I have a list,
A = ['A','B','C','D','E','F','G','H']
if user input x = 4, then I need an output that shows every value that is 4 distance away from each other.
If starting from 'A' after printing values that are 4 distance away from each other ie: {'A', 'E'}, the code should iterate back and start from 'B' to print all values from there ie: {'B', 'F'}
No number can be in more than one group
Any help is going to be appreciated since I am very new to python.
this is what I have done
x = input("enter the number to divide with: ")
A = ['A','B','C','D','E','F','G','H']
print("Team A is divided by " +x+ " groups")
print("---------------------")
out = [A[i] for i in range(0, len(A), int(x))]
print(out)
My code is printing only the following when user input x =4
{'A', 'E'}
But I need it to look like the following
{'A', 'E'}
{'B', 'F'}
{'C', 'G'}
{'D', 'H'}
what am I doing wrong?
Use zip:
out = list(zip(A, A[x:]))
For example:
x = 4 # int(input("enter the number to divide with: "))
A = ['A','B','C','D','E','F','G','H']
print(f"Team A is divided by {x} groups")
print("---------------------")
out = list(zip(A, A[x:]))
print(out)
Outputs:
[('A', 'E'), ('B', 'F'), ('C', 'G'), ('D', 'H')]
Here you have the live example
If you want to keep the comprehension:
out = [(A[i], A[i+x]) for i in range(0, len(A)-x)]
**You can find my answer below.
def goutham(alist):
for passchar in range(0,len(alist)-4):
i = alist[passchar]
j = alist[passchar+4]
print("{"+i+","+j+"}")
j = 0
alist = ['a','b','c','d','e','f','g','h']
goutham(alist)
Related
I'm trying to find a way to iterate through a text file and list to find character frequency. I understand that I could use Count() for this. But Count() gives everything including spaces periods and whatnots. Also it does not show the character frequency in alphabetical order. I found a way to do it and it works but not really. I'll explain later. Also when I try to put the frequency I get a KeyError. I'll also explain.
I don't want to put my whole project on here so I'll explain some stuff first. I have a separate list called alphabet_list which includes the alphabet. There's a text file that is already read through and converted into uppercase called new_text.
Character frequency Code:
for i in range(len(alphabet_list)):
for c in new_text:
if c == alphabet_list[i]:
count += 1
else:
count = 0
print(alphbet_list[i] + " " + str(count)
i += 1
Output
A 0
A 0
.
.
.
A 1
A 0
.
.
.
B 0
.
.
.
B 1
B 2
B 0
.
.
.
Z 0
P.S the str(count) is temporarily there because I want to see how it looks like print out, I needed to store the result in dictionary
My output would be that, like I said it works but not really. It will iterate but it iterates through every letter and prints out the result already and does not iterate the whole text file and just print final result. It will add to the result if there is another letter same as before right next to each other. Ex (... bb...) it will be B 1, B 2 like shown in my output. And for some reason when I use return it doesn't work. It returns nothing and just ends the program.
Second Code with KeyError:
I skipped the problem on top because I couldn't find the answer and didn't want to waste my time but ran into another problem lol*
for i in range(len(alphabet_list)):
for c in new_text:
if c == alphabet_list[i]:
count += 1
else:
count = 0
c_freq[alphabet_list[i]] == count
print(c_freq)
i += 1
This one was pretty simple I got a KeyError: 'A'.
I tried only doing the
i = 3 #just random number to test
count = 50
c_freq[alphabet_list[i]] == count
print(c_freq)
and it works, so I'm thinking that problem is also related to the problem above(? maybe). Anyways any help would be great. Thanks!
Sorry for long question but I really needed help.
This should help you:
lst = ['A', 'Z', 'H', 'A', 'B', 'N', 'H', 'Y', '.' , ',','Z'] #Initial list. Note: The list also includes characters such as commas and full stops.
