Good evening,
I'm wondering how i can exclude certain characters from a frequency table?
first i read the file, creates a frequency table. after this i change it to a tuple to be able to get a percentage of occourence for each letter.
however i am wondering how i can implement that it does not count certain letters.
ie. an exclude list.
with open('test.txt', 'r') as file:
data = file.read().replace('\n', '')
frequency_table = {char : data.count(char) for char in set(data)}
x0= ("Character frequency table for '{}' is :\n {}".format(data, str(frequency_table)))
from collections import Counter
res = [(*key, val) for key, val in Counter(frequency_table).most_common()]
print("Frequency Tuple list : " + str(res))
print(res[1][1]/res[1][1])#
Sounds like you need an if at the end of your dictionary comprension:
frequency_table = {char : data.count(char) for char in set(data) if char not in EXCLUDE}
You can then set your EXCLUDE as, for example:
a list, i.e. ['a', 'b', 'c', 'd'] or list('abcd')
or, you can just use a string of characters directly, such as 'abcd'.
>>> data = 'aaaabcdefededefefedf'
>>> EXCLUDE_LIST = 'a'
>>> frequency_table = {char : data.count(char) for char in set(data) if char not in EXCLUDE_LIST}
>>> frequency_table
{'b': 1, 'c': 1, 'e': 6, 'f': 4, 'd': 4}
Add an if to your comprehension:
exclude = {'a', 'r'}
frequency_table = {char: data.count(char) for char in set(data) if char not in exclude}
Alternatively, use the difference between two sets:
exclude = {'a', 'r'}
frequency_table = {char: data.count(char) for char in set(data) - exclude}
'L134e2t1C1o1d1e1'
the original string is "LeetCode"
but I need to separate strings from digits, digits can be not only single-digit but also 3-4 digits numbers like 345.
My code needs to separate into dict of key values; keys are characters and numbers is the digit right after the character. Also create 2 lists of separate digits, letters only.
expected output:
letters = ['L', 'e', 't', 'C', 'o', 'd', 'e']
digits = [134,2,1,1,1,1,1]
This code is not properly processing this.
def f(s):
d = dict()
letters = list()
# letters = list(filter(lambda x: not x.isdigit(), s))
i = 0
while i < len(s):
print('----------------------')
if not s[i].isdigit():
letters.append(s[i])
else:
j = i
temp = ''
while j < len(s) and s[j].isdigit():
j += 1
substr = s[i:j]
print(substr)
i += 1
print('----END -')
print(letters)
With the following modification your function separates letters from digits in s:
def f(s):
letters = list()
digits = list()
i = 0
while i < len(s):
if not s[i].isdigit():
letters.append(s[i])
i += 1
else:
j = i
temp = ''
while j < len(s) and s[j].isdigit():
j += 1
substr = s[i:j]
i = j
digits.append(substr)
print(letters)
print(digits)
f('L134e2t1C1o1d1e1')
As said in my comments you didn't update i after the inner loop terminates which made i go back to a previous and already processed index.
If I would be limited to not use regex I would do it following way
text = 'L134e2t1C1o1d1e1'
letters = [i for i in text if i.isalpha()]
digits = ''.join(i if i.isdigit() else ' ' for i in text).split()
print(letters)
print(digits)
output
['L', 'e', 't', 'C', 'o', 'd', 'e']
['134', '2', '1', '1', '1', '1', '1']
Explanation: for letters I use simple list comprehension with condition, .isalpha() is str method which check if string (in this consisting of one character) is alphabetic. For digits (which should be rather called numbers) I replace non-digits using single space, turn that into string using ''.join then use .split() (it does split on one or more whitespaces). Note that digits is now list of strs rather than ints, if that is desired add following line:
digits = list(map(int,digits))
Your string only had two e's, so I've added one more to complete the example. This is one way you could do it:
import re
t = 'L1e34e2t1C1o1d1e1'
print(re.sub('[^a-zA-Z]', '', t))
Result:
LeetCode
I know you cannot use regex, but to complete this answer, I'll just add a solution:
def f(s):
d = re.findall('[0-9]+', s)
l = re.findall('[a-zA-Z]', s)
print(d)
print(l)
f(t)
Result:
['134', '2', '1', '1', '1', '1', '1']
['L', 'e', 't', 'C', 'o', 'd', 'e']
You edited your question and I got a bit confused, so here is a really exhaustive code giving you a list of letters, list of the numbers, the dict with the number associated with the number, and finally the sentence with corresponding number of characters ...
