I have two lists:
a - dictionary which contains keywords such as ["impeccable", "obvious", "fantastic", "evident"] as elements of the list
b - sentences which contains sentences such as ["I am impeccable", "you are fantastic", "that is obvious", "that is evident"]
The goal is to use the dictionary list as a reference.
The process is as follows:
Take an element for the sentences list and run it against each element in the dictionary list. If any of the elements exists, then spit out that sentence to a new list
Repeating step 1 for each of the elements in the sentences list.
Any help would be much appreciated.
Thanks.
Below is the code:
sentences = "The book was awesome and envious","splendid job done by those guys", "that was an amazing sale"
dictionary = "awesome","amazing", "fantastic","envious"
##Find Matches
for match in dictionary:
if any(match in value for value in sentences):
print match
Now that you've fixed the original problem, and fixed the next problem with doing the check backward, and renamed all of your variables, you have this:
for match in dictionary:
if any(match in value for value in sentences):
print match
And your problem with it is:
The way I have the code written i can get the dictionary items but instead i want to print the sentences.
Well, yes, your match is a dictionary item, and that's what you're printing, so of course that's what you get.
If you want to print the sentences that contain the dictionary item, you can't use any, because the whole point of that function us to just return True if any elements are true. It won't tell you which ones—in fact, if there are more than one, it'll stop at the first one.
If you don't understand functions like any and the generator expressions you're passing to them, you really shouldn't be using them as magic invocations. Figure out how to write them as explicit loops, and you will be able to answer these problems for yourself easily. (Note that the any docs directly show you how to write an equivalent loop.)
For example, your existing code is equivalent to:
for match in dictionary:
for value in sentences:
if match in value:
print match
break
Written that way, it should be obvious how to fix it. First, you want to print the sentence instead of the word, so print value instead of match (and again, it would really help if you used meaningful variable names like sentence and word instead of meaningless names like value and misleading names like match…). Second, you want to print all matching sentences, not just the first one, so don't break. So:
for match in dictionary:
for value in sentences:
if match in value:
print value
And if you go back to my first answer, you may notice that this is the exact same structure I suggested.
You can simplify or shorten this by using comprehensions and iterator functions, but not until you understand the simple version, and how those comprehensions and iterator functions work.
First translate your algorithm into psuedocode instead of a vague description, like this:
for each sentence:
for each element in the dictionary:
if the element is in the sentence:
spit out the sentence to a new list
The only one of these steps that isn't completely trivial to convert to Python is "spit out the sentence to a new list". To do that, you'll need to have a new list before you get started, like a_new_list = [], and then you can call append on it.
Once you convert this to Python, you will discover that "I am impeccable and fantastic" gets spit out twice. If you don't want that, you need to find the appropriate please to break out of the inner loop and move on to the next sentence. Which is also trivial to convert to Python.
Now that you've posted your code… I don't know what problem you were asking about, but there's at least one thing obviously wrong with it.
sentences is a list of sentences.
So, for partial in sentences means each partial will be a sentence, like "I am impeccable".
dictionary is a list of words. So, for value in dictionary means each value will be a word, like "impeccable".
Now, you're checking partial in value for each value for each partial. That will never be true. "I am impeccable" is not in "impeccable".
If you turn that around, and check whether value in partial, it will give you something that's at least true sometimes, and that may even be what you actually want, but I'm not sure.
As a side note, if you used better names for your variables, this would be a lot more obvious. partial and value don't tell you what those things actually are; if you'd called them sentence and word it would be pretty clear that sentence in word is never going to be true, and that word in sentence is probably what you wanted.
Also, it really helps to look at intermediate values to debug things like this. When you use an explicit for statement, you can print(partial) to see each thing that partial holds, or you can put a breakpoint in your debugger, or you can step through in a visualizer like this one. If you have to break the any(genexpr) up into an explicit loop to do, then do so. (If you don't know how, then you probably don't understand what generator expressions or the any function do, and have just copied and pasted random code you didn't understand and tried changing random things until it worked… in which case you should stop doing that and learn what they actually mean.)
Related
why does the first use of "sorted_words" return the sorted_words WITHOUT the words "All" and "wait
why does the second use of "sorted_words" return the sorted_words WITHOUT the words "All" and "who"
BUT if we look below
When I use "sorted_words" one more time at the bottom,
all the words are there. What is different when I call on this again this time
that is different from the previous two times where the words were not shown, but
now they are shown here?
The methods you're using on 'sorted_words' are 'popping' the first/last words of the list. That means they are removing them from the list, so they won't appear again.
