', '*//-', '*/*/', '*/**', '*/*+', '*/*-', '*/+/', '*/+*', '*/++', '*/+-', '*/-/', '*/-*', '*/-+', '*/--', '**//', '**/*', '**/+', '**/-', '***/', '****', '***+', '***-', '**+/', '**+*', '**++', '**+-', '**-/', '**-*', '**-+', '**--', '*+//', '*+/*', '*+/+', '*+/-', '*+*/', '*+**', '*+*+', '*+*-', '*++/', '*++*', '*+++', '*++-', '*+-/', '*+-*', '*+-+', '*+--', '*-//', '*-/*', '*-/+', '*-/-', '*-*/', '*-**', '*-*+', '*-*-', '*-+/', '*-+*', '*-++', '*-+-', '*--/', '*--*', '*--+', '*---', '+///', '+//*', '+//+', '+//-', '+/*/', '+/**', '+/*+', '+/*-', '+/+/', '+/+*', '+/++', '+/+-', '+/-/', '+/-*', '+/-+', '+/--', '+*//', '+*/*', '+*/+', '+*/-', '+**/', '+***', '+**+', '+**-', '+*+/', '+*+*', '+*++', '+*+-', '+*-/', '+*-*', '+*-+', '+*--', '++//', '++/*', '++/+', '++/-', '++*/', '++**', '++*+', '++*-', '+++/', '+++*', '++++', '+++-', '++-/', '++-*', '++-+', '++--', '+-//', '+-/*', '+-/+', '+-/-', '+-*/', '+-**', '+-*+', '+-*-', '+-+/', '+-+*', '+-++', '+-+-', '+--/', '+--*', '+--+', '+---', '-///', '-//*', '-//+', '-//-', '-/*/', '-/**', '-/*+', '-/*-', '-/+/', '-/+*', '-/++', '-/+-', '-/-/', '-/-*', '-/-+', '-/--', '-*//', '-*/*', '-*/+', '-*/-', '-**/', '-***', '-**+', '-**-', '-*+/', '-*+*', '-*++', '-*+-', '-*-/', '-*-*', '-*-+', '-*--', '-+//', '-+/*', '-+/+', '-+/-', '-+*/', '-+**', '-+*+', '-+*-', '-++/', '-++*', '-+++', '-++-', '-+-/', '-+-*', '-+-+', '-+--', '--//', '--/*', '--/+', '--/-', '--*/', '--**', '--*+', '--*-', '--+/', '--+*', '--++', '--+-', '---/', '---*', '---+', '----']
['///', '//*', '//+', '//-', '/*/', '/**', '/*+', '/*-', '/+/', '/+*', '/++', '/+-', '/-/', '/-*', '/-+', '/--', '*//', '*/*', '*/+', '*/-', '**/', '***', '**+', '**-', '*+/', '*+*', '*++', '*+-', '*-/', '*-*', '*-+', '*--', '+//', '+/*', '+/+', '+/-', '+*/', '+**', '+*+', '+*-', '++/', '++*', '+++', '++-', '+-/', '+-*', '+-+', '+--', '-//', '-/*', '-/+', '-/-', '-*/', '-**', '-*+', '-*-', '-+/', '-+*', '-++', '-+-', '--/', '--*', '--+', '---']
How can I do that?
Here's a generic solution for any value of 'repeat'.
def pr(l, repeat=1):
if repeat == 1:
return [[x] for x in l]
sub_prod = pr(l, repeat-1)
return [ [x] + y for x in l for y in sub_prod ]
o = ['/','*','+','-']
pr(o, 3)
Result:
[['/', '/', '/'],
['/', '/', '*'],
['/', '/', '+'],
['/', '/', '-'],
['/', '*', '/'],
['/', '*', '*'],
['/', '*', '+'],
['/', '*', '-'],
['/', '+', '/'],
['/', '+', '*'],
['/', '+', '+'],
...
