I have written a code which outputs the sub tokens a word. I have defined many cases for which it works perfectly except for one case for which it returns :
['']
instead of :
[]
Is there a way that i can ask python to print the square brackets, without the apostrophe's whenever that case occurs?
Assuming you have a list containing an empty string you can check for it and print the [] symbols (I really don't see a reason for this other than making your code "nicer"(?)):
>>> a = [""]
>>> print(a)
['']
>>> if a==[""]:
print("[]")
[]
Related
I'm writing a program to check if a given user input is a palindrome or not. if it is the program should print "Yes", if not "no". I realize that this program is entirely too complex since I actually only needed to check the whole word using the reversed() function, but I ended up making it quite complex by splitting the word into two lists and then checking the lists against each other.
Despite that, I'm not clear why the last conditional isn't returning the expected "Yes" when I pass it "racecar" as an input. When I print the lists in line 23 and 24, I get two lists that are identical, but then when I compare them in the conditional, I always get "No" meaning they are not equal to each other. can anyone explain why this is? I've tried to convert the lists to strings but no luck.
def odd_or_even(a): # function for determining if odd or even
if len(a) % 2 == 0:
return True
else:
return False
the_string = input("How about a word?\n")
x = int(len(the_string))
odd_or_even(the_string) # find out if the word has an odd or an even number of characters
if odd_or_even(the_string) == True: # if even
for i in range(x):
first_half = the_string[0:int((x/2))] #create a list with part 1
second_half = the_string[(x-(int((x/2)))):x] #create a list with part 2
else: #if odd
for i in range(x):
first_half = the_string[:(int((x-1)/2))] #create a list with part 1 without the middle index
second_half = the_string[int(int(x-1)/2)+1:] #create a list with part 2 without the middle index
print(list(reversed(second_half)))
print(list(first_half))
if first_half == reversed(second_half): ##### NOT WORKING BUT DONT KNOW WHY #####
print("Yes")
else:
print("No")
Despite your comments first_half and second_half are substrings of your input, not lists. When you print them out, you're converting them to lists, but in the comparison, you do not convert first_half or reversed(second_half). Thus you are comparing a string to an iterator (returned by reversed), which will always be false.
So a basic fix is to do the conversion for the if, just like you did when printing the lists out:
if list(first_half) == list(reversed(second_half)):
A better fix might be to compare as strings, by making one of the slices use a step of -1, so you don't need to use reversed. Try second_half = the_string[-1:x//2:-1] (or similar, you probably need to tweak either the even or odd case by one). Or you could use the "alien smiley" slice to reverse the string after you slice it out of the input: second_half = second_half[::-1].
There are a few other oddities in your code, like your for i in range(x) loop that overwrites all of its results except the last one. Just use x - 1 in the slicing code and you don't need that loop at all. You're also calling int a lot more often than you need to (if you used // instead of /, you could get rid of literally all of the int calls).
I just started to use python 3. I want to find specific characters inside a string that is part of a list. Here is my code:
num = ["one","two","threex"]
for item in num:
if item.find("x"):
print("found")
So, I want to print "found" if the character "x" is inside of one of the elements of the list. But when I run the code, it prints 3 times instead of one.
Why is printing 3 times? Can someone help me?
find() returns -1 if the character is not found in the string. Anything that is not zero is equal to True. try if item.find("x") > -1.
You can use in again for strings:
num = ["one","two","threex"]
for item in num:
if "x" in item:
print("found")
Think in Strings as a list of chars like "ext" -> ['e', 'x', 't']
so "x" in "extreme" is True
find returns Index if found and -1 otherwise.
num = ["one","two","threex"]
for item in num:
if item.find("x"):
print item.find("x")
i hope that you got the solution from above post ,here you know the reason why
You need to break out of looping through the strings if 'x' is found as otherwise, it may be found in other strings. Also, when checking if 'x' is in the string, use in instead.
num = ["one","two","threex"]
for item in num:
if "x" in item:
print("found")
break
which outputs:
found
And if I modify the num list so that it has no x in any of the elements:
num = ["one","two","three"]
then there is no output when running the code again.
But why was it printing 3 times before?
Well simply, using item.find("x") will return an integer of the index of 'x' in the string. And the problem with evaluating this with an if-statement is that an integer always evaluates to True unless it is 0. This means that every string in the num list passed the test: if item.find("x") and so for each of the 3 strings, found was printed. In fact, the only time that found wouldn't be printed would be if the string began with an 'x'. In which case, the index of 'x' would be 0 and the if would evaluate to False.
