Here is my task to solve.
For a list of integers, find and print the items that appear in the list
only once. Items must be printed in the order in which they are
are in the incoming list.
I wrote the code but it also counts two and three digit numbers as a single digits. What's the problem?
x = []
a = input()
a.split(" ")
for i in a:
if a.count(i) == 1:
x.append(i)
print(x)
User Mechanical Pig provides the answer, split does not work in-place, since you're discarding its result what you're looking at is the count of each character in the string, not each space-separated sub-string.
The standard library also contains a collection which is ready-made for this use-case: collections.Counter. You can just give it a sequence of items and it'll collate the counts, which you can then filter:
print([k for k, v in collections.Counter(a.split()).items() if v == 1])
a.split(" ") does not change the value of a, so it is still a string and not a list. Hence when you iterate over a, you only get single characters.
str.split is a method that returns a list, but does not magically turn the string into a list. So you need to assign the value it returns to a variable (for example, to a, if you don't want to hold on to the input string).
You can assign the variable as mentioned above or you can directly use the split function in the loop.
x = []
a = input()
for i in a.split(" "):
if a.count(i) == 1:
x.append(i)
print(x)
or split when you are reading the input.
x = []
a = input().split(" ")
for i in a:
if a.count(i) == 1:
x.append(i)
print(x)
Using the Counter class from the collections module is the right way to do this but here's another way without any imports where you effectively have your own limited implementation of a Counter:
a = '1 2 3 4 4 5 6 7 7 7'
d = {}
for n in a.split():
d[n] = d.get(n, 0) + 1
for k, v in d.items():
if v == 1:
print(k)
Output:
1
2
3
5
6
Related
I was attempting to do one kata problem in Python, where I have two lists of input, let's say weight and value, which was getting input in the order (value1, weight1, value2, weight2,....)
If it were C++, I could just take one element at a time with cin to my arrays. But with Python I don't know how to take this input.
If the input is like
60 10 100 20 120 30 (in a single line)
I want my two lists val and w have values
val=[60,100,120]
w=[10,20,30]
How to take this kind of inputs in Python? Is it possible to read only one input at a time, like cin does in C++?
You can read space-separated input to a list using splitand then use slicing to get the odd/even indexed elements:
val = input().split()
val = [int(i) for i in val] #to integer
w = val[1::2] # get only odd indices
val = val[0::2] # get only even indices
print(val) # [60,100,120]
print(w) # [10,20,30]
You can then use regular indexing to get individual elements in the lists.
This should do it.
values = input("Enter values: ").split(" ")
if(len(values) % 2 != 0):
print("Error, uneven number of values and weights")
else:
val = []
w = []
for i in range(0, len(values) - 1, 2):
val.append(values[i])
w.append(values[i+1])
print("val: ", val)
print("w: ", w)
No you cannot as far as I know. input() is a function that reads one line at a time as an entire string.
A popular way to make a list out of an entire line of space separated integers is using the in-built map() function and split() method of a string object:
numbers = map(int, input.split())
numbers will now contain the list of numbers on that line.
Firstly, input() reads the entire line as a string including spaces.
split() function splits the input at space, you can also pass in a string argument to make it split anywhere you want, i.e a custom delimiter like , and creates a list of strings of those numbers
map() calls int() on each of the numbers in that list.
In C++, just like you need to know the size of input in advance, you can run a for loop on the list to now split values and weights into two lists.
val, w = [], []
for i in range(len(numbers)):
if i % 2 == 0: val.append(numbers[i])
else: w.append(numbers[i])
For better performance(relevant for large input size), you can even skip the map() step, and do something like this:
numbers = input().split()
val, w = [], []
for i in range(len(numbers)):
if i % 2 == 0: val.append(int(numbers[i]))
else: w.append(int(numbers[i]))
Here's a simple solution:
a = "60 10 100 20 120 30"
split_a = a.split(" ")
val = [e for (i, e) in enumerate(split_a) if i % 2 == 0]
w = [e for (i, e) in enumerate(split_a) if i % 2 == 1]
# or shorter with slicing (start:stop:step)
val = [e for e in split_a[0::2]]
w = [e for e in split_a[1::2]]
I am not sure what cin does but I assume the input you have is a string. One way you could do is to split it into a list of ints. And then create your val and w lists and append your values into the lists. (The first, third, fifth, etc would be your val list)
my_input_list = my_input.split(' ')
Read more here.
I want to remove single letter words such as a, i e, e g, b f f, y o l o, c y l using a Python function.
