Function return 'None' Python - 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 ')))

Related

Best approach for converting lowerCamelCase to snake_case

I came across below mentioned scenario:
Input:-
parselTongue
Expected Output:-
parsel_tongue
My code:-
empty_string = ""
word = input()
if word.islower() == 1:
empty_string = empty_string + word
print(empty_string)
else:
for char in word:
char = str(char)
if char.isupper() == 1:
x = char
y = word.find(x)
print(char.replace(char, word[0:y] + "_" + char.lower() + word[y:]))
My output:-
parsel_tTongue
Please advice where i am going wrong as my output is coming as "parsel_tTongue" and not "parsel_tongue"
The more elegant solution would be just to implement the logic using comprehension.
word = input()
output= ''.join(c if not c.isupper() else f'_{c.lower()}' for c in word)
#output: 'parsel_tongue'
I believe that this approach could be better.
It prevents from situations where word contains not only letters but also special characters or numbers.
word = "camelCaseWord"
res = "" # sanke case word
# handle 1st upper character
if word[0].isupper():
word = word[0].lower() + word[1:]
for w in word:
# Only letter can be upper
if w.isupper():
res += "_" + w.lower()
else:
res += w
print(res)
>>> camel_case_word
if word = "camelCase3Wor& - > >>> camel_case3_wor&
no need for loop use regex
import re
name = 'parselTongue'
name = re.sub(r'(?<!^)(?=[A-Z])', '_', name).lower()
print(name) # camel_case_name
Adjust the slice on word
empty_string = ""
word = input()
if word.islower() == 1:
empty_string = empty_string + word
print(empty_string)
else:
for char in word:
char = str(char)
if char.isupper() == 1:
x = char
y = word.find(x)
print(char.replace(char, word[0:y] + "_" + char.lower()+ word[y+1:]))
prints the following for the input parselTongue
praselTongue
prasel_tongue
The best practice may be using regex:
fooBarBaz -> foo_bar_baz
re.sub(r'([A-Z])',lambda match:'_'+match.group(1).lower(),'fooBarBaz')
foo_bar_baz -> fooBarBaz
re.sub(r'_([a-z])',lambda match:match.group(1).upper(),'foo_bar_baz')
import re
camel_case = 'miaBau'
snake_case = re.sub(r'([A-Z])', r'_\1', camel_case).lower()

printing out or returning strings between brackets

I'm trying to write a function in similar manner as started, so that I will get what it's doing. I'm assuming this can be done with one line of code, with some fancy functions, but for the sake of practice and understanding I'm trying to come up with similar solution.
The task is the following: the function takes a text once it encounters enclosed square brackets [ word ] It should print out or return all words which are between square brackets. For example, if the text string would be "[a]n example[ string]", you are expected to print out "a string".
def string():
text = "some random text [and I need this bit of txt] but I don't know how to continue [to get this bit as well]"
for i in text:
for j in range(len(text)):
if text[j] == '[':
new = text.find(']')
return(text[j+1:new])
print(string())
Try this:
def extract(text, skip_chars=("\n", )):
output = ""
flag = False
for c in text:
if c == "]":
flag = False
if flag and not c in skip_chars:
output += c
if c == "[":
flag = True
return output
print(extract("""[a]n example[
stri
ng]"""))
# -> "a string"
def string():
result = []
text = "some random text [and I need this bit of txt] but I don't know how to continue [to get this bit as well]"
for i in text:
if i == '[':
new = text.find(']')
result.append(text[text.index(i) + 1:new])
return " ".join(result)
print(string())
def parse(source):
i = source.index("[") # throw an exception
result = ""
while i < len(source):
if s[i] == "[":
i += 1
while i < len(source):
temp = ""
if source[i] == "]":
result += temp
break;
temp += source[i]
i += 1
i += 1
return result

Turning random.choice into a mutable

I'm very new to python and I'm trying to make a hangman game. I currently have a line of code saying word = (random.choice(open("Level1py.txt").readline())).
I'm getting the error 'str' object does not support item assignment.
Here is the rest of my code (sorry for the mess):
import random
def checkLetter(letter, word, guess_word):
for c in word:
if c == letter:
guess_word[word.index(c)] = c
word[word.index(c)] = '*'
print(guess_word)
word = (random.choice(open("Level1py.txt").readline().split()))
guess_word = ['_' for x in word]
print(guess_word)
while '_' in guess_word:
guess = input('Letter: ')
print(checkLetter(guess, word, guess_word))
Strings are immutable in python. An easy workaround is to use lists, which are mutable:
st = "hello"
ls = list(st)
ls[3] = 'r'
st = ''.join(ls)
print(st)
Outputs
helro
Edit: here's how you would implement it in your own code
import random
def checkLetter(letter, word, guess_word):
for c in word:
if c == letter:
guess_word[word.index(c)] = c
word_list = list(word)
word_list[word.index(c)] = "*"
word = ''.join(word_list)
print(guess_word)
word = 'test'
guess_word = ['_' for x in word]
print(guess_word)
while '_' in guess_word:
guess = input('Letter: ')
print(checkLetter(guess, word, guess_word))
Note that there are still other issues that have nothing to do with this, like printing None and duplicate printing
You can also solve your problem using a dictionary:
word = "my string" #replace this with your random word
guess_word = {i: '_' for i in set(word)} # initially assign _ to all unique letters
guess_word[' '] = ' ' # exclude white space from the game
wrong_attempts = 0
while '_' in guess_word.values():
guess = input('Letter: ')
if guess in guess_word.keys():
guess_word[guess] = guess
else:
wrong_attempts += 1
if wrong_attempts > 11:
break
printable = [guess_word[i] for i in word]
print(' '.join(printable))
if '_' in guess_word.values():
print('you lost')
else:
print('congratulation, you won')

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)

Reading a (compressed) file

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()

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