I'm practicing my python and trying to format the print statement of the below code for match_string so it will print in this format:
There are 5 numbers, 4 letters and 2 other characters.
I've tried to do:
print("There are:", + x.nums, + "numbers", +x.letters,+"letter", + x.other, +"other characters")
and I get the error:
AttributeError: 'tuple' object has no attribute 'nums'
I also think I have an issue with the getDupes part as well but i can't figure out what, it just prints the same as.
Here's my code:
def match_string(words):
nums = 0
letter = 0
other = 0
for i in words :
if i.isalpha():
letter+=1
elif i.isdigit():
nums+=1
else:
other+=1
return nums,letter,other
def getDupes(x):
d = {}
for i in x:
if i in d:
if d[i]:
yield i
d[i] = False
else:
d[i] = True
x = match_string(input("enter a sentence"))
c = getDupes(x)
print(x)
#print("There are:", + str(x.nums), + "numbers", +x.letters,+"letter", + x.other, +"other characters")
print("PRINT",x)
funtion match_string returns a tuple hence cannot be accessed by using x.nums
instead try the below code
nums,letter,other = match_string(input("enter a sentence"))
print("There are:", + nums, + "numbers", + letters,+"letter", + other, +"other characters")
Related
I'm trying to create a function that -given a string- will return the count of non-allowed characters ('error_char'), like so: 'total count of not-allowed / total length of string'.
So far I've tried:
def allowed_characters(s):
s = s.lower()
correct_char = 'abcdef'
error_char = 'ghijklmnopqrstuvwxyz'
counter = 0
for i in s:
if i in correct_char:
no_error = '0'+'/'+ str(len(s))
return no_error
elif i in error_char:
counter += 1
result = str(sum(counter)) + '/' + str(len(s))
return result
but all I get is '0/56' where I'm expecting '22/56' since m,x,y,z are 'not allowed' and m repeats 19 times
allowed_characters('aaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbmmmmmmmmmmmmmmmmmmmxyz')
'0/56'
Then I've tried:
def allowed_characters(s):
s = s.lower()
correct_char = 'abcdef'
error_char = 'ghijklmnopqrstuvwxyz'
counter = 0
for i in s:
if i in correct_char:
no_error = '0'+'/'+ str(len(s))
return no_error
elif i in error_char:
import regex as re
rgx_pattern = re.compile([error_char])
count_e = rgx_pattern.findall(error_char, s)
p_error = sum([count_e.count(i) for i in error_char])
result = str(p_error) + '/' + str(len(s))
But I get the same result...
I've also tried these other ways, but keep getting the same:
def allowed_characters1(s):
s = s.lower()
correct_char = 'abcdef'
for i in s:
if i not in correct_char:
counter = sum([s.count(i) for i in s])
p_error = str(counter) + '/' + str(len(s))
return p_error
elif i in correct_char:
no_error = '0'+'/'+ str(len(s))
return no_error
and...
def allowed_characters2(s):
s = s.lower()
correct_char = 'abcdef'
for i in s:
if i not in correct_char:
counter = sum(s.count(i))
p_error = str(counter) + '/' + str(len(s))
return p_error
elif i in correct_char:
no_error = '0'+'/'+ str(len(s))
return no_error
I've even tried changing the logic and iterating over 'correct/error_char' instead, but nothing seems to work... I keep getting the same result over and over. It looks as though the loop stops right after first character or doesn't run the 'elif' part?
Whenever it comes to do quicker counting - it's always good to think about Counter You can try to simplify your code like this:
Notes - please don't change your Problem Description during the middle of people's answering posts. That make it very hard to keep in-sync.
There is still room to improve it though.
from collections import Counter
def allowed_char(s):
s = s.lower()
correct_char = 'abcdef'
error_char = 'ghijklmnopqrstuvwxyz'
ok_counts = Counter(s)
print(f' allowed: {ok_counts} ')
correct_count = sum(count for c, count in ok_counts.items() if c in correct_char)
error_count = sum(count for c, count in ok_counts.items() if c in error_char)
#return sum(not_ok.values()) / total
return correct_count, error_count # print both
s =('aaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbmmmmmmmmmmmmmmmmmmmxyz')
print(allowed_char(s)) # (34, 22)
print(allowed_char('abdmmmmxyz')) # (3, 7)
Alternatively, you really want to use for-loop and learn to process the string of characters, you could try this:
def loop_count(s):
s = s.lower()
correct_count = error_count = 0
for c in s:
if c in correct_char:
correct_count += 1
else:
error_count += 1
return correct_count, error_count
I would use a regex replacement trick here using len():
def allowed_characters(s):
return len(s) - len(re.sub(r'[^ghijklmnopqrstuvwxyz]+', '', s))
The above returns the length of the input string minus the length of the input with all allowed characters removed (alternatively minus the length of the string with only non allowed characters).
How would I print an element inside a for loop using its index?
flag = "T"
while flag == "T":
x = []
y = []
x1 = int(input())
y1 = int(input())
x.append(x1)
y.append(y1)
for i in range(len(x)):
v = len(x)-1
print(x[v])
print("First value: " + x[v] + "\r\n" + "Second value: " + y[v])
The main question is that printing x with the index of v works in the first print call, but not the second. Why?
You are trying to print integers. Wrap it in str() to be able to print them:
flag = "T"
while flag == "T":
x = []
y = []
x1 = int(input())
y1 = int(input())
x.append(x1)
y.append(y1)
for i in range(len(x)):
v = len(x)-1
print(x[v])
print("First value: " + str(x[v]) + "\r\n" + "Second value: " + str(y[v]))
Don’t bother trying to concatenate elements together only to print them. Use the ability of print to print all its arguments.
print("First value: ", x[v], "\r\nSecond value: ", y[v])
I have to edit the code i previously wrote to make it work with a input, I will attach the problem. I can not seem to get it to work.
it has to be some sort of input,
string =" "
reversed_string = string[::-1]
result_string = " ".join(string)
for a in range (0 ,3):
result_string += chr(ord('a')+a)
for a in range(0 , 2)[::-1]:
result_string += chr(ord('a')+a)
print result_string
string =" "
add this to input from keyboard
input( 'type a letter from a to z')
This is the shortest I could make the answer
character = 97
char_list = []
con = True
a = 0
for interger in range(26):
char = chr(character)
char_list.append(char)
character = character+1
get_char = input("Enter a letter from a-z")
while con == True:
if get_char == char_list[a]:
b = char_list[:a+1]
char_list.reverse()
print(b+char_