My current code looks follows.
def remove_single_letters(concept):
mystring = "valid"
if len(concept) == 1:
mystring = "invalid"
if len(concept)>1:
validation = []
splits = concept.split()
for item in splits:
if len(item) > 1:
validation.append("valid")
if len(validation) != 1:
mystring = "invalid"
return mystring
print(remove_single_letters("b f f"))
It works fine. However, I am wondering if there is a more efficient way (with lesser time) of doing it in python.
Here is a single line solution:
def remove_single_letters(concept):
return ["valid", "invalid"][concept.count(" ") >= len(concept) // 2]
Update: Note that this is shorter and cool looking but does not necessarily run faster.
Explanation:
concept.count(" "): Returns the number of spaces in string
>= len(concept) // 2: Returns True if more than half of the string is spaces (it fails when there are multiple spaces between legitimate words as #user202729 mentioned)
["valid", "invalid"][result]: This part is just for fun: It returns the first element if the result is False and the second element if the result is True (because False is equal to 0 and True is equal to 1).
I would go for a more concise solution (yet not faster as both solutions are of O(n)) if you want to check if any 1 letter character exists in the string:
remove_single_letters = lambda concept:"invalid" if 1 in [len(item) for item in concept.split()] else "valid"
print(remove_single_letters("a b c"))
#prints invalid
An ordinary function would be like this:
def remove_single_letters(concept):
return "invalid" if 1 in [len(item) for item in concept.split()] else "valid"
They both check the length of elements in the split input to find any item with length of 1 and is insensitive to multiple spaces thanks to split().
If you want to check the strings that are totally made up of single characters:
def remove_single_letters(concept):
u = set(len(item) for item in concept.split())
return "invalid" if len(u) == 1 and 1 in u else "valid"
I've encountered a small problem with a simple shift deciphering.
N,K = [int(s) for s in input().split()]
myres = []
alph = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
myalph = [a for a in alph]
for i in range(N):
s = input()
mylist = [d for d in str(s)]
for b in range(len(mylist)):
for c in range(len(myalph)):
if mylist[b] == myalph[c]:
mylist[b] = myalph[c-K]
print(myalph[c-K], c-K, b, c)
myres = myres + mylist
Res = [str(i) for i in myres]
print("".join(Res))
The idea is for every character of my input string to be replaced with a different character from the alphabet that's been shifted by a given key (K).
The problem occurs when c-K < 0 and the replacing key is taken from the back of the list. Then the loop is being iterated twice.
If the key is 3 and I input A instead of getting X, I'm getting U as the first iteration gives X but then X is also iterated and becomes U.
Your mistake is looping over all the letters in alph:
The for loop tests all the letters of the alphabet, in order, and 'A' is matched. You set mylist[b] to 'X' (0 - 3 is -3 and myalph[-3] is 'X'.
The loop then continues to test all the other letters of the alphabet against mylist[b], so eventually it gets to 'X', sees that the letter matches and sets mylist[b] to 'U'.
The loop continues to test the remaining letters of the alphabet against mylist[b], and reaches the end without further matches.
At the very least you need to break out of the loop when you have shifted a letter.
But rather than loop, you could use the str.find() method (directly on the alph string) to find a matching index for the letter; it'll be set to -1 if the letter is not found at all:
for b in range(len(mylist)):
c = alph.find(s[b])
if c > -1: # the letter exists
s[b] = alph[c - K]
Aside from that, there are some other improvements you could make:
You can loop over and index into strings directly, there is no need to turn alph into a list here. When you do need to to turn a string into a list of individual characters, you should use list(stringobject). So mylist = list(s) would suffice.
myres is already a list of strings, there is no need to convert each to a string again.
Rather than put all the letters from s into a list, then adding the whole mylist list to res to myres, you could just directly append each letter you processed to myres; that also removes the need to alter myres.
Python variable names do not need to be limited to single characters. Use more descriptive names so that it is easier to understand what your code does when you return to it later.
Taken together, that'd lead to:
alphabet = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
parts, key = [int(s) for s in input().split()]
results = []
for part in range(parts):
characters = input()
for character in enumerate(characters):
letter_idx = alphabet.index(character)
if letter_idx > -1:
# this is a letter in the alphabet, shift it with the key
character = alphabet[letter_idx - key]
results.append(character)
print("".join(results))
I am learning Python and while practicing, I have this type of input:
1 2 3 4 S
3 4 5 6 N
3 4 1 8 S
It will always be a set of numbers separeted by space and end with a letter (char).
I am reading it with input() and making it a list like this
data = input()
myList = data.split()
What I want is to take the list and put only the numbers to a new list.