If you want to turn each sublist to a string, use:
["".join(x) for x in pr(o, 3)]
Just wrap your command in a function
def f(l):
return [(x, y) for x in l for y in l]
or if you want the same generator behavior as itertools has:
def f(l):
return ((x, y) for x in l for y in l)
or a more verbose generator syntax:
def f(l):
for x in l:
for y in l:
yield x, y
So, after a lot of attempts, search and research I give up.
I have a webpage where all employees name, phone, mail and userid can be query. The way you do that is that the request to the server needs to have at least 4 digits, with all 26 ascll character + 0-9 numbers. I was able to do it with Selenium in Python...but it whould take 20 days to go through - see code.
from selenium import webdriver
import csv
alphanum = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm',
'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'x', 'y', 'z', '1',
'2', '3', '4', '5', '6', '7', '8', '9', '0']
driver = webdriver.Firefox()
driver.get('http://brnet.intra.corpintra.net/Quem/pessoas2/Default.asp')
list_base = dict()
i = 0
data_str = []
found = False
for first_chr in alphanum:
for second_chr in alphanum:
for third_chr in alphanum:
text = first_chr + second_chr + third_chr
element_name = driver.find_element_by_name('nome').clear()
element_name = driver.find_element_by_name('nome')
element_name.send_keys(text)
element_search = driver.find_element_by_name('B1')
element_search.click()
if driver.find_elements_by_class_name('dados'):
for table_data in driver.find_elements_by_class_name('dados'):
cells_table = table_data.find_elements_by_tag_name('td')
for cell_data in cells_table:
data_str.append(cell_data.text.strip())
if list_base:
for key, value in list_base.items():
for data in data_str:
if data in value:
found = False
else:
found = True
else:
found = False
if found is False:
list_base[i] = data_str
i = i+1
data_str = []
found = False
driver.back()
w = csv.writer(open("output.csv", "w"))
for key, value in list_base.items():
w.writerow([key, value])
driver.quit()
Is there a way to reduce the time?
I made a program which convert infix to postfix in python. The problem is when I introduce the arguments.
If i introduce something like this: (this will be a string)
( ( 73 + ( ( 34 - 72 ) / ( 33 - 3 ) ) ) + ( 56 + ( 95 - 28 ) ) )
it will split it with .split() and the program will work correctly.
But I want the user to be able to introduce something like this:
((73 + ( (34- 72 ) / ( 33 -3) )) + (56 +(95 - 28) ) )
As you can see I want that the blank spaces can be trivial but the program continue splitting the string by parentheses, integers (not digits) and operands.
I try to solve it with a for but I don't know how to catch the whole number (73 , 34 ,72) instead one digit by digit (7, 3 , 3 , 4 , 7 , 2)
To sum up, what I want is split a string like ((81 * 6) /42+ (3-1)) into:
[(, (, 81, *, 6, ), /, 42, +, (, 3, -, 1, ), )]
Tree with ast
You could use ast to get a tree of the expression :
import ast
source = '((81 * 6) /42+ (3-1))'
node = ast.parse(source)
def show_children(node, level=0):
if isinstance(node, ast.Num):
print(' ' * level + str(node.n))
else:
print(' ' * level + str(node))
for child in ast.iter_child_nodes(node):
show_children(child, level+1)
show_children(node)
It outputs :
<_ast.Module object at 0x7f56abbc5490>
<_ast.Expr object at 0x7f56abbc5350>
<_ast.BinOp object at 0x7f56abbc5450>
<_ast.BinOp object at 0x7f56abbc5390>
<_ast.BinOp object at 0x7f56abb57cd0>
81
<_ast.Mult object at 0x7f56abbd0dd0>
6
<_ast.Div object at 0x7f56abbd0e50>
42
<_ast.Add object at 0x7f56abbd0cd0>
<_ast.BinOp object at 0x7f56abb57dd0>
3
<_ast.Sub object at 0x7f56abbd0d50>
1
As #user2357112 wrote in the comments : ast.parse interprets Python syntax, not mathematical expressions. (1+2)(3+4) would be parsed as a function call and list comprehensions would be accepted even though they probably shouldn't be considered a valid mathematical expression.
