Okay, below is my issue:
this program reads from a file, makes a list without using rstrip('\n'), which I did on purpose. From there, it prints the list, sorts it, prints it again, saves the new, sorted list to a text file, and allows you to search the list for a value.
The problem I am having is this:
when I search for a name, no matter how I type it in, it tells me that its not in the list.
the code worked til I changed the way I was testing for the variable. Here is the search function:
def searchNames(nameList):
another = 'y'
while another.lower() == 'y':
search = input("What name are you looking for? (Use 'Lastname, Firstname', including comma: ")
if search in nameList:
print("The name was found at index", nameList.index(search), "in the list.")
another = input("Check another name? Y for yes, anything else for no: ")
else:
print("The name was not found in the list.")
another = input("Check another name? Y for yes, anything else for no: ")
For the full code, http://pastebin.com/PMskBtzJ
For the content of the text file: http://pastebin.com/dAhmnXfZ
Ideas? I feel like I should note that I have tried to add ( + '\n') to the search variable
You say you explicitly did not strip off the newlines.
So, your nameList is a list of strings like ['van Rossum, Guido\n', 'Python, Monty\n'].
But your search is the string returned by input, which will not have a newline. So it can't possibly match any of the strings in the list.
There are a few ways to fix this.
First, of course, you could strip the newlines in your list.
Alternatively, you could strip them on the fly during the search:
if search in (name.rstrip() for name in nameList):
Or you could even add them onto the search string:
if search+'\n' in nameList:
If you're doing lots of searches, I would do the stripping just once and keep a list of stripped names around.
As a side note, searching the list to find out if the name is in the list, and then searching it again to find the index, is a little silly. Just search it once:
try:
i = nameList.index(search)
except ValueError:
print("The name was not found in the list.")
else:
print("The name was found at index", i, "in the list.")
another = input("Check another name? Y for yes, anything else for no: ")
Reason for this error is that any input in your list ends with a "\n". SO for example "john, smith\n". Your search function than uses the input which does NOT include "\n".
You've not given us much to go on, but maybe using sys.stdin.readline() instead of input() would help? I don't believe 2.x input() is going to leave a newline on the end of your inputs, which would make the "in" operator never find a match. sys.stdin.readline() does leave the newline at the end.
Also 'string' in list_ is slow compared to 'string' in set_ - if you don't really need indices, you might use a set instead, particularly if your collection is large.
Related
I have the following string: "https://www.instagram.com/paula.mtzm/"
I want to put the user "paula.mtzm" to a variable.
Anyone know how to do this ? Maybe you can somehow delete a part of the string like "https://www.instagram.com/" and then delete the last character "/" ?
"https://www.instagram.com/paula.mtzm/".split(".com/")[-1].replace("/", "")
This should do what you want. Effectively it splits the string into a list using the separator .com/, gets the last item of that list ("paula.mtzm/"), and finally removes any remaining /s
I'm not sure how specific your use-case is so I don't know how suitable this is in general.
This is actually pretty easy:
Strings are indexed in Python just like a list. So:
string = "potato"
print string[0] #this will print "p" to the console
#we can 'slice' in the index too
print string[0:3] #this will print "pot" to the console
So for your specific problem you could have your code search for the 3rd
forward slash and grab everything after that.
If you always know the web address you can just start your index at the end of
the address and where the user begins:
string = "https://www.instagram.com/paula.mtzm/"
string_index = 26 # the 'p' in paula begins here
user_name = string[string_index:string.len]
print user_name #outputs paula.mtzm
I need to create a box with parameters that prints any input the user puts in. I figured that the box should be the length of the string, but I'm stuck with empty code, because I don't know where to start.
It should look like this:
I agree with Daniel Goldfarb comments. Don't look for help without trying.
If you still couldn't get how to do that, then only read my remaining comment.
Just print :
str = string entered
len(str) = string length
+-(len(str) * '-')-+
| str |
+-(len(str) * '-')-+
So hopefully you can learn, don't want to just write the code for you. Basically break it into steps. First you need to accept user input. If you don't know how to do that, try googling, "python accept user input from stdin" or here is one of the results from that search: https://www.pythonforbeginners.com/basics/getting-user-input-from-the-keyboard
Then, as you mentioned, you need the length of the string that was input. You can get that with the len function. Then do the math: It looks like you want "|" and two spaces on each side of the string, giving the length plus 6 ("| " on either side). This new length is what you should make the "+---+" strings. Use the print() function to print out each line. I really don't want to say much more than that because you should exercise your brain to figure it out. If you have a question on how to generate "+---+" of the appropriate length (appropriate number of "-" characters) you can use string concatenation and a loop, or just use the python string constructor (hint: google "construct python string of len repeat characters"). HTH.
