First post to site so I apologize if I do something wrong. I have looked for an appropriate answer, but could not find one.
I am new to python and have been playing around trying to take a long string (passage in a book,) and printing all but the first letter of each word while keeping the punctuation marks (Though not apostrophe marks.) and have been unsuccessful so far.
Example:
input = "Hello, I'm writing a sentence. (Though not a good one.)"
Code....
output = H, I W A S. (T N A G O.)
--Note the ",", ".", "()", but not the " ' ".
Any tips? Thank you all so much for taking the time to look
To help you on your adventure, I'll give you like a step-by-step logic of it
In python first use the .split() to seperate it by spaces
Go through each string in the list
Go through every char in the string
Print any punctuation marks that you specify and the first alphabetical character you find
Related
This is the first time I am learning python. I have had two C programming classes during my undergraduate days (many years back). I usually understand basic algorithms, but struggle to write codes out.
Currently doing a course in UDEMY, and the problem requires us to capitalize the first and third letters of the string. I have written the code (took me a while) and it works, but I know it is not pretty.
Please note: Trying to code it without using the enumerate function.
def wordplay(text):
first = text[0:1] #isolate the first letter
third = text[2:3] #isolate the third letter
firstc = first.capitalize() #capitalize the first letter
thirdc = third.capitalize() #capitalize the third letter
changedword = firstc + text[1:2] + thirdc + text[3:] #change the first and third letter to capital in the string
print(changedword)
The code worked, but looking to improve my logic (without using enumerate)
Here is one option which uses the capitalize() function:
inp = "hello"
output = inp[0:2].capitalize() + inp[2:].capitalize()
print(output) # HeLlo
The idea here is to just capitalize two substrings, one for the first two letters and the other for the remainder of the string.
I have a list of sentences, where some of them contain only one word but it is split into characters. How can I either merge the characters to make it one word or drop the whole row?
list = ['What a rollercoaster', 'y i k e s', 'I love democracy']
I try to avoid writing regular expressions as much as I can, but from what you told me, this one could work :
import re
a = ['What a rollercoaster', 'y i k e s', 'I love democracy']
regex = re.compile(r'^(\w ){2,}.')
result = list(filter(regex.search, a))
This captures strings having at least two groups of character and space, followed by anything else. This is assuming you wouldn't have a sentence beginning with something like 'a a foo'.
I'm trying to re-arrange long sentence from a puzzle that is encoded using a Caesar Cipher.
Here is my code.
sentence="g fmnc wms bgblr rpylqjyrc gr zw fylb. rfyrq ufyr amknsrcpq ypc dmp. bmgle gr gl zw fylb gq glcddgagclr ylb rfyr'q ufw rfgq rcvr gq qm jmle. sqgle qrpgle.kyicrpylq() gq pcamkkclbcb. lmu ynnjw ml rfc spj."
import string
a=string.ascii_lowercase[]
b=a[2:]+a[:2]
for i in range(26):
sentence.replace(sentence[sentence.find(a[i])],b[i])
Am I, missing anything in replace function?
When I tried sentence.replace(sentence[sentence.find(a[0])],b[0])
it worked but why I can't loop through?
Thanks.
sentence.replace
returns a new string, which you are immediately throwing away. Note that replacing each character repeatedly will cause duplicate replacements in your cipher. See #RemcoGerlich's answer for a better-detailed explanation of what is wrong. As for the solution, what about
import string
letters = string.ascii_lowercase
shifted = {l: letters[(i + 2) % len(letters)] for i, l in enumerate(letters)}
sentence = ''.join(shifted.get(c, c) for c in sentence.lower())
or if you really want the tabled way:
from string import ascii_lowercase
rotated_lowercase = ascii_lowercase[2:] + ascii_lowercase[:2]
translation_table = str.maketrans(ascii_lowercase, rotated_lowercase)
sentence = sentence.translate(translation_table)
There are a few problems:
One, sentence[sentence.find(a[i])] is strange. It tries to look up where in the sentence the character a[1] occurs, and then looks up which character is there. Well, you already know -- a[1]. Unless that character doesn't occur in the string, then .find will return -1, and sentence[-1] is the last character in the sentence. Probably not what you meant. So instead you meant sentence.replace(a[i], b[i]).
But, you don't save the result anywhere. You meant sentence = sentence.replace(a[i], b[i]).