List with a regex
If you want a flat structure, a regex could work :
import re
number_or_symbol = re.compile('(\d+|[^ 0-9])')
print(re.findall(number_or_symbol, source))
# ['(', '(', '81', '*', '6', ')', '/', '42', '+', '(', '3', '-', '1', ')', ')']
It looks for either :
multiple digits
or any character which isn't a digit or a space
Once you have a list of elements, you could check if the syntax is correct, for example with a stack to check if parentheses are matching, or if every element is a known one.
You need to implement a very simple tokenizer for your input. You have the following types of tokens:
(
)
+
-
*
/
\d+
You can find them in your input string separated by all sorts of white space.
So a first step is to process the string from start to finish, and extract these tokens, and then do your parsing on the tokens, rather than on the string itself.
A nifty way to do this is to use the following regular expression: '\s*([()+*/-]|\d+)'. You can then:
import re
the_input='(3+(2*5))'
tokens = []
tokenizer = re.compile(r'\s*([()+*/-]|\d+)')
current_pos = 0
while current_pos < len(the_input):
match = tokenizer.match(the_input, current_pos)
if match is None:
raise Error('Syntax error')
tokens.append(match.group(1))
current_pos = match.end()
print(tokens)
This will print ['(', '3', '+', '(', '2', '*', '5', ')', ')']
You could also use re.findall or re.finditer, but then you'd be skipping non-matches, which are syntax errors in this case.
If you don't want to use re module, you can try this:
s="((81 * 6) /42+ (3-1))"
r=[""]
for i in s.replace(" ",""):
if i.isdigit() and r[-1].isdigit():
r[-1]=r[-1]+i
else:
r.append(i)
print(r[1:])
Output:
['(', '(', '81', '*', '6', ')', '/', '42', '+', '(', '3', '-', '1', ')', ')']
It actual would be pretty trivial to hand-roll a simple expression tokenizer. And I'd think you'd learn more that way as well.
So for the sake of education and learning, Here is a trivial expression tokenizer implementation which can be extended. It works based upon the "maximal-much" rule. This means it acts "greedy", trying to consume as many characters as it can to construct each token.
Without further ado, here is the tokenizer:
class ExpressionTokenizer:
def __init__(self, expression, operators):
self.buffer = expression
self.pos = 0
self.operators = operators
def _next_token(self):
atom = self._get_atom()
while atom and atom.isspace():
self._skip_whitespace()
atom = self._get_atom()
if atom is None:
return None
elif atom.isdigit():
return self._tokenize_number()
elif atom in self.operators:
return self._tokenize_operator()
else:
raise SyntaxError()
def _skip_whitespace(self):
while self._get_atom():
if self._get_atom().isspace():
self.pos += 1
else:
break
def _tokenize_number(self):
endpos = self.pos + 1
while self._get_atom(endpos) and self._get_atom(endpos).isdigit():
endpos += 1
number = self.buffer[self.pos:endpos]
self.pos = endpos
return number
def _tokenize_operator(self):
operator = self.buffer[self.pos]
self.pos += 1
return operator
def _get_atom(self, pos=None):
pos = pos or self.pos
try:
return self.buffer[pos]
except IndexError:
return None
def tokenize(self):
while True:
token = self._next_token()
if token is None:
break
else:
yield token
Here is a demo the usage:
tokenizer = ExpressionTokenizer('((81 * 6) /42+ (3-1))', {'+', '-', '*', '/', '(', ')'})
for token in tokenizer.tokenize():
print(token)
Which produces the output:
(
(
81
*
6
)
/
42
+
(
3
-
1
)
)
Quick regex answer:
re.findall(r"\d+|[()+\-*\/]", str_in)
Demonstration:
>>> import re
>>> str_in = "((81 * 6) /42+ (3-1))"
>>> re.findall(r"\d+|[()+\-*\/]", str_in)
['(', '(', '81', '*', '6', ')', '/', '42', '+', '(', '3', '-', '1',
')', ')']
For the nested parentheses part, you can use a stack to keep track of the level.
