Input : abcdABCD
Output : AaBbCcDd
ms=[]
n = input()
for i in n:
ms.append(i)
ms.sort()
print(ms)
It gives me ABCDabcd.
How to sort this in python?
Without having to import anything, you could probably do something like this:
arr = "abcdeABCDE"
temp = sorted(arr, key = lambda i: (i.lower(), i))
result = "".join(temp)
print(result) # AaBbCcDdEe
The key will take in each element of arr and sort it first by lower-casing it, then if it ties, it will sort it based on its original value. It will group all similar letters together (A with a, B with b) and then put the capital first.
Use a sorting key:
ms = "abcdABCD"
sorted_ms = sorted(ms, key=lambda letter:(letter.upper(), letter.islower()))
# sorted_ms = ['A', 'a', 'B', 'b', 'C', 'c', 'D', 'd']
sorted_str = ''.join(sorted_ms)
# sorted_str = 'AaBbCcDd'
Why this works:
You can specify the criteria by which to sort by using the key argument in the sorted function, or the list.sort() method - this expects a function or lambda that takes the element in question, and outputs a new criteria by which to sort it. If that "new criteria" is a tuple, then the first element takes precedence - if it's equal, then the second argument, and so on.
So, the lambda I provided here returns a 2-tuple:
(letter.upper(), letter.islower())
letter.upper() as the first element here means that the strings are going to be sorted lexigraphically, but case-insensitively (as it will sort them as if they were all uppercase). Then, I use letter.islower() as the second argument, which is True if the letter was lowercase and False otherwise. When sorting, False comes before True - which means that if you give a capital letter and a lowercase letter, the capital letter will come first.
Try this:
>>>s='abcdABCD'
>>>''.join(sorted(s,key=lambda x:x.lower()))
'aAbBcCdD'
Related
I have an alphabets list:
alpha_list = ['a', 'b', 'c', 'd', 'e']
For a given alphabet(considering it will always be present in alpha_list) I want to get an alphabet whose index is garter by a given number, consider below function for example:
def get_replacing_letter(alphabet, number):
index = alpha_list.index(alphabet)
return alpha_list[index + number]
get_replacing_letter('a', 2) will give me 'c'
what I want is get_replacing_letter('d', 2) should give 'a'
similarly get_replacing_letter('e', 2) should give 'b'
So the alph_list should work in a chaining sequence or cyclic manner. I am wondering how to achieve this in Python?
You can make new index take the modulo of the length of the list:
return alpha_list[(index + number) % len(alpha_list)]
At the moment I achieved it by extending the alpha_list with itself:
alpha_list.extend(alpha_list)
This way as index method gives first index of occurance, I could manage to get it working.
However I am wondering if there is any better way to achieve this.
Problem to solve:
Write a function that will find all the anagrams of a word from a list. You will be given two inputs a word and an array with words. You should return an array of all the anagrams or an empty array if there are none.
Solution Tested:
a = ['aabb', 'abcd', 'bbaa', 'dada']
b = ['abab']
listA = []
sorted_defaultword = sorted(b[0])
print (sorted_defaultword)
for i in range (len(a)):
#print(a[i])
sorted_word = sorted(a[i])
#print (sorted_word)
if (sorted_word == sorted_defaultword):
listA.append(a[i])
print (listA)
Test Output:
['a', 'a', 'b', 'b']
['aabb', 'bbaa']
Using the test, I then tried to write my function but apparently it will not work. Can someone please suggest why:
def anagrams(word, words):
sorted_defaultword = sorted(word[0])
anagram_List = []
for i in range (len(words)):
sorted_word = sorted(words[i])
if (sorted_word == sorted_defaultword):
anagram_List.append(words[i])
return anagram_List
Why is this failing when I put it in a function?
You are passing wrong arguments to the function.
Test.assert_equals(anagrams('abba', ['aabb', 'abcd', 'bbaa', 'dada']), ['aabb', 'bbaa']
here you are passing the first parameter as a string. while the function expects a list.
Change your code to:
Test.assert_equals(anagrams(['abba'], ['aabb', 'abcd', 'bbaa', 'dada']), ['aabb', 'bbaa']
note that I have just passed 'abba' in a list, because your function expects it to be a list.
If you want to use your previous code, from your function change this line sorted_defaultword = sorted(word[0]) to sorted_defaultword = sorted(word)
And this should do the job...
Problem:
So I was trying to alphabetically sort my list of strings maybe I overlooked something very minor. I have tried both .sort and sorted() but maybe I didn't do it correctly?
Here is my Code:
words = input("Words: ")
list1 = []
list1.append(words.split())
print(sorted(list1))
Expected output-
Input: "a b d c"
Output: ['a', 'b', 'c', 'd']
Current output-
Input: "a b d c"
Output: [['a', 'b', 'd', 'c']]
Your code is not working because you are trying to sort a list inside a list.
When you call words.split() it returns a list. So when you do list1.append(words.split()) it is appending a list into list1.