alpha_dict = {}
for ch in lst:
if ch.isalpha(): #Checks if the character is an alphabet
if ch in alpha_dict.keys():
alpha_dict[ch] += 1 #If key already exists, value is incremented by 1
else:
alpha_dict[ch] = 1 #If key does not exist, a new key is created with value 1
print(alpha_dict)
Output:
{'A': 2, 'Z': 2, 'H': 2, 'B': 1, 'N': 1, 'Y': 1}
Since you want the output to be sorted in alphabetical order, add these lines to your code:
key_list = list(alpha_dict.keys()) #Creates a list of all the keys in the dict
key_list.sort() #Sorts the list in alphabetical order
final_dict = {}
for key in key_list:
final_dict[key] = alpha_dict[key]
print(final_dict)
Output:
{'A': 2, 'B': 1, 'H': 2, 'N': 1, 'Y': 1, 'Z': 2}
Thus, here is the final code:
lst = ['A', 'Z', 'H', 'A', 'B', 'N', 'H', 'Y', '.' , ',','Z']
alpha_dict = {}
for ch in lst:
if ch.isalpha():
if ch in alpha_dict.keys():
alpha_dict[ch] += 1
else:
alpha_dict[ch] = 1
key_list = list(alpha_dict.keys())
key_list.sort()
final_dict = {}
for key in key_list:
final_dict[key] = alpha_dict[key]
print(final_dict)
Output:
{'A': 2, 'B': 1, 'H': 2, 'N': 1, 'Y': 1, 'Z': 2}
Here in the program how can you find the second repetitive character in the string. for ex:abcdaabdefaggcbd"
Output : d (because 'd' occurred 3 times where 'a' occurred 4 times)
how can I get the output, please help me.
Given below is my code:
s="abcdaabdefaggcbd"
d={}
for i in s:
d[i] = d.get(i,0)+1
print(d,"ddddd")
max2 = 0
for k,v in d.items():
if(v>max2 and v<max(d.values())):
max2=v
if max2 in d.values():
print k,"kkk"
The magnificent Python Counter and its most_common() method are very handy here.
import collections
my_string = "abcdaabdefaggcbd"
result = collections.Counter(my_string).most_common()
print(result[1])
Output
('b', 3)
In case you need to capture all the second values (if you have more than one entry) you can use the following:
import collections
my_string = "abcdaabdefaggcbd"
result = collections.Counter(my_string).most_common()
second_value = result[1][1]
seconds = []
for item in result:
if item[1] == second_value:
seconds.append(item)
print(seconds)
Output
[('b', 3), ('d', 3)]
I also wanted to add an example of solving the problem using a methodology more similar to the one that you showed in your question:
my_string="abcdaabdefaggcbd"
result={}
for character in my_string:
if character in result:
result[character] = result.get(character) + 1
else:
result[character] = 1
sorted_data = sorted([(value,key) for (key,value) in result.items()])
second_value = sorted_data[-2][0]
result = []
for item in sorted_data:
if item[0] == second_value:
result.append(item)
print(result)
Output
[(3, 'b'), (3, 'd')]
Ps
Please forgive me if I took the freedom to change variable names but I think that in this way my answer will be more readable for a broader audience.
Sort the dict's items on their values (descending) and get the second item:
>>> from collections import Counter
>>> c = Counter("abcdaabdefaggcbd")
>>> vals = sorted(c.items(), key=lambda item:item[1], reverse=True)
>>> vals
[('a', 4), ('b', 3), ('d', 3), ('c', 2), ('g', 2), ('e', 1), ('f', 1)]
>>> print(vals[1])
('b', 3)
>>>
EDIT:
or just use Counter.most_common():
>>> from collections import Counter
>>> c = Counter("abcdaabdefaggcbd")
>>> print(c.most_common()[1])
Both b and d are second most repetitive. I would think that both should be displayed. This is how I would do it:
Code:
s="abcdaabdefaggcbd"
d={}
for i in s:
ctr=s.count(i)
d[i]=ctr
fir = max(d.values())
sec = 0
for j in d.values():
if(j>sec and j<fir):
sec = j
for k,v in d.items():
if v == sec:
print(k,v)
Output:
b 3
d 3
in order to find the second most repetitive character in string you can very well use collections.Counter()
Here's an example:
import collections
s='abcdaabdefaggcbd'
count=collections.Counter(s)
print(count.most_common(2)[1])
Output: ('b', 3)
You can do a lot with Counter(). Here's a link for a further read:
More about Counter()
I hope this answers your question. Cheers!
A string = 1 2 3 4
Program should return = [[1,2],[3,4]]
in python
I want the string to be converted into a list of every two element from string
You could go for something very simple such as:
s = "10 2 3 4 5 6 7 8"
l = []
i = 0
list_split_str = s.split() # splitting the string according to spaces
while i < len(s) - 1:
l.append([s[i], s[i + 1]])
i += 2
This should output:
[['10', '2'], ['3', '4'], ['5', '6'], ['7', '8']]
You could also do something a little more complex like this in a two-liner:
list_split = s.split() # stripping spaces from the string
l = [[a, b] for a, b in zip(list_split[0::2], list_split[1::2])]
The slice here means that the first list starts at index zero and has a step of two and so is equal to [10, 3, 5, ...]. The second means it starts at index 1 and has a step of two and so is equal to [2, 4, 6, ...]. So we iterate over the first list for the values of a and the second for those of b.
zip returns a list of tuples of the elements of each list. In this case, [('10', '2'), ('3', '4'), ('5', '6'), ...]. It allows us to group the elements of the lists two by two and iterate over them as such.