def f(s):
letters = [c for c in s if c.isalpha()]
numbers = [c for c in s if c.isdigit()]
mydict = {}
currentKey = ""
for c in s:
print(c)
if c.isalpha():
mydict[c] = [] if c not in mydict.keys() else mydict[c]
currentKey = c
elif c.isdigit():
mydict[currentKey].append(c)
sentence = ""
for i in range(len(letters)):
count = int(numbers[i])
while count > 0:
sentence += letters[i]
count -= 1
print(letters)
print(numbers)
print(mydict)
print(sentence)
letters = []
digits = []
dig = ""
for letter in 'L134e2t1C1o1d1e1':
if letter.isalpha():
# do not add empty string to list
if dig:
# append dig to list of digits
digits.append(dig)
dig = ""
letters.append(letter)
# if it is a actual letter continue
continue
# add digits to `dig`
dig = dig + letter
Try this. The idea is to skip all actual letters and add the digits to dig.
I know there's an accepted answer but I'll throw this one in anyway:
letters = []
digits = []
lc = 'L134e2t1C1o1d1e1'
n = None
for c in lc:
if c.isalpha():
if n is not None:
digits.append(n)
n = None
letters.append(c)
else:
if n is None:
n = int(c)
else:
n *= 10
n += int(c)
if n is not None:
digits.append(n)
for k, v in zip(letters, digits):
dct.setdefault(k, []).append(v)
print(letters)
print(digits)
print(dct)
Output:
['L', 'e', 't', 'C', 'o', 'd', 'e']
[134, 2, 1, 1, 1, 1, 1]
{'L': [134], 'e': [2, 1], 't': [1], 'C': [1], 'o': [1], 'd': [1]}
I have a string as follows:
f = 'ATCTGTCGTYCACGT'
I want to check whether the string contains any characters except: A, C, G or T, and if so, print them.
for i in f:
if i != 'A' and i != 'C' and i != 'G' and i != 'T':
print(i)
Is there a way to achieve this without looping through the string?
You can use set to achieve the desired output.
f = 'ATCTGTCGTYCACGTXYZ'
not_valid={'A', 'C', 'G' , 'T'}
unique=set(f)
print(unique-not_valid)
output
{'Y','X','Z'} #characters in f which are not equal to 'A','C','G','T'
Depending on the size of your input string, the for loop might be the most efficient solution.
However, since you explicitly ask for a solution without an explicit loop, this can be done with a regex.
import re
f = 'ABCDEFG'
print(*re.findall('[^ABC]', f), sep='\n')
Outputs
D
E
F
G
Just do
l = ['A', 'C', 'G', 'T']
for i in f:
if i not in l:
print(i)
It checks whether the list contains a char of the list
If you don't want to loop through the list you can do:
import re
l = ['A', 'C', 'G', 'T']
contains = bool(re.search("%s" % "[" + "".join(l) + "]", f))
Technically this loops but we convert your input string to a set which removes duplicate values
accepted_values = ['a','t','c','g']
input = 'ATCTGTCGTYCACGT'
print([i for i in set(input.lower()) if i not in accepted_values])
I am very new to python and trying to find the solution to this for a class.
I need the function missing_letters to take a list, check the letters using histogram and then loop over the letters in alphabet to determine which are missing from the input parameter. Finally I need to print the letters that are missing, in a string.
alphabet = "abcdefghijklmnopqrstuvwxyz"
test = ["one","two","three"]
def histogram(s):
d = dict()
for c in s:
if c not in d:
d[c] = 1
else:
d[c] += 1
return d
def missing_letter(s):
for i in s:
checked = (histogram(i))
As you can see I haven't gotten very far, at the moment missing_letters returns
{'o': 1, 'n': 1, 'e': 1}
{'t': 1, 'w': 1, 'o': 1}
{'t': 1, 'h': 1, 'r': 1, 'e': 2}
I now need to loop over alphabet to check which characters are missing and print. Any help and direction will be much appreciated. Many thanks!