When they reappear later, it's because you're acting on your original sentence string, not the sorted_words list.
I am writing a simple cryptogram solver and am having trouble 'unrolling' a recursive function. I must unroll it for other reasons, otherwise I would leave it recursive.
Here's the idea: I have a variable number of lists, each with words in them. The function's job is to go through each list and, after checking that the word fits in the current alphabet setup, find it's score. So if you have the following lists:
LIST1: [the, and, can,...]
LIST2: [kids, cars, knee,...]
LIST3: [talks, walks, music,...]
...
and the function needs to go through each list (in order) and try to find the best sentence. (I have a scoring algorithm that it calls to compare.) It starts with the first word in the first list, then iterates the second list until it finds a word that works, then starts iterating the third list until it finds a word in that list that works, etc. Once it exhausts the words in the 3rd list, it should then go back to the second and find the next word that works, continuing the process until it's done.
I tried using the Product function, but that doesn't work the right way...that just gives me all possible combinations, and technically works, but is not very efficient.
def find_sentence():
cycle through first list:
cycle through second list:
if word works:
start cycling through third word list.
else:
keep cycling through 2nd word list.
...
Keep going until we have gone through all word lists, finding a score that is above a threshold..
Any help?
From Bakuriu's response:
Thanks for your fast reply! I'm not that great at Python, but I don't think this is working the way I need it to. Your solution is similar to the Product method in that it's goal is to find all words that will work (or fit a score.) The method I need to use is : 1. Start with the 1st word in the 1st list. 2. Start iterating the next list of words. 3. As soon as one of those words works, start going through the 3rd list, etc. 4. When you've reached the end (to the last list of words) and find a candidate, you now have a solution, as you have one word in each list that works. 5. If, say, a word in list 3 does not fit, you must go back to list 2 and CONTINUE searching through that list, finding the next word that works, moving on to start list 3 OVER AGAIN, and continuing until nothing works or you've reached the end. I hope this is clear. Please let me know if I can clarify anything.
You really don't need recursion here at all, actually.
def find_sentence(*variable_number_of_lists):
out = []
for eachlist in variable_number_of_lists:
for word in eachlist:
if scoreword(out, word) > threshhold:
# presumably, your 'scoreword' function would take in the current
# list of okayed words in order to find the most recent one for use
# in your scoring, if I've understood the problem correctly
out.append(word)
break
return out
I have an assignment that I am supposed to find each word in a line, and add that word to a list. Then there is also another list corresponding to the list of word, but that list will tell the amount of times the word appear in the text.
I have finished that part. However, I cannot find a way to compare the new found word to the word in the list, and find the index to insert it in the list in an alphabetical order. I know that I am supposed to write a function that will find that index in the list, so i can insert that item in both lists. I am not allowed to use the sort operator, so I am having a little trouble. Can anyone help me writing that one function using only conditions operators.
If I am not clear, please let me know.
Homework hint: look at Python's source code for the bisect module. That shows how to find indexes and make insertions in a sorted list.
Exercise problem: "given a word list and a text file, spell check the
contents of the text file and print all (unique) words which aren't
found in the word list."
I didn't get solutions to the problem so can somebody tell me how I went and what the correct answer should be?:
As a disclaimer none of this parses in my python console...
My attempt:
a=list[....,.....,....,whatever goes here,...]
data = open(C:\Documents and Settings\bhaa\Desktop\blablabla.txt).read()
#I'm aware that something is wrong here since I get an error when I use it.....when I just write blablabla.txt it says that it can't find the thing. Is this function only gonna work if I'm working off the online IVLE program where all those files are automatically linked to the console or how would I do things from python without logging into the online IVLE?
for words in data:
for words not in a
print words
wrong = words not in a
right = words in a
print="wrong spelling:" + "properly splled words:" + right
oh yeh...I'm very sure I've indented everything correctly but I don't know how to format my question here so that it doesn't come out as a block like it has. sorry.
What do you think?
There are many things wrong with this code - I'm going to mark some of them below, but I strongly recommend that you read up on Python control flow constructs, comparison operators, and built-in data types.
a=list[....,.....,....,whatever goes here,...]
data = open(C:\Documents and Settings\bhaa\Desktop\blablabla.txt).read()
# The filename needs to be a string value - put "C:\..." in quotes!
for words in data:
# data is a string - iterating over it will give you one letter
# per iteration, not one word
for words not in a
# aside from syntax (remember the colons!), remember what for means - it
# executes its body once for every item in a collection. "not in a" is not a
# collection of any kind!
print words
wrong = words not in a
# this does not say what you think it says - "not in" is an operator which
# takes an arbitrary value on the left, and some collection on the right,
# and returns a single boolean value
right = words in a
# same as the previous line
print="wrong spelling:" + "properly splled words:" + right
I don't know what you are trying to iterate over, but why don't you just first iterate over your words (which are in the variable a I guess?) and then for every word in a you iterate over the wordlist and check whether or not that word is in the wordslist.