Hope this clears up why your code wasn't working.
Oh, and some examples of testing the if:
>>> if 0:
... print("yes")
...
>>> if 1:
... print("yes")
...
yes
>>> if -1:
... print("yes")
...
yes
You know when you find a solution via trial and error but you stumbled so much thtat you can't understand the answer now?
Well this is happening to me with this piece:
entr = [list(int(x) for x in input().split()) for i in range(int(input()))]
The input is done by copying and pasting this whole block:
9
8327 0
0070 0
2681 2
1767 0
3976 0
9214 2
2271 2
4633 0
9500 1
What is my list comprehension exactly doing in each step? And taking into account this: How can I rewrite it using for loops?
In fact, your code is not a nested list-comprehension, beause you use list construtor rather than mere list-comprehension.
This line serves as same as your code:
entr = [[int(x) for x in input().split()] for i in range(int(input()))]
To understand this line, you must remember the basic structure of list-comprehension in python, it consists of two component obj and condition with a square brackets surrounding:
lst = [obj condition]
it can be converted to a loop like this:
lst = []
condition:
lst.append(obj)
So, back to this question.
What you need to do now is to break the nested list-comprehension into loop in loop, usually you begin from the condition in latter part, from outer space to inner space. You got:
entr = []
for i in range(int(input())):
entr.append([int(x) for x in input().split()])) # the obj is a list in this case.
And now, you can break the list-comprehension in line 3.
entr = []
for i in range(int(input())):
entry = []
for x in input().split():
entry.append(int(x))
entr.append(entry)
So, now the stuff the original line can be easily understand.
the program construct a entry list named entr;
the program ask for user input and convert the input string into an int, which is the number of the entrys you want to input(assume it is num);
the program ask for user input for num times, each time you should input something seperate with spaces.
The program split every string into a list (named entry in above code) you input with str.split() method (with parameter sep default to space). And append each entry list in every loop.
for every element in the entry list, it converted to int.
My English may be poor, feel free to improve my answer:)
That is equivalent to this:
entr = []
for i in range(int(input())):
row = []
for x in input().split():
row.append(int(x))
entr.append(row)
You can copy-paste that into a list comprehension in a couple steps. First the inner loop/list:
entr = []
for i in range(int(input())):
row = [int(x) for x in input().split()]
entr.append(row)
Without the row variable:
entr = []
for i in range(int(input())):
entr.append([int(x) for x in input().split()])
Then the outer loop/list (copied over multiple lines for clarity):
entr = [
[int(x) for x in input().split()]
for i in range(int(input()))
]
You have that same nested comprehension except that the inner one has been written as a generator passed to the list constructor so it looks like list(int(x) for x in input().split()) instead of [int(x) for x in input().split()]. That's a little more confusing than using a list comprehension.
I hope that explanation helps!
Dear public recently I have been receiving an unusual error to do with tuples. I think it has something to do with the spaces. Thanks for the help Adam FYI this program generates all combination and puts numbers together.
Does anybody know whats going on
import intercools
list1 = []
stuff = [1, 2, 3]
for L in range(0, len(stuff+1):
for subset in itertools.combinations(stuff, L):
list1.append(subset)
print(list1)
sep = [map(str,l)for l in list1]
nl = [int(''.join(s)) for s in sep]
print(nl)
There are a number of syntax errors present in your code. To start,
Your outer loop is missing an right parenthesis after stuff
The name of the module is itertools not intercools
After getting through those,
nl = [int(''.join(s)) for s in sep]
ValueError: invalid literal for int() with base 10: ''
So it appears that sep is occasionally empty. It's not clear what your goal is, but it looks like list1 does not contain what you expect.
I use Python3 and write result into a file like this:
with open(output,'w') as f:
f.write('Line count of the log files is: ' + str(line_count) + '. \n')
f.write() automatically returns # of characters written, is there a way to do not output it? I ask this because I do not want it output.
Thanks.
This is not unique to file.write(). The interactive interpreter prints the result of any evaluated expression that does not result in None.
>>> for i in range(3):
... i # expression evaluates to the value of i
...
0
1
2
>>>
Two things to note. First, these won't be displayed when you are not using the interactive interpreter, so it's safe to ignore.
Second, you can make the display go away by assigning the result. That turns the expression into a statement.
>>> for i in range(3):
... _ = i # underscore is a nice meaningless variable name
...
>>>