You should do this:
words = input("Words: ")
list1 = words.split()
print(sorted(list1))
You can try a simple method as follows:
list1 = [i for i in input('Words: ').split(' ')]
print(sorted(list1))
I've tested it. And it is working
Without deviating from your current effort, the only modification you need to do to fix your code is :
words = input("Words: ")
list1 = []
list1.append(words.split())
print(sorted(list1[0]))
Explanation of what you were doing wrong:
The root cause of your confusion is append() .According to python docs,append() takes exactly one argument.
So when you do this,
words.split()
You are trying to append more than 1 element into the list1 and when you append() something more than 1 in a list, it appends as a nested list (i.e a list inside another list.)
To support my explanation you can see that your code fixed by a simple [0]
print(sorted(list1[0]))
That is because your input is stored as a list of list, AND it is stored in the first index (Point to note - 1st index in a python list is 0, hence the usage of list1[0])
Please let me know if I could have explained it in a more simpler way or if you have any other confusions that aid from the above explanation.
I have a dictionary as follows:
input = {1:'a', 2:'b', 3:'c', 4:'b', 5:'c', 6:'b'}
a - 1 time
b - 3 times
c - 2 times
I want to sort the dictionary in such a way that the value which repeats maximum times will come at fist followed by the value which repeats second most times and so on...
Desired Output
output = {1:'b', 2:'c', 3:'a'}
Please help
Thanks in advance
First of all, you need to realize you do not even care about input dict keys, only about input dict values.
real_input = input.values()
Second of all, you need to count occurences:
counted_items = collections.Counter(real_input)
Third of all, you want to iterate over them in order from most common to least common. .most_common() returns list of (key, count) tuples in expected order.
most_common_in_order = counted_items.most_common()
After that you want to convert that list to dict, with consecutive inegers as keys
result = {i: v for i, (v, _) in zip(itertools.count(1), most_common_in_order)}
Or, in concise form:
result = {i: v for i, (v, _) in zip(itertools.count(1),
collections.Counter(input.values()).most_common())}
Note that dictionaries are inherently not ordered, so actual order is implementation detail, and it's not guaranteed.
from collections import Counter
input = {1: 'a', 2: 'b', 3: 'c', 4: 'b', 5: 'c', 6: 'b'}
output = {index: value for index, (value, _)
in enumerate(Counter(input.values()).most_common(),
start=1)}
print(output)
Edited for Python 3 compatibility.
from collections import Counter
i = {1:'a', 2:'b', 3:'c', 4:'b', 5:'c', 6:'b'}
print (Counter(i.values()))
#Output
Counter({'b': 3, 'c': 2, 'a': 1})
i have the task to get the String 'AAAABBBCCDAABBB' into a list like this: ['A','B','C','D','A','B']
I am working on this for 2 hours now, and i can't get the solution. This is my code so far:
list = []
string = 'AAAABBBCCDAABBB'
i = 1
for i in string:
list.append(i)
print(list)
for element in list:
if list[element] == list[element-1]:
list.remove(list[element])
print(list)
I am a newbie to programming, and the error "TypeError: list indices must be integers or slices, not str" always shows up...
I already changed the comparison
if list[element] == list[element-1]
to
if list[element] is list[element-1]
But the error stays the same. I already googled a few times, but there were always lists which didn't need the string-format, but i need it (am i right?).
Thank you for helping!
NoAbL
First of all don't name your variables after built in python statements or data structures like list, tuple or even the name of a module you import, this also applies to files. for example naming your file socket.py and importing the socket module is definitely going to lead to an error (I'll leave you to try that out by yourself)
in your code element is a string, indexes of an iterable must be numbers not strings, so you can tell python
give me the item at position 2.
but right now you're trying to say give me the item at position A and that's not even valid in English, talk-less of a programming language.
you should use the enumerate function if you want to get indexes of an iterable as you loop through it or you could just do
for i in range(len(list))
and loop through the range of the length of the list, you don't really need the elements anyway.
Here is a simpler approach to what you want to do
s = string = 'AAAABBBCCDAABBB'
ls = []
for i in s:
if ls:
if i != ls[-1]:
ls.append(i)
else:
ls.append(i)
print(ls)
It is a different approach, but your problem can be solved using itertools.groupby as follows:
from itertools import groupby
string = 'AAAABBBCCDAABBB'
answer = [group[0] for group in groupby(string)]
print(answer)
Output
['A', 'B', 'C', 'D', 'A', 'B']
According to the documentation, groupby:
Make an iterator that returns consecutive keys and groups from the iterable
In my example we use a list comprehension to iterate over the consecutive keys and groups, and use the index 0 to extract just the key.
You can try the following code:
list = []
string = 'AAAABBBCCDAABBB'
# remove the duplicate character before append to list
prev = ''
for char in string:
if char == prev:
pass
else:
list.append(char)
prev = char
print(list)
Output:
['A', 'B', 'C', 'D', 'A', 'B']
In your loop, element is the string. You want to have the index.
Try for i, element in enumerate(list).
EDIT: i will now be the index of the element you're currently iterating through.