You can use set functions in python, which is very fast and efficient:
alphabet = set('abcdefghijklmnopqrstuvwxyz')
s1 = 'one'
s2 = 'two'
s3 = 'three'
list_of_missing_letters = set(alphabet) - set(s1) - set(s2) - set(s3)
print(list_of_missing_letters)
Or like this:
from functools import reduce
alphabet = set('abcdefghijklmnopqrstuvwxyz')
list_of_strings = ['one', 'two', 'three']
list_of_missing_letters = set(alphabet) - \
reduce(lambda x, y: set(x).union(set(y)), list_of_strings)
print(list_of_missing_letters)
Or using your own histogram function:
alphabet = "abcdefghijklmnopqrstuvwxyz"
test = ["one", "two", "three"]
def histogram(s):
d = dict()
for c in s:
if c not in d:
d[c] = 1
else:
d[c] += 1
return d
def missing_letter(t):
test_string = ''.join(t)
result = []
for l in alphabet:
if l not in histogram(test_string).keys():
result.append(l)
return result
print(missing_letter(test))
Output:
['a', 'b', 'c', 'd', 'f', 'g', 'i', 'j', 'k', 'l', 'm', 'p', 'q', 's', 'u', 'v', 'x', 'y', 'z']
from string import ascii_lowercase
words = ["one","two","three"]
letters = [l.lower() for w in words for l in w]
# all letters not in alphabet
letter_str = "".join(x for x in ascii_lowercase if x not in letters)
Output:
'abcdfgijklmpqsuvxyz'
It is not the easiest question to understand, but from what I can gather you require all the letters of the alphabet not in the input to be returned in console.
So a loop as opposed to functions which have been already shown would be:
def output():
output = ""
for i in list(alphabet):
for key in checked.keys():
if i != key:
if i not in list(output):
output += i
print(output)
Sidenote: Please either make checked a global variable or put it outside of function so this function can use it
I have a 26-digit list. I want to print out a list of alphabets according to the numbers. For example, I have a list(consisting of 26-numbers from input):
[0,0,0,0,2,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0]
I did like the output to be like this:
[e,e,l,s]
'e' is on the output 2-times because on the 4-th index it is the 'e' according to the English alphabet formation and the digit on the 4-th index is 2. It's the same for 'l' since it is on the 11-th index and it's digit is 1. The same is for s. The other letters doesn't appear because it's digits are zero.
For example, I give another 26-digit input. Like this:
[1,2,2,3,4,0,3,4,4,1,3,1,4,4,1,0,0,0,0,0,4,2,3,2,2,1]
The output should be:
[a,b,b,c,c,d,d,d,e,e,e,e,g,g,g,h,h,h,h,i,i,i,i,j,k,k,k,l,m,m,m,m,n,n,n,n,o,u,u,u,u,v,v,w,w,w,x,x,y,y,z]
Is, there any possible to do this in Python 3?
You can use chr(97 + item_index) to get the respective items and then multiply by the item itself:
In [40]: [j * chr(97 + i) for i, j in enumerate(lst) if j]
Out[40]: ['ee', 'l', 's']
If you want them separate you can utilize itertools module:
In [44]: from itertools import repeat, chain
In [45]: list(chain.from_iterable(repeat(chr(97 + i), j) for i, j in enumerate(lst) if j))
Out[45]: ['e', 'e', 'l', 's']
Yes, it is definitely possible in Python 3.
Firstly, define an example list (as you did) of numbers and an empty list to store the alphabetical results.
The actual logic to link with the index is using chr(97 + index), ord("a") = 97 therefore, the reverse is chr(97) = a. First index is 0 so 97 remains as it is and as it iterates the count increases and your alphabets too.
Next, a nested for-loop to iterate over the list of numbers and then another for-loop to append the same alphabet multiple times according to the number list.
We could do this -> result.append(chr(97 + i) * my_list[i]) in the first loop itself but it wouldn't yield every alphabet separately [a,b,b,c,c,d,d,d...] rather it would look like [a,bb,cc,ddd...].
my_list = [1,2,2,3,4,0,3,4,4,1,3,1,4,4,1,0,0,0,0,0,4,2,3,2,2,1]
result = []
for i in range(len(my_list)):
if my_list[i] > 0:
for j in range(my_list[i]):
result.append(chr(97 + i))
else:
pass
print(result)
An alternative to the wonderful answer by #Kasramvd
import string
n = [0,0,0,0,2,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0]
res = [i * c for i, c in zip(n, string.ascii_lowercase) if i]
print(res) # -> ['ee', 'l', 's']
Your second example produces:
['a', 'bb', 'cc', 'ddd', 'eeee', 'ggg', 'hhhh', 'iiii', 'j', 'kkk', 'l', 'mmmm', 'nnnn', 'o', 'uuuu', 'vv', 'www', 'xx', 'yy', 'z']
Splitting the strings ('bb' to 'b', 'b') can be done with the standard schema:
[x for y in something for x in y]
Using a slightly different approach, which gives the characters individually as in your example:
import string
a = [0,0,0,0,2,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0]
alphabet_lookup = np.repeat(np.arange(len(a)), a)
letter_lookup = np.array(list(string.ascii_lowercase))
res = letter_lookup[alphabet_lookup]
print(res)
To get
['e' 'e' 'l' 's']