Reading a (compressed) file - python

My code :
sent = str(input("Please input a sentence: "))
dl = [0]
for count , v in enumerate (splitsent):
if splitsent.count(v) < 2:
dl.append(max(dl) +1)
else:
dl.append(splitsent.index(v) +1)
dl.remove(0)
print(sent, "\n",dl)
gives the output :
"1,2,3,4,1,2"
with the input:
"To be or not to be"
This is it in it's "compressed" form. How would I take the output,"1,2,3,4,1,2" from an external file and turn it into the "To be or not to be"?

Your method really not an efficient way of compressing a text file, just use the existing zlib.
But, for the academic exercise, you will want to use pickle to store your dictionary keys such that when you recover it you get the same values. As you want the 'compressed' form to exist between invocations, so that you can successfully decompress a previously 'compressed' file, you will need to allocate an index to each word.
If you want a 'standard' python method, OrderedDict from collections can be used to create an index in this way, new words are added to the end, but unlike conventional dict objects, old ones keep their position. A better method is an OrderedSet, but this is not in standard python, see this recipe.
Case
You also have to decide if 'THIS', 'this' and 'ThIs' are different words or the same word. Perhaps each word token needs a bitfield to indicate if each character is lower or upper case, e.g. 'ThIs' gets a token 15, but a bitfield of 5 "0x1010", producing a tuple of (15,5) in the compressed file.
Punctuation
You will also need to consider punctuation, where a word is thus punctuated you will need a way to represent this in the compressed form, a token for the punctuation character.
But there is a problem with this.
Then when you decompress you will need to recreate the original exactly, so handle punctuation. e.g. "Is this correct?" -> [1,2,3,4] -> "Is this correct ?" or "Is this correct?" without the space.
So for each punctuation you need to indicate how it joins to the previous and next character, e.g.
As punctuation is only ever one character (i.e. one 8 bit number), you may want to consider just putting the character as-is.
Multiple spaces
You will also need to handle multiple spaces.
Example code
This code is incomplete, mostly untested and probably does not handle all use cases, but it illustrates one possible solution to the question.
To use it, create a file called in.txt containing the text you want to compress, then run
python compdict.py -c in.txt out.comp
or
python compdict.py -d out.comp out.txt
or
python compdict.py --list
from ordered_set import OrderedSet #pip install ordered_set
import os
import cPickle as pickle
import string
import argparse
class CompDecomp(object):
__DEFAULT_PICKLE_FN__ = "my.dict"
printable_non_chars = set(string.printable) - set(string.digits) - set(string.ascii_letters)
def __init__(self, fn=None, *args, **kw):
if fn is None:
self.fn = self.__DEFAULT_PICKLE_FN__
else:
self.fn = fn
self.dict = self.loaddict()
def loaddict(self):
if os.path.exists(self.fn):
pkl = open(self.fn, "rb")
d = pickle.load(pkl)
pkl.close()
else:
d = OrderedSet()
return d
def savedict(self):
pkl = open(self.fn, "wb")
pickle.dump(self.dict, pkl)
pkl.close()
def compressword(self, word, conjoin=False):
if word.lower() not in self.dict:
self.dict.append(word.lower())
print "New word: \'%s\'" % word
self.savedict()
index, flag, _ = self.__caseflag__(word, conjoin)
#print index, bin(flag)[2:].zfill(len(word)), conjoin
return index, flag, conjoin
def decompressword(self, index, caseflag=0, conjoin=False):
if isinstance(index, int):
word = self.dict[index]
else:
word = index
if caseflag == 0:
return word, conjoin
flag = bin(caseflag)[2:].zfill(len(word))
res = ""
for n, c in enumerate(word):
if flag[n] == '1':
res += c.upper()
else:
res += c.lower()
return res, conjoin
def __caseflag__(self, word, conjoin):
index = self.dict.index(word.lower())
if word.lower() == word:
#Word is all lowercase
return (index,0, conjoin)
if word.upper() == word:
#Word is all uppercase
return index, int("1" * len(word), 2), conjoin
res = ""
for c in word:
if c in string.uppercase:
res += "1"
else:
res += "0"
return index, int(res, 2), conjoin
def compressfile(self, fileobj):
with fileobj as f:
data = f.read(-1)
f.close()
words = data.split(" ")
compress = []
for word in words:
#Handle multiple spaces
if word == "":
compress.append(" ")
continue
#Handle puntuation, treat apostrophied words as new words
substr = []
p1 = 0
csplit = word.translate(None, string.ascii_letters+'\'')
for n, c in enumerate(csplit):
subword, word = word.split(c, 1)
compress.append(self.compressword(subword, True if n > 0 else False))
compress.append((c, 0, True))
#Handle words
if len(word) and not len(csplit):
compress.append(self.compressword(word))
return compress
def decompressfile(self, fileobj):
data = pickle.load(fileobj)
decomp = ""
for v in data:
if not isinstance(v,tuple):
print "Bad data %s" % v
continue
if len(v) > 0 and len(v) <= 3:
d, conjoin = self.decompressword(*v)
if len(decomp):
decomp += "" if conjoin else " "
decomp += d
else:
print "Bad data %s (length %d)" % (v, len(v))
return decomp
if __name__ == "__main__":
parser = argparse.ArgumentParser(description='Test file compress / decompress')
group = parser.add_mutually_exclusive_group()
parser.add_argument('infile', nargs='?', default=None)
parser.add_argument('outfile', nargs='?', default=None)
group.add_argument('-compress', action='store_true')
group.add_argument('-decompress', action='store_true')
group.add_argument('--list', action='store_true')
args = parser.parse_args()
cd = CompDecomp()
#Invocation
#python dictcompress.py [-h|-c|-d|--list] [<infile>] [<outfile>]
infile, outfile = args.infile, args.outfile
if infile is not None and not os.path.exists(infile):
print "Input file missing"
if outfile is not None:
of = open(outfile, "wb")
else:
of = None
if not args.list:
if args.compress:
print "Compress"
pickle.dump(cd.compressfile(open(infile, "r")), of)
if args.decompress:
print "Decompress"
of.write(cd.decompressfile(open(infile, "r")))
else:
for k in cd.dict:
print k
if of is not None:
of.close()