I won't paste code since it seems like homework to me (if so, please add the homework tag).
Btw the first argument to open() should be a string.
It's simple really. Turn both lists into sets then take the difference. Should take like 10 lines of code. You just have to figure out the syntax on your own ;) You aren't going to learn anything by having us write it for you.
I'm studing Python for one month and I'm trying to make a keygen application by using the dictionary. The idea was to compare each letter in name = input('Name: ') to dict.keys() and print as result dict.values() for each letter of name equal to dict.keys(). That's what I wrote:
name = input('Name: ')
kalg = dict()
kalg['a'] = '50075'
kalg['b'] = '18099'
kalg['c'] = '89885'
etc...
I tryed writing this...
for x in kalg.keys():
print(x)[/code]
...but i need to keep print(x) result but i don't know how to do it! If i do this:
for x in kalg.keys():
a = x
'a' keeps only the last key of the dictionary :(. I thought it was because print(x) prints each key of dict.keys() on a new line but i don't know how to solve it (I tryed by converting type etc... but it didn't work).
Please can you help me solve this? I also don't know how to compare each letter of a string with another string and print dict.values() as result and in the right position.
Sorry for this stupid question but i'm too excited in writing python apps :)
# Karl
I'm studing Python over two differt books: 'Learning Python' by Mark Luts which covers Python
2 and a pocket which covers Python 3. I examined the list comprehension ón the pocket one and Imanaged to write three other variants of this keygen. Now i want to ask you how can I implementthe source code of this keygen in a real application with a GUI which verify if name_textbox andkey_textbox captions match (i come from basic so that was what i used to write, just to give youan idea) as the keygen output result. I know i can try to do this by my own (I did but with nosuccess) but I would like to first complete the book (the pocket one) and understand all the mainaspects of Python. Thank you for the patience.
Calling print can't "keep" anything (since there is no variable to store it in), and repeatedly assigning to a variable replaces the previous assignments. (I don't understand your reasoning about the problem; how print(x) behaves has nothing to do with how a = x behaves, as they're completely different things to be doing.)
Your question boils down to "how do I keep a bunch of results from several similar operations?" and on a conceptual level, the answer is "put them into a container". But explicitly putting things into the container is more tedious than is really necessary. You have an English description of the data you want: "dict.values() for each letter of name equal to dict.keys()". And in fact the equivalent Python is shockingly similar.
Of course, we don't actually want a separate copy of dict.values() for each matching letter; and we don't actually want to compare the letter to the entire set of dict.keys(). As programmers, we must be more precise: we are checking whether the letter is a key of the dict, i.e. if it is in the set of dict.keys(). Fortunately, that test is trivial to write: for a given letter, we check letter in dict. When the letter is found, we want the corresponding value; we get that by looking it up normally, thus dict[letter].
Then we wrap that all up with our special syntax that gives us what we want: the list comprehension. We put the brackets for a list, and then inside we write (some expression that calculates a result from the input element) for (a variable name for the input elements, so we can use it in that first expression) in (the source of input elements); and we can additionally filter the input elements at the same time, by adding if (some condition upon the input element).
So that's simple enough: [kalg[letter] for letter in name if letter in kalg]. Notice that I have name as the "source of elements", because that's what it should be. You explained that perfectly clearly in your description of the problem - why are you iterating over dict.keys() in your existing for-loops? :)
Now, this expression will give us a list of the results, so e.g. ['foo', 'bar', 'baz']. If we want one continuous string (I assume all the values in your dict are strings), then we'll need to join them up. Fortunately, that's easy as well. In fact, since we're going to pass the results to a function taking one argument, there is a special syntax rule that will let us drop the square brackets, making things look quite a bit neater.
It's also easier than you're making it to initialize the dict in the first place; idiomatic Python code rarely actually needs the word dict.
Putting it all together:
kalg = {'a': '50075', 'b': '18099', 'c': '89885'} # etc.
name = input('Name: ')
print(''.join(kalg[letter] for letter in name if name in kalg))
I can only guess, but this could be what you want:
name = input('Name: ')
kalg = {'a':'50075', 'b': '18099', 'c': '89885'}
keylist = [kalg[letter] for letter in name]
print(" ".join(keylist))