I am trying to write python code that takes two strings as input and checks if all the characters of the second string are present in the first string. If so, then the output is the second string. If not, the output is the string "This does not work". I have tested this procedure for a variety of inputs, and it typically works, but not always. For example, if my two inputs are "helo" and "olhe", respectively, the output is "This does not work," whereas it should be "olhe" since all the characters in "olhe" are present in "helo."
Here is my code
def fix_machine(debris,product):
n = 1
while True:
first_element = product[0:n]
find_first_element = debris.find(first_element)
if first_element == product:
return product
break
n = n + 1
if find_first_element == -1:
return "This does not work"
break
So why does this not work?
You could do this as a more elegant solution, assuming that you strictly want all of the second string characters to be present in first.
def fix_machine(first, second):
for character in second:
if character not in first:
return False
return second
This returns correct input for your code. Not entirely sure what is wrong with your code as I have not stepped through.
EDIT: Albert has a much more elegant solution, reference his
As far as I understood you just want to compare two strings containing different characters.
For doing so, I would suggest converting both strings to a set which then provides (in contrast to a list) order-less comparison of its elements.
def check_string(first_str, second_str):
# convert strings to sets in order to compare characters
if set(first_string) == set(second_string):
return second_str
else:
return 'This does not work.'
first_string = 'helo'
second_string = 'olhe'
print(check_string(first_string, second_string))
# prints True
first_string = 'helo'
second_string = 'hello'
print(check_string(first_string, second_string))
# prints True
first_string = 'helo'
second_string = 'helofoo'
print(check_string(first_string, second_string))
# prints 'This does not work.'
I agree that you should step through that code with a debugger (maybe try PyCharm if you don't have an IDE set up yet) to see what went wrong. Would be hard to explain whats going wrong but I think it has something to do with first_element = product[0:n]. This will return increasingly larger segments of the string. i.e. 'ol' on the second run through.
Here's an alternative way of writing it
def fix_machine(debris, product):
all_present = all(letter in debris for letter in product)
return product if all_present else 'This does not work'
Related
Create a function in python that replaces at least four different words or phrases with internet slang acronyms such as LOL, OMG, TBH. For example, if the user enters a sentence "Oh my god, I am scared to be honest." The output should be "OMG I am scared TBH". The program must not use any built-in find, replace, encode, index, or translate functions. The program can use indexing (i.e., [ ] ), slicing (i.e., :), the in operator, and the len() function.
This is what I have so far:
user_string = (input("Please enter a string: ")).lower()
punctuations = '''.,!##$%^&*()[]{};:-'"\|<>/?_~'''
new_string = ""
list = []
for i in range(0, len(user_string)):
if (user_string[i] not in punctuations):
new_string = new_string + user_string[i]
print(new_string)
slang = "to be honest"
for i in range(0, len(slang)):
for j in range(0, len(new_string)):
if (new_string[j] == slang[i]):
list.append(j)
if (i < len(slang)):
i = i + 1
elif (new_string[j] != slang[i]):
if (len(list) > 0):
list.pop()
print(list)
First I am getting the sentence from the user and removing all the punctuations from the sentence. Then I have created a variable called slang which holds the slang that I want to replace in the sentence with the acronym "TBH".
I have nested for loops which compare the string that the user has entered to the first letter of the slang variable. If the letters are the same, it compares the next letter of the string with the next letter of the slang.
I'm getting an error from the last part. How do I check if "to be honest" is in the string that the user has entered? And if it is in the string, how do I replace it with "TBH"?
I cannot see any python errors that your code will actually produce, given the number of guard clauses, so I will assume what you mean by error is actually the program not working as you intended.
With that in mind, the main problem with your code is that you have nested for loops. This means that for any one character in slang, you check it against every character in new_string.
If you run through your code with this in mind, you will see that for every character in slang, you are attempting to add one value to the list and remove len(slang) - 1 values from it. Your clause, however, prevents this from causing an python error.
I would also like to mention that the statement
if (i < Len(slang)):
i = i + 1
is completely unnecessary because i is already automatically incremented by the for loop, which could cause issues later. It is guarded by a clause though, which is why it isn't a problem yet.