This does not provide quite the result you want but might be of interest to others who view this question. It makes use of the pyparsing library.
# Stolen from http://pyparsing.wikispaces.com/file/view/simpleArith.py/30268305/simpleArith.py
# Copyright 2006, by Paul McGuire
# ... and slightly altered
from pyparsing import *
integer = Word(nums).setParseAction(lambda t:int(t[0]))
variable = Word(alphas,exact=1)
operand = integer | variable
expop = Literal('^')
signop = oneOf('+ -')
multop = oneOf('* /')
plusop = oneOf('+ -')
factop = Literal('!')
expr = operatorPrecedence( operand,
[("!", 1, opAssoc.LEFT),
("^", 2, opAssoc.RIGHT),
(signop, 1, opAssoc.RIGHT),
(multop, 2, opAssoc.LEFT),
(plusop, 2, opAssoc.LEFT),]
)
print (expr.parseString('((81 * 6) /42+ (3-1))'))
Output:
[[[[81, '*', 6], '/', 42], '+', [3, '-', 1]]]
Using grako:
start = expr $;
expr = calc | value;
calc = value operator value;
value = integer | "(" #:expr ")" ;
operator = "+" | "-" | "*" | "/";
integer = /\d+/;
grako transpiles to python.
For this example, the return value looks like this:
['73', '+', ['34', '-', '72', '/', ['33', '-', '3']], '+', ['56', '+', ['95', '-', '28']]]
Normally you'd use the generated semantics class as a template for further processing.
To provide a more verbose regex approach that you could easily extend:
import re
solution = []
pattern = re.compile('([\d\.]+)')
s = '((73 + ( (34- 72 ) / ( 33 -3) )) + (56 +(95 - 28) ) )'
for token in re.split(pattern, s):
token = token.strip()
if re.match(pattern, token):
solution.append(float(token))
continue
for character in re.sub(' ', '', token):
solution.append(character)
Which will give you the result:
solution = ['(', '(', 73, '+', '(', '(', 34, '-', 72, ')', '/', '(', 33, '-', 3, ')', ')', ')', '+', '(', 56, '+', '(', 95, '-', 28, ')', ')', ')']
Similar to #McGrady's answer, you can do this with a basic queue implementation.
As a very basic implementation, here's what your Queue class can look like:
class Queue:
EMPTY_QUEUE_ERR_MSG = "Cannot do this operation on an empty queue."
def __init__(self):
self._items = []
def __len__(self) -> int:
return len(self._items)
def is_empty(self) -> bool:
return len(self) == 0
def enqueue(self, item):
self._items.append(item)
def dequeue(self):
try:
return self._items.pop(0)
except IndexError:
raise RuntimeError(Queue.EMPTY_QUEUE_ERR_MSG)
def peek(self):
try:
return self._items[0]
except IndexError:
raise RuntimeError(Queue.EMPTY_QUEUE_ERR_MSG)
Using this simple class, you can implement your parse function as:
def tokenize_with_queue(exp: str) -> List:
queue = Queue()
cum_digit = ""
for c in exp.replace(" ", ""):
if c in ["(", ")", "+", "-", "/", "*"]:
if cum_digit != "":
queue.enqueue(cum_digit)
cum_digit = ""
queue.enqueue(c)
elif c.isdigit():
cum_digit += c
else:
raise ValueError
if cum_digit != "": #one last sweep in case there are any digits waiting
queue.enqueue(cum_digit)
return [queue.dequeue() for i in range(len(queue))]
Testing it like below:
exp = "((73 + ( (34- 72 ) / ( 33 -3) )) + (56 +(95 - 28) ) )"
print(tokenize_with_queue(exp)")
would give you the token list as:
['(', '(', '73', '+', '(', '(', '34', '-', '72', ')', '/', '(', '33', '-', '3', ')', ')', ')', '+', '(', '56', '+', '(', '95', '-', '28', ')', ')', ')']