One more thing, after looking at your code, in addition to my comment about printing the string itself within the box, I see some minor logic errors in your code (for example, why are you subtracting 2 from the width). THE POINT i want to me here is, if you ware going to break this into multiple small functions (a bit overkill here, but definitely a good idea if you are just learning as it teaches you an important skill) then YOU SHOULD TEST EACH FUNCTION individually to make sure it does what you think and expect it to do. I think you will see your logic errors that way.
Here is the solution, but I recommend to try it out by yourself, breakdown the problem into smaller pieces and start from there.
def format(word):
#It declares all the necessary variables
borders =[]
result = []
# First part of the result--> it gives the two spaces and the "wall"
result.append("| ")
# Second part of the result (the word)
for letter in word:
result.append(letter)
# Third part of the result--> Ends the format
result.append(" |")
#Transforms the list to a string
result = "".join(result)
borders.append("+")
borders.append("--"+"-"*len(word)+"--")
borders.append("+")
borders="".join(borders)
print(borders)
print(result)
print(borders)
sentence = input("Enter a word: ")
format(sentence)
I'm new to Python, and I've found this solution. Maybe is not the best solution, but it works!
test = input()
print("+-", end='')
for i in test:
print("-", end='')
print("-+")
print("| " + test + " |")
print("+-", end='')
for i in test:
print("-", end='')
print("-+")
print "YOU HAVE CHOSEN TO REARRANGE YOUR THE WORD THAT YOU ARE ABOUT TO ENTER..."
word = raw_input ("FIRSTLY YOU MUST ENTER A WORD TO BE REARRANGED, ENTER IT HERE:")
character_save = word[1]
def anagram(word):
if len(word)>1:
print str.replace('a','b')
word = str.replace(word[1],word[3])
word= str.replace(word[3], character_save,1)
print word
anagram(word)
I tried to fix this on numerous occasions, the problem with the first time was that it would just replicate characters instead of replacing the positions, the second time I tried to store the position that I was going to replace in a variable but now it mentions that I have only one argument given (when it should be 2).
Would it be easier to do this with a list instead of a string?
The replace message that you are using is called on the string that you want to replace and not on the str type itself.
In your case that is the word parameter that you are providing.
So if you replace the instances of str.replace with word.replace your code will run. However, it doesn't create an anagram yet. The algorithm is still lacking.
I am writing a program that loads an external txt file of movies. This part works fine. I then have a function that searches a list of movies generated from the file. The function should print out all movies that start with the search string.
def startsWithSearch(movieList):
searchString = input("Enter search string: ")
for movie in movieList:
if(movie.startswith(searchString) == True):
print(movie)
However, no movies are printed when I enter a search string, even though there are movies in the list that start with that string.
if given correct input data your function does work as expected:
def startsWithSearch(movieList):
searchString = "test4"
for movie in movieList:
if(movie.startswith(searchString)):
print(movie)
startsWithSearch(["test1","testnomatch","test4","test4should","not_test4"])
output is:
test4
test4should
so all correct... must be your input data
i know you want a StartsWith solution as your function name says, but actually searching for movies, it is a lot more convenient, if you find any match inside the string, so if i search for "mentalist" i will find "the mentalist", then you could just use:
if searchString in movie:
print(movie)
And as suggested by Anna to ignore case:
if searchString.lower() in movie.lower():
or even fancier with regular expressions (need import re at first line):
if re.match(".*" + searchString,movie,re.I):
or if you really just want match on beginning of name:
if re.match(searchString,movie,re.I):
that should be enough alternatives :)
I would think it might be that the input() function is returning with a new line at the end. Try adding searchString = searchString.strip() after collecting the input data.
Also you might want to try converting both to lower case before comparing.
Also the line if(movie.startswith(searchString) == True): can just be written as if movie.startswith(searchString):
I had same problem when I was reading from files. startswith was failing due to newline character which was read from file. Use rstrip() to remove newline character from both strings before you use startswith function.
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))