If you're still stuck on this problem, here's my version on how to solve this exercise:
# This is a dictionary so we can automate the replacement on the `__main__` scope
targets = {'to be honest': 'TBH', 'oh my god': 'OMG'}
# Returns a list of intervals that tells where all occurences of the
# `sequence` passed as parameter resides inside `source`.
#
# If `sequence` is not present, the list will be empty.
def findSequences(source, sequence):
# This is our return value.
intervals = []
# len is O(1). But if you need to implement your own len function,
# this might be handy to save for the linear complexity.
srcLength = len(source)
seqLength = len(sequence)
# If the sequence is larger than source, it's not inside
if (seqLength > srcLength):
return intervals
# If it's smaller or equal than source, it might be
else:
buffer = ''
for i in range(srcLength):
buffer = ''
# From a starting character on index `i`, we will create
# a temporary buffer with the length of sequence.
for j in range(seqLength):
# We must take care to not go out of the source string
# otherwise, there's no point in continuing on building
# buffer.
if (i+j >= srcLength):
break
else:
buffer += source[i+j]
# If this temporary buffer equals sequence, we found the
# substring!
if (buffer == sequence):
# Return the interval of the substring
intervals.append((i, i+j))
# Out of the for-loop.
return intervals
# Takes out any characters inside `punctuation` from source.
#
# Uses the `in` keyword on the if-statement. But as the post says,
# it's allowed.
def takeOutPunctuation(source, punctuation='.,!##$%^&*()[]{};:-\'"\\|<>/?_~'):
buffer = ''
for char in source:
if (char not in punctuation):
buffer += char
return buffer
# A naive approach would not to find all intervals, but to find the first
# `phrase` occurence inside the `source` string, and replace it. If you do
# that, it will get replacing "TBH" to "TBH2" infinitelly, always append "2"
# to the string.
#
# This function is smart enough to avoid that.
#
# It replaces all occurences of the `phrase` string into a `target` string.
#
# As `findSequences` returns a list of all capture's intervals, the
# replacement will not get stuck in an infinite loop if we use
# parameters such as: myReplace(..., "TBH", "TBH2")
def myReplace(source, phrase, target):
intervals = findSequences(source, phrase)
if (len(intervals) == 0):
return source
else:
# Append everything until the first capture
buffer = source[:intervals[0][0]]
# We insert this first interval just for writting less code inside the for-loop.
#
# This is not a capture, it's just so we can access (i-1) when the iteration
# starts.
intervals.insert(0, (0, intervals[0][0]))
# Start a the second position of the `intervals` array so we can access (i-1)
# at the start of the iteration.
for i in range(1, len(intervals)):
# For every `phrase` capture, we append:
# - everything that comes before the capture
# - the `target` string
buffer += source[intervals[i-1][1]+1:intervals[i][0]] + target
# Once the iteration ends, we must append everything that comes later
# after the last capture.
buffer += source[intervals[-1][1]+1:]
# Return the modified string
return buffer
if __name__ == '__main__':
# Note: I didn't wrote input() here so we can see what the actual input is.
user_string = 'Oh my god, I am scared to be honest and to be honest and to be honest!'.lower()
user_string = takeOutPunctuation(user_string)
# Automated Replacement
for key in targets:
user_string = myReplace(user_string, key, targets[key])
# Print the output:
print(user_string)
# -> OMG i am scared TBH and TBH and TBH
Note: I used Python 3.10.2 to run this script.
Below I have some strings in a list:
some_list = ['a','l','p','p','l','l','i','i','r',i','r','a','a']
Now I want to take the word april from this list. There are only two april in this list. So I want to take that two april from this list and append them to another extract list.
So the extract list should look something like this:
extract = ['aprilapril']
or
extract = ['a','p','r','i','l','a','p','r','i','l']
I tried many times trying to get the everything in extract in order, but I still can't seems to get it.
But I know I can just do this
a_count = some_list.count('a')
p_count = some_list.count('p')
r_count = some_list.count('r')
i_count = some_list.count('i')
l_count = some_list.count('l')
total_count = [a_count,p_count,r_count,i_count,l_count]
smallest_count = min(total_count)
extract = ['april' * smallest_count]
Which I wouldn't be here If I just use the code above.
