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ValueError when splitting lines in a text file [closed]

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I'd like to split each line of a text file into two by " - ", but I keep getting this error:
File "quiz.py", line 21, in Vocab
questions, answers = line.split("-")
ValueError: too many values to unpack (expected 2)
I'm quite new to coding and could use some help. All tips are welcome as well!
import hashlib
testFile = ""
def qSearch():
options = input ("Vocab/Grammar/or Special? (v/g/s)")
if options == "v":
testFile = "Vocabtest"
Vocab()
elif options == "g":
Grammar()
testFile = "Grammartest"
elif options == "s":
Special()
testFile = "Specialtest"
else:
qSearch()
def Vocab():
with open('Vocabtest.txt','r') as f:
for line in f:
questions, answers = line.split("-") ### error
print (questions)
qSearch()
The text in my text file is formatted like so:
Magandang umaga - Good Morning
Magandang hapon - Good Afternoon
Magandang gabi - Good evening
Magandang umaga sa’yo - Good Morning to you
Magandang hapon din sa’yo - Good Afternoon to you to

"Unpacking" is the name for what you're doing when you write
value1, value2 = a_list
When you do an assignment like that, you're implicitly making an assumption about how many values are contained in a_list -- here it's 2. If there's more or less than 2, there's no good way to give value1 and value2 values without doing very surprising and unhelpful things (like leaving one empty, or leaving some elements of the list unassigned).
So too many values to unpack means that there's at least one line in your file where line.split('-') results in more than 2 elements -- that is, there's at least one line with more than one -.

The problem is because on line 21 in your input text (.txt) file you have more than one - but you only expect one.
A safer way to do it would be to only split once:
questions, answers = line.split("-", 1)

Python - SyntaxError: can't assign to literal [closed]

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I want to read from a file the variables (they are links) and then open them with urlopen in a while loop so that every link is opened.
My code is:
# Variables
from config import *
# Imports
import urllib
i = 0
url = 100
while i < 25:
page = urllib.urlopen( url );
page.close();
i = i + 1
url = 100
url = url + i
The error i get is SyntaxError: can't assign to literal. I kind of understand why, but i don't know how to bypass it!
config.py
100 = 'https:link'
101 = 'https:link'
102 = 'https:link'

The error is telling you exactly what the error is. You can't assign to a literal. 100 is a literal because it literally has the value 100. Your config.py is trying to change the value of the integer 100.
If you're trying to iterate over a list of variables or values starting with 100, one solution would be to create a dictionary and use the numbers for the keys. For example:
# config.py
urls = {
100: "https:link",
101: "https:link",
102: "https:link"
}
Then, in your code you can do something like this:
i = 0
while i < 25:
url_number = 100 + i
page = urllib.urlopen( urls[url_number] );

python code to generate password list [closed]

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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()

make a global condition break

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)

How to rewrite the code more elegantly [closed]

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The code below reads lines from a file, then it executes the custom function (My_Function) and return values to the variables(e.g. condition_A)
for line in input_file:
if condition_A:
condition_A = My_Function(A_tuple[0], B_tuple[0])
if condition_B:
condition_B = My_Function(A_tuple[1], B_tuple[1])
if condition_C:
condition_C = My_Function(A_tuple[2], B_tuple[2])
if condition_D:
condition_D = My_Function(A_tuple[3], B_tuple[3])
if condition_E:
condition_E = My_Function(A_tuple[4], B_tuple[4])
...
My question is: can the code be modified to more elegant version? After all, many code is similar(I don't want to define another function to simplify it because the code is still similar after the new function is defined). thanks.

Instead of having 5 variables condition_*, use a list, conditions:
conditions=[1]*5 # initialize conditions as you wish
for line in input_file:
for i,condition in enumerate(conditions):
if condition:
conditions[i]=My_Function(A_tuple[i],B_tuple[i])

What about something like
conditions = [condition_A, condition_B, condition_C, condition_D, condition_E]
condition_test = lambda c, i: My_Function(A_tuple[i], B_tuple[i]) if c else c
for line in input_file:
conditions = [condition_test(c, i) for i, c in enumerate(conditions)]

'line' is not referenced in teh loop, is that an error in simplifying it for posting?
How about
condition=1 #or 2 or...
for line in input_file:
My_Function(A_tuple[condition],B_tuple[condition])

Before refactoring your code on a purely syntactic level (which is covered in examples above), it might be useful to evaluate what you're doing with the code on a functional level
Check out your condition_x variables. I think you might be using the same variable for two different things (both type-wise and logically) - usually a bad idea in a weakly typed language. It looks to me as if the user sets a condition to true or false, and then that condition is assigned the output - is the output boolean? is it related to the original value of that variable? Rethinking this might lead to more understandable code.
It is also difficult to evaluate how this can be refactored without seeing what goes in to condition_x - since these might have commonalities.

One more sample(not solution) based on unutbu's:
data = [1,2,3,'',4,5,6, '', 0]
for i in (i for i in xrange(len(data)) if data[i] not in ['',0]):
data[i] += 1
Sorry if duplicate

Here is a generic solution where you can have custom index and you can also access conditions by name if need be and it can be easily extended to add any new complexities
class Condition(object):
def __init__(self, active, index1, index2):
self.active = active
self.index1 = index1
self.index2 = index2
conditions = {
'A': Condition(True,0,0),
'B': Condition(True,1,1),
'C': Condition(True,2,2),
'D': Condition(True,3,3),
'E': Condition(True,4,4),
}
for line in input_file:
for condition in conditions.itervalues():
if condition.active:
condition.active = My_Function(A_tuple[condition.active.index1], B_tuple[condition.active.index2])