I tried this, but it only works if the list only has int values:
myList = [int(i) for i in myList]
How can I only take the int values and put them in a new list.
NOTE: My data input always ends with a letter but it will be nice if I someone gives a solution where there are letters at any random index of the list. Thanks in advance
You just need to add a filter to the comprehension.
msj = [int(i) for i in msg if i.isdigit()]
I would do it like this
new_list = []
for item in myList:
if item.isdigit():
new_list.append(int(item))
You can take advantage of the fact that the letter will always appear at the end of the string:
newList = [int(n) for n in myList[:-1]]
I just started learning Python and want to create a simple script that will read integer numbers from user input and print their sum.
The code that I wrote is
inflow = list(map(int, input().split(" ")))
result = 1
for i in inflow:
result += inflow[i]
print(result)
It gives me an error
IndexError: list index out of range
pointing to result += inflow[i] line. Can't see what am I doing wrong?
BTW, is there more elegant way to split input flow to the list of integers?
You can also avoid the loop altogether:
inflow = '1 2 34 5'
Then
sum(map(int, inflow.split()))
will give the expected value
42
EDIT:
As you initialize your result with 1 and not 0, you can then also simply do:
sum(map(int, input.split()), 1)
My answer assumes that the input is always valid. If you want to catch invalid inputs, check Anton vBR's answer.
Considering: I just started learning Python
I'd suggest something like this, which handle input errors:
inflow = raw_input("Type numbers (e.g. '1 3 5') ") #in py3 input()
#inflow = '1 2 34 5'
try:
nums = map(int, inflow.split())
result = sum(nums)
print(result)
except ValueError:
print("Not a valid input")
for i in list gives the values of the list, not the index.
inflow = list(map(int, input().split(" ")))
result = 1
for i in inflow:
result += i
print(result)
If you wanted the index and not the value:
inflow = list(map(int, input().split(" ")))
result = 1
for i in range(len(inflow)):
result += inflow[i]
print(result)
And finally, if you wanted both:
for index, value in enumerate(inflow):
script that will read integer numbers from user input and print their
sum.
try this :
inflow = list(map(int, input().split(" ")))
result = 0
for i in inflow:
result += i
print(result)
If you were going to index the list object you would do:
for i in range(len(inflow)):
result += inflow[i]
However, you are already mapping int and turning the map object into a list so thus you can just iterate over it and add up its elements:
for i in inflow:
result += i
As to your second question, since you arent doing any type testing upon cast (i.e. what happens if a user supplies 3.14159; in that case int() on a str that represents a float throws a ValueError), you can wrap it all up like so:
inflow = [int(x) for x in input().split(' ')] # input() returns `str` so just cast `int`
result = 1
for i in inflow:
result += i
print(result)
To be safe and ensure inputs are valid, I'd add a function to test type casting before building the list with the aforementioned list comprehension:
def typ(x):
try:
int(x)
return True # cast worked
except ValueError:
return False # invalid cast
inflow = [x for x in input().split(' ') in typ(x)]
result = 1
for i in inflow:
result += i
print(result)
So if a user supplies '1 2 3 4.5 5 6.3 7', inflow = [1, 2, 3, 5, 7].
What your code is doing, in chronological order, is this:
Gather user input, split per space and convert to integer, and lastly store in inflow
Initialize result to 1
Iterate over every item in inflow and set item to i
Add the current item, i to result and continue
After loop is done, print the result total.
The broken logic would be at step 4.
To answer your question, the problem is that you're not gathering the index from the list as expected-- you're iterating over each value in the list opposed to the index, so it's really undefined behavior based on what the user inputs. Though of course, it isn't what you want.
What you'd want to do is:
inflow = list(map(int, input().split(" ")))
result = 1
for i in range(len(inflow)):
result += inflow[i]
print(result)
And on your last regard; there are two real ways to do it, one being the way you're doing right now using list, map and:
inflow = [int(v) for v in input().split()]
IMO the latter looks better. Also a suggestion; you could call stdlib function sum(<iterable>) over a list of integers or floats and it'll add each number and return the summation, which would appear more clean over a for loop.
Note: if you're splitting per whitespace you can just call the str.split() function without any parameters, as it'll split at every whitespace regardless.
>>> "hello world!".split()
['hello', 'world!']
Note: also if you want extra verification you could add a bit more logic to the list comprehension:
inflow = [int(v) for v in input().split() if v.isdigit()]
Which checks whether the user inputted valid data, and if not, skip past the data and check the following.