Because I made some rules for solving this problem
Each of the characters (a,p,r,i and l) are some magical code elements, these code elements can't be created out of thin air; they are some unique code elements, that has some uniquw identifier, like a secrete number that is associated with them. So you don't know how to create this magical code elements, the only way to get the code elements is to extract them to a list.
Each of the characters (a,p,r,i and l) must be in order. Imagine they are some kind of chains, they will only work if they are together. Meaning that we got to put p next to and in front of a, and l must come last.
These important code elements are some kind of top secrete stuff, so if you want to get it, the only way is to extract them to a list.
Below are some examples of a incorrect way to do this: (breaking the rules)
import re
word = 'april'
some_list = ['aaaaaaappppppprrrrrriiiiiilll']
regex = "".join(f"({c}+)" for c in word)
match = re.match(regex, text)
if match:
lowest_amount = min(len(g) for g in match.groups())
print(word * lowest_amount)
else:
print("no match")
from collections import Counter
def count_recurrence(kernel, string):
# we need to count both strings
kernel_counter = Counter(kernel)
string_counter = Counter(string)
effective_counter = {
k: int(string_counter.get(k, 0)/v)
for k, v in kernel_counter.items()
}
min_recurring_count = min(effective_counter.values())
return kernel * min_recurring_count
This might sounds really stupid, but this is actually a hard problem (well for me). I originally designed this problem for myself to practice python, but it turns out to be way harder than I thought. I just want to see how other people solve this problem.
If anyone out there know how to solve this ridiculous problem, please help me out, I am just a fourteen-year-old trying to do python. Thank you very much.
I'm not sure what do you mean by "cannot copy nor delete the magical codes" - if you want to put them in your output list you will need to "copy" them somehow.
And btw your example code (a_count = some_list.count('a') etc) won't work since count will always return zero.
That said, a possible solution is
worklist = [c for c in some_list[0]]
extract = []
fail = False
while not fail:
lastpos = -1
tempextract = []
for magic in magics:
if magic in worklist:
pos = worklist.index(magic, lastpos+1)
tempextract.append(worklist.pop(pos))
lastpos = pos-1
else:
fail = True
break
else:
extract.append(tempextract)
Alternatively, if you don't want to pop the elements when you find them, you may compute the positions of all the occurences of the first element (the "a"), and set lastpos to each of those positions at the beginning of each iteration
May not be the most efficient way, although code works and is more explicit to understand the program logic:
some_list = ['aaaaaaappppppprrrrrriiiiiilll']
word = 'april'
extract = []
remove = []
string = some_list[0]
for x in range(len(some_list[0])//len(word)): #maximum number of times `word` can appear in `some_list[0]`
pointer = i = 0
while i<len(word):
j=0
while j<(len(string)-pointer):
if string[pointer:][j] == word[i]:
extract.append(word[i])
remove.append(pointer+j)
i+=1
pointer = j+1
break
j+=1
if i==len(word):
for r_i,r in enumerate(remove):
string = string[:r-r_i] + string[r-r_i+1:]
remove = []
elif j==(len(string)-pointer):
break
print(extract,string)
I've been write this practice program for while now, the whole purpose of the code is to get user input and generate passwords, everything almost works, but the replace statements are driving me nuts. Maybe one of you smart programmers can help me, because I'm kinda new to this whole field of programming. The issue is that replace statement only seems to work with the first char in Strng, but not the others one. The other funcs blower the last run first and then the middle one runs.
def Manip(Strng):
#Strng = 'jayjay'
print (Strng.replace('j','h',1))
#Displays: 'hayjay'
print (Strng.replace('j','h',4))
#Displays: 'hayhay'
return
def Add_nums(Strng):
Size=len(str(Strng))
Total_per = str(Strng).count('%')
# Get The % Spots Position, So they only get replaced with numbers during permutation
currnt_Pos = 0
per = [] # % position per for percent
rGen = ''
for i in str(Strng):
if i == str('%'):
per.append(currnt_Pos)
currnt_Pos+=1
for num,pos in zip(str(self.ints),per):
rGen = Strng.replace(str(Strng[pos]),str(num),4);
return rGen
for pos in AlphaB: # DataBase Of The Positions Of Alphabets
for letter in self.alphas: #letters in The User Inputs
GenPass=(self.forms.replace(self.forms[pos],letter,int(pos)))
# Not Fully Formatted yet; you got something like Cat%%%, so you can use another function to change % to nums
# And use the permutations function to generate other passwrds and then
# continue to the rest of this for loop which will generate something like cat222 or cat333
Add_nums(GenPass) # The Function That will add numbers to the Cat%%%
print (rGen);exit()
Am kind of new to coding,please help me out with this one with explanations:
songs is an array of objects which are formatted as follows:
{artist: 'Artist', title: 'Title String', playback: '04:30'}
You can expect playback value to be formatted exactly like above.