This also works on lists with odd lengths.
For example, with s = "10 2 3 4 5 6 7 ", the above code would output:
[['10', '2'], ['3', '4'], ['5', '6']]
disregarding the 7 since it doesn't have a buddy.
here is the solution if the numbers exact length is divisible by 2
def every_two_number(number_string):
num = number_string.split(' ')
templist = []
if len(num) % 2 == 0:
for i in range(0,len(num),2):
templist.append([int(num[i]),int(num[i+1])])
return templist
print(every_two_number('1 2 3 4'))
you can remove the if condition and enclosed the code in try and except if you want your string to still be convert even if the number of your list is not divisible by 2
def every_two_number(number_string):
num = number_string.split(' ')
templist = []
try:
for i in range(0,len(num),2):
templist.append([int(num[i]),int(num[i+1])])
except:
pass
return templist
print(every_two_number('1 2 3 4 5'))
I've been trying to solve this homework problem for days, but can't seem to fix it. I started my study halfway through the first semester, so I can't ask the teacher yet and I hope you guys can help me. It's not for grades, I just want to know how.
I need to write a program that reads a string and converts the triplets abc into bca. Per group of three you need to do this. For examplekatzonbecomesatkonz`.
The closest I've gotten is this:
string=(input("Give a string: "))
for i in range(0, len(string)-2):
a = string[i]
b = string[i + 1]
c = string[i + 2]
new_string= b, c, a
i+=3
print(new_string)
The output is:
('a', 't', 'k')
('t', 'z', 'a')
('z', 'o', 't')
('o', 'n', 'z')
The code below converts for example "abc" to "bca". It works for any string containing triplets. Now, if input is "abcd", it is converted to "bcad". If you input "katzon", it is converted to "atkonz". This is what I understood from your question.
stringX = input()
# create list of words in the string
listX = stringX.split(" ")
listY = []
# create list of triplets and non-triplets
for word in listX:
listY += [[word[i:i+3] for i in range(0, len(word), 3)]]
# convert triplets, for example: "abc" -> "bca"
for listZ in listY:
for item in listZ:
if len(item)==3:
listZ[listZ.index(item)] = listZ[listZ.index(item)][1:] + listZ[listZ.index(item)][0]
listY[listY.index(listZ)] = "".join(listZ)
# create final string
stringY = " ".join(listY)
print(stringY)
I have a list that contains multiple sets of strings, and I would like to find the symmetric difference between each string and the other strings in the set.
For example, I have the following list:
targets = [{'B', 'C', 'A'}, {'E', 'C', 'D'}, {'F', 'E', 'D'}]
For the above, desired output is:
[2, 0, 1]
because in the first set, A and B are not found in any of the other sets, for the second set, there are no unique elements to the set, and for the third set, F is not found in any of the other sets.
I thought about approaching this backwards; finding the intersection of each set and subtracting the length of the intersection from the length of the list, but set.intersection(*) does not appear to work on strings, so I'm stuck:
set1 = {'A', 'B', 'C'}
set2 = {'C', 'D', 'E'}
set3 = {'D', 'E', 'F'}
targets = [set1, set2, set3]
>>> set.intersection(*targets)
set()
The issue you're having is that there are no strings shared by all three sets, so your intersection comes up empty. That's not a string issue, it would work the same with numbers or anything else you can put in a set.
The only way I see to do a global calculation over all the sets, then use that to find the number of unique values in each one is to first count all the values (using collections.Counter), then for each set, count the number of values that showed up only once in the global count.
from collections import Counter
def unique_count(sets):
count = Counter()
for s in sets:
count.update(s)
return [sum(count[x] == 1 for x in s) for s in sets]
Try something like below:
Get symmetric difference with every set. Then intersect with the given input set.
def symVal(index,targets):
bseSet = targets[index]
symSet = bseSet
for j in range(len(targets)):
if index != j:
symSet = symSet ^ targets[j]
print(len(symSet & bseSet))
for i in range(len(targets)):
symVal(i,targets)
Your code example doesn't work because it's finding the intersection between all of the sets, which is 0 (since no element occurs everywhere). You want to find the difference between each set and the union of all other sets. For example:
set1 = {'A', 'B', 'C'}
set2 = {'C', 'D', 'E'}
set3 = {'D', 'E', 'F'}
targets = [set1, set2, set3]
result = []
for set_element in targets:
result.append(len(set_element.difference(set.union(*[x for x in targets if x is not set_element]))))
print(result)
(note that the [x for x in targets if x != set_element] is just the set of all other sets)