Related

Function return 'None' Python

Hi i am learning python on my own.
Task:
Reverse word without affecting special characters
Example "abcd efgh" => "dcba hgfe"
Example "a1bcd efg!h" => "d1cba hgf!e"
My problem: the function return None
Then i added this line:
return reverse_text but it still return None
Can anyone show me where is my mistake is, please?
My code:
from string import punctuation
from string import digits
def reverse_text(str_smpl):
sp = set.union(set(punctuation), set(digits))
reverse_text.lst = []
for word in str_smpl.split(' '):
letters = [c for c in word if c not in sp]
for c in word:
if c not in sp:
reverse_text.lst.append(letters.pop())
continue
else:
reverse_text.lst.append(c)
reverse_text.lst.append(' ')
return reverse_text
if __name__ == '__main__':
cases = [
("abcd efgh", "dcba hgfe"),
("a1bcd efg!h", "d1cba hgf!e"),
("", "")
]
for text, reversed_text in cases:
assert reverse_text(str_smpl) == reversed_text
reverse_text(input('Input string '))
print("".join(reverse_text.lst))
The issue is that you are returning reverse_text which is the name of the function, so the function is returning a reference to itself (not what you want!).
Assigning properties to functions like you have with reverse_text.lst is not something I have really come across in Python and I would suggest you just use a new local variable named something like reversed_text_list to avoid confusion.
I think you also want to join the characters in the list together and return a string.
The following seems to be doing what I think you are trying to do:
def reverse_text(str_smpl):
sp = set.union(set(punctuation), set(digits))
reversed_text_list = []
for word in str_smpl.split(' '):
letters = [c for c in word if c not in sp]
for c in word:
if c not in sp:
reversed_text_list.append(letters.pop())
continue
else:
reversed_text_list.append(c)
reversed_text_list.append(' ')
reversed_text = ''.join(reversed_text_list)
return reversed_text
It returned error because you had defined reverse_text.lst but returned only reverse_text, the following code will work:-
from string import punctuation
from string import digits
def reverse_text(str_smpl):
sp = set.union(set(punctuation), set(digits))
lst = []
for word in str_smpl.split(' '):
letters = [c for c in word if c not in sp]
for c in word:
if c not in sp:
lst.append(letters.pop())
continue
else:
lst.append(c)
lst.append(' ')
return "".join(lst[:len(lst)-1])
if __name__ == '__main__':
cases = [
("abcd efgh", "dcba hgfe"),
("a1bcd efg!h", "d1cba hgf!e"),
("", "")
]
for text, reversed_text in cases:
assert reverse_text(text) == reversed_text
print(reverse_text(input('Input string ')))