But that still doesn't work! What if a should be changed into b, and then b into c? Then the original as are also changed into c! That's a fundamental problem with your approach.
Better solutions are given by modesitt. Mine would have been something like
lookupdict = {a_char: b_char for (a_char, b_char) in zip(a, b)}
sentence_translated = [lookupdict.get(s, '') for s in sentence]
sentence = ''.join(sentence_translated)
I am building a puzzle word game in Python. I have the correct puzzle word, and the guessed puzzle word. I want to build a third string which shows the correct letters in the guessed puzzle in the correct puzzle word, and _ at the position of the incorrect letters.
For example, say the correct word is APPLE and the guessed word is APTLE
then i want to have a third string: AP_L_
The guessed word and correct word are guaranteed to be 3 to 5 characters long, but the guessed word is not guaranteed to be the same length as the correct word
For example, correct word is TEA and the guessed word is TEAKO, then the third string should be TEA__ because the players guessed the last two letters incorrectly.
Another example, correct word is APPLE and guessed word is POP, the third string should be:
_ _ P_ _ (without space separation)
I can successfully get the matched indexes of the correct and guessed word; however, I am having problems building the third string. I just learned that strings in Python are immutable and that i cannot assign something like str1[index] = str2[index]
I have tried many things, including using lists, but i am not getting the correct answer. The attached code is my most recent attempt, would you please help me solve this?
Thank you
find the match between puzzle_word and guess
def matcher(str_a, str_b):
#find indexes where letters overlap
matched_indexes = [i for i, (a, b) in enumerate(zip(str_a, str_b)) if a == b]
result = []
for i in str_a:
result.append('_')
for value in matched_indexes:
result[value].replace('_', str_a[value])
print(result)
matcher("apple", "allke")
the output result right now is list of five "_"
cases:
correct word is APPLE and the guessed word is APTLE third
string: AP_L_
correct word is TEA and the guessed word is TEAKO,
third string should be TEA__
correct word is APPLE and guessed
word is POP, third string should be _ _ P_ _
You can use itertools.zip_longest here to always make sure you pad out to the longest word provided and then create a new string by joining the matching characters or otherwise a _. eg:
from itertools import zip_longest
correct_and_guess = [
('APPLE', 'APTLE'),
('TEA', 'TEAKO'),
('APPLE', 'POP')
]
for correct, guess in correct_and_guess:
# If characters in same positions match - show character otherwise `_`
new_word = ''.join(c if c == g else '_' for c, g in zip_longest(correct, guess, fillvalue='_'))
print(correct, guess, new_word)
Will print the following:
APPLE APTLE AP_LE
TEA TEAKO TEA__
APPLE POP __P__
Couple of things here.
str.replace() does not replace inline; as you noted strings are immutable, so you have to assign the result of replace:
result[value] = result[value].replace('_', str_a[value])
However, there's no point doing this since you can just assign to the list element:
result[value] = str_a[value]
And finally you can assign a list of the length of str_a without the for loop, which might be more readable:
result = ['_'] * len(str_a)
As part of my code, I need to align things like the pound sign to the left of a string. For example my code starts with:
"A price of £ 8 is roughly the same as $ 10.23!"
and needs to end with:
"A price of £8 is roughly the same as $10.23!"
I've created the following function to solve this however I feel that it is very inefficient and was wondering if there was a way to do this with regular expressions in Python?
for i in sentence:
if i == "(" or i == "{" or i == "[" or i == "£" or i == "$":
if i != len(sentence):
corrected_sentence.append(" ")
corrected_sentence.append(i)
else:
corrected_sentence.append(i)
What this is doing right now is going through the 'sentence' list where I have split up all of the words and punctuation and t then reforming this followed by a space EXPECT where the listed characters are used and adding to another list to be made into a single string again.
I only want to do this with the characters I have listed above (so I need to ignore things like full stops or exclamation marks etc).
Thanks!
I'm not sure what you want to do with the brackets, but from the description you can use a regex to find and replace whitespace preceded by the characters (lookbehind) and followed by a digit (lookahead).
>>> print(re.sub(r"(?<=[\{\[£\$])\s+(?=\d)", "", "A price of £ 8 is roughly the same as $ 10.23!"))
A price of £8 is roughly the same as $10.23!