Output should be a title of the longest song from the database that matches the criteria of not being longer than specified time. If there's no songs matching criteria in the database, return false.
Either you could change playback, so that instead of a string, it's an integer (for instance, the length of the song in seconds) which you convert to a string for display, and test from there, or, during the test, you could take playback and convert it to its length in seconds, like so:
def songLength(playback):
seconds = playback.split(':')
lengthOfSong = int(seconds[0]) * 60 + int(seconds[1])
return lengthOfSong
This will give the following result:
>>> playback = '04:30'
>>> songLength(playback)
270
I'm not as familiar with the particular data structure you're using, but if you can iterate over these, you could do something like this:
def longestPossible(array, maxLength):
longest = 0
songName = ''
for song in array:
lenSong = songLength(song.playback) # I'm formatting song's playback like this because I'm not sure how you're going to be accessing it.
if maxLength >= lenSong and (maxLength - lenSong) < (maxLength - longest):
longest = lenSong
songName = song.title
if longest != 0:
return songName
else:
return '' # Empty strings will evaluate to False.
I haven't tested this, but I think this should at least get you on the right track. There are more Pythonic ways of doing this, so never stop improving your code. Good luck!
i have a programm that generate the list and then i ask them to tell me what they want to do from the menu and this is where my problem start i was able to get the input form the user to different function but when i try to use the if else condition it doesn't check, below are my code
def menu(x,l):
print (x)
if x == 1:
return make_table(l)
if x == 2:
y= input("enter a row (as a number) or a column (as an uppercase letter")
if y in [ "1",'2','3']:
print("Minmum is:",minimum(y,l))
if x== 3:
print ('bye')
def main():
bad_filename = True
l =[]
while bad_filename == True:
try:
filename = input("Enter the filename: ")
fp = open(filename, "r")
for f_line in fp:
f_str=f_line.strip()
f_str=f_str.split(',')
for unit_str in f_str:
unit=float(unit_str)
l.append(unit)
bad_filename = False
except IOError:
print("Error: The file was not found: ", filename)
#print(l)
condition=True
while condition==True:
print('1- open\n','2- maximum')
x=input("Enter the choice")
menu(x,l)
main()
from the bottom function i can get list and i can get the user input and i can get the data and move it in second function but it wont work after that.thank you
I think your problem is simple, and has nothing to do with how you're passing values between functions.
In main, you're reading a value from the user like this:
x=input("Enter the choice")
The input function:
… reads a line from input, converts it to a string (stripping a trailing newline), and returns that.
So, if the user types 1 at the prompt, you get back the string "1".
Now, you pass that value—perfectly correctly—to menu.
In menu, you then try to compare it to various numbers, like this:
if x == 1:
But this will never be true. A string, like "1", is never equal to a number, like 1. They're not even the same kind of value, much less the same value.
So, you need to do one of two things:
Convert the input to an number. For example, change menu(x,l) to menu(int(x), l). OR…
Write menu to expect strings. For example, change if x == 1: to if x == "1":.
You may be wondering why that print (x) didn't help you debug the problem.
print(x) prints out the end-user-friendly string representation of whatever you give it. That is, it automatically calls the str function for you. For debugging purposes, you often want to use repr instead of str, to get the programmer-friendly string representation instead of the end-user-friendly string representation.
For example, print(str("10")) will print out 10—just like print(str(10)), so you can't tell them apart. But print(repr("10")) will print out '10', unlike print(repr(10)), while prints 10, so you can tell them apart. repr can also help you spot things like strings with invisible characters in them, having special "node" objects from a parser instead of just strings, etc.