Markov Analysis, Formatting

I have a program that reads in a big chunk of text from a text file and then randomizes the content to display back as a short story based on the content of the text. The program works but the last part, where I am displaying the material is super clunky and not efficient and I was wondering if anyone had any ideas on how I could more efficiently take in the text and then display it as a string to the user but allow it to span multiple lines (wrap text, essentially) so that it is not just a giant string of text continuing to the right of the console.
from __future__ import print_function, division
import sys
import random
# global variables
suffix_map = {} # map from prefixes to a list of suffixes
prefix = () # current tuple of words
big_list = []
def process_file(filename, order=2):
"""Reads a file and performs Markov analysis.
filename: string
order: integer number of words in the prefix
returns: map from prefix to list of possible suffixes.
"""
fp = open(filename)
for line in fp:
for word in line.rstrip().split():
process_word(word, order)
def process_word(word, order=3):
"""Processes each word.
word: string
order: integer
During the first few iterations, all we do is store up the words;
after that we start adding entries to the dictionary.
"""
global prefix
if len(prefix) < order:
prefix += (word,)
return
try:
suffix_map[prefix].append(word)
except KeyError:
# if there is no entry for this prefix, make one
suffix_map[prefix] = [word]
prefix = shift(prefix, word)
def random_text(n=300):
"""Generates random wordsfrom the analyzed text.
Starts with a random prefix from the dictionary.
n: number of words to generate
"""
global big_list
# choose a random prefix (not weighted by frequency)
start = random.choice(list(suffix_map.keys()))
for i in range(n):
suffixes = suffix_map.get(start, None)
if suffixes == None:
random_text(n-i)
return
# choose a random suffix
word = random.choice(suffixes)
big_list.append(word + " ")
start = shift(start, word)
def shift(t, word):
"""Forms a new tuple by removing the head and adding word to the tail.
t : tuple of strings
word: string
Returns: tuple of strings
"""
return t[1:] + (word,)
def list_to_str_format():
global big_list
whole = " ".join(str(i) for i in big_list)
# 25 words per line
l1 = big_list[:25]
l2 = big_list[26:50]
l3 = big_list[51:75]
l4 = big_list[76:100]
l5 = big_list[101:125]
l6 = big_list[126:150]
l7 = big_list[151:175]
l8 = big_list[176:200]
l9 = big_list[201:225]
l10 = big_list[226:250]
l11 = big_list[256:275]
l12 = big_list[276:300]
str_1 = " ".join(str(i) for i in l1).capitalize()
str_2 = " ".join(str(i) for i in l2)
str_3 = " ".join(str(i) for i in l3)
str_4 = " ".join(str(i) for i in l4)
str_5 = " ".join(str(i) for i in l5)
str_6 = " ".join(str(i) for i in l6)
str_7 = " ".join(str(i) for i in l7)
str_8 = " ".join(str(i) for i in l8)
str_9 = " ".join(str(i) for i in l9)
str_10 = " ".join(str(i) for i in l10)
str_11 = " ".join(str(i) for i in l11)
str_12 = " ".join(str(i) for i in l12)
print(str_1)
print(str_2)
print(str_3)
print(str_4)
print(str_5)
print(str_6)
print(str_7)
print(str_8)
print(str_9)
print(str_10)
print(str_11)
print(str_12)
def main(filename, n=300, order=3):
try:
n = int(n)
order = int(order)
except ValueError as e:
print('Usage: %d filename [# of words] [prefix length]' % e)
else:
process_file(filename, order)
random_text(n)
list_to_str_format()
print()
main('C:\\Users\\Desktop\\TheBrothersKaramazov.txt')
i allowed myself to change your joining pattern which made a double space. you must import module re
def list_to_str_format(line_length=80):
global big_list
whole = "".join(str(i) for i in big_list)
regex = re.compile('(.*?(\s))*')
while whole != "":
break_pos = regex.match(whole[:line_length]).end()
print(whole[:break_pos])
whole = whole[break_pos:]

Error in anagram code: python

This function will search for anagrams in a list from a .txt file, I want to be able to check for anagrams and return all anagrams of the word that I input, and if it's not an anagram it will return the input, when I do it in the code below, it iterates through the for loop then ignores my first if statement and heads directly to my else statement. How can I fix this?
def find_in_dict():
input_word = input("Enter input string)")
sorted_word = ''.join(sorted(input_word.strip()))
a_word = ''.join((input_word.strip()))
word_file = open("filename", "r")
word_list = {}
for text in word_file:
simple_text = ''.join(sorted(text.strip()))
word_list.update({text.strip(): simple_text})
alist = []
for key, val in word_list.items():
if val == sorted_word:
alist.append(key)
return alist
else:
return "No words can be formed from:" + a_word
you are making a return statement in the if and else branch, that will break the for (because return invoked inside a function do exactly that, interrupt the execution and return the value) , so, don't do that, just ask if the word is equal, and in the end, check if there is none occurrences (empty list)
for text in word_file:
simple_text = ''.join(sorted(text.strip()))
word_list.update({text.strip(): simple_text})
alist = []
for key, val in word_list.items():
if val == sorted_word:
alist.append(key)
if alist == []: print("No words can be formed from: " + a_word)

Python: Create multiple dictionaries of letter transitions

So me and my groupmates are trying to make a Markov Model that finds the probability of letter transitions in a text file. In the text file we have a group of words "Steam, Teams, Meets, Teems, Eat, Ate, State, Tease, Test, Mast, Mates". In the code we have spaces added to the beginning of the first letter and after the last letter in each word. So the problem we are having is making a function that puts the letter transitions into separate dictionaries. For example all the e transitions(ex: "_e", "ea"...etc, the _ is a space) would go into a dictionary and then the t, s, a, and m.
This is the code we have so far:
import random
import re
inFile = open("markov.txt",'r')
file = inFile.read().lower()
inFile.close()
file=re.sub('[^[a-z\ \']+', " ", file)
fileTuple=tuple(file.split())
fileList=list(fileTuple)
fileString=file
def addSpaces(atuple):
theString=''
for i in atuple:
theString=theString+' '+i+' '
return(theString)
print('The words in the text file:',addSpaces(fileTuple))
fileDict = { }
for i in fileList:
fileDict['_'+i+'_']=''
print("This is a dictionary of the words in the text file with underscores as spaces:",fileDict)
def countTotalWords(atuple):
count=0
for i in atuple:
count=count+1
return(count)
print('Total amount of words:',countTotalWords(fileTuple))
def findFirstLetter(aDict):
for i in aDict:
aDict[i]=i[0:2]
return(aDict)
print('The first letters of each word in the file:',findFirstLetter(fileDict))
valueList=list(fileDict.values())
keyList=list(fileDict.keys())
def countFirstLetters(alist):
d={}
count = 0
for character in alist:
if character in d:
d[character] += 1
else:
d[character] = 1
return d
print('Total amount of occurences of each first letter:',countFirstLetters(valueList))
def countFirstLettersProbability(alist):
d={}
count = 0
for character in alist:
if character in d:
d[character] += (1/countTotalWords(fileTuple))
else:
d[character] = (1/countTotalWords(fileTuple))
return d
print('Probility that each letter is the first in the word:',countFirstLettersProbability(valueList))
def countAllLetters(alist):
d={}
for word in alist:
for char in word:
if char in d:
d[char] += 1
else:
d[char] = 1
return d
print('Total amount of occurences of each letter:',countFirstLetters(fileString))
Here is a solid start; I've rewritten your code as a Markov class.
from random import choice
import re
from collections import defaultdict
from itertools import chain, tee, izip
def strip_non_alpha(text, reg=re.compile('[^a-z\']+', re.IGNORECASE)):
return reg.sub(' ', text.strip())
def nwise(iterable, n):
"s -> (s0,s1, ... sn-1), (s1,s2, ... sn), (s2, s3, ... sn+1), ..."
args = tee(iterable, n)
for i,t in enumerate(args):
for j in range(i):
next(t, None)
return izip(*args)
class Markov():
CHAINLEN = 3
PRE = ' '*(CHAINLEN - 1)
#classmethod
def from_file(cls, fname):
with open(fname) as inf:
return Markov(inf)
def __init__(self, text):
"""
Create a new Markov chain model
text
Either a string or a sequence of strings
"""
self.lookup = defaultdict(list)
self.words = 0
self.strings = 0
if hasattr(text, '__iter__'):
for s in text:
self.add_text(s)
else:
self.add_text(text)
def add_text(self, text):
"""
Add a string to the lookup table
text
string to add
"""
text = strip_non_alpha(text).lower()
self.words += len(text.split())
self.strings += 1
for chars in nwise(chain(Markov.PRE, text, Markov.PRE), Markov.CHAINLEN):
stem = ''.join(chars[:-1])
self.lookup[stem].append(chars[-1])
def gen_text(self, upto=200):
"""
Generate a string
upto
maximum length of string to be generated
"""
s = Markov.PRE
res = []
for i in range(upto + Markov.CHAINLEN):
ch = choice(self.lookup[s])
res.append(ch)
s = s[1:] + ch
if s == Markov.PRE: # terminal string
break
return ''.join(res[:-(Markov.CHAINLEN - 1)])
def __str__(self):
return '\n'.join("'{}': {}".format(k, self.lookup[k]) for k in sorted(self.lookup))
def main():
# mc = Markov.from_file('markov.txt')
mc = Markov('Steam,Teams,Meets,Teems,Eat,Ate,State,Tease,Test,Mast,Mates'.split(','))
print mc.strings, mc.words
print mc
for i in range(10):
print(mc.gen_text())
if __name__=="__main__":
main()

How to ensure all string literals are unicode in python

I have a fairly large python code base to go through. It's got an issue where some string literals are strings and others are unicode. And this causes bugs. I am trying to convert everything to unicode. I was wondering if there is a tool that can convert all literals to unicode. I.e. if it found something like this:
print "result code %d" % result['code']
to:
print u"result code %d" % result[u'code']
If it helps I use PyCharm (in case there is an extension that does this), however I am would be happy to use a command like too as well. Hopefully such a tool exists.
You can use tokenize.generate_tokens break the string representation of Python code into tokens. tokenize also classifies the tokens for you. Thus you can identify string literals in Python code.
It is then not hard to manipulate the tokens, adding 'u' where desired:
import tokenize
import token
import io
import collections
class Token(collections.namedtuple('Token', 'num val start end line')):
#property
def name(self):
return token.tok_name[self.num]
def change_str_to_unicode(text):
result = text.splitlines()
# Insert a dummy line into result so indexing result
# matches tokenize's 1-based indexing
result.insert(0, '')
changes = []
for tok in tokenize.generate_tokens(io.BytesIO(text).readline):
tok = Token(*tok)
if tok.name == 'STRING' and not tok.val.startswith('u'):
changes.append(tok.start)
for linenum, s in reversed(changes):
line = result[linenum]
result[linenum] = line[:s] + 'u' + line[s:]
return '\n'.join(result[1:])
text = '''print "result code %d" % result['code']
# doesn't touch 'strings' in comments
'handles multilines' + \
'okay'
u'Unicode is not touched'
'''
print(change_str_to_unicode(text))
yields
print u"result code %d" % result[u'code']
# doesn't touch 'strings' in comments
u'handles multilines' + u'okay'
u'Unicode is not touched'
Try this (uses regex), and it's shorter than #unutbu's solution.
But there's s loop hole, the strings containing # won't work with this.
import re
scode = '''
print "'Hello World'" # prints 'Hello World'
u'Unicode is unchanged'"""
# so are "comments"'''
x1 = re.compile('''(?P<unicode>u?)(?P<c>'|")(?P<data>.*?)(?P=c)''')
def repl(m):
return "u%(c)s%(data)s%(c)s" % m.groupdict()
fcode = '\n'.join(
[re.sub(x1,repl,i)
if not '#' in i
else re.sub(x1,repl,i[:i.find('#')])+i[i.find('#'):]
for i in scode.splitlines()])
print fcode
Outputs:
print u"'Hello World'" # prints 'Hello World'
u'Unicode is unchanged'
# so are "comments"
For # I have this (and it's longer than #unutbu's solution :| )
import re
scode = '''print "'Hello World'" # prints 'Hello World'
u'Unicode is unchanged'
# so are "comments"
'#### Hi' # 'Hi' '''
x1 = re.compile('''(?P<unicode>u?)(?P<c>'|")(?P<data>.*?)(?P=c)''')
def in_string(text,index):
curr,in_l,in_str,level = '',0,False,[]
for c in text[:index+1]:
if c == '"' or c == "'":
if in_str and curr == c:
instr = False
curr = ''
in_l -= 1
else:
instr = True
curr = c
in_l += 1
level.append(in_l)
return bool(level[index])
def repl(m):
return "u%(c)s%(data)s%(c)s" % m.groupdict()
def handle_hashes(i):
if i.count('#') == 1:
n = i.find('#')
else:
n = get_hash_out_of_string(i)
return re.sub(x1,repl,i[:n]) + i[n:]
def get_hash_out_of_string(i):
n = i.find('#')
curr = i[:]
last = (len(i)-1)-''.join(list(reversed(i))).find('#')
while in_string(curr,n) and n < last:
curr = curr[:n]+' '+curr[n+1:]
n = curr.find('#')
return n
fcode = '\n'.join(
[re.sub(x1,repl,i)
if not '#' in i
else handle_hashes(i)
for i in scode.splitlines()])
print fcode
Output:
print u"'Hello World'" # prints 'Hello World'
u'Unicode is unchanged'
# so are "comments"
u'#### Hi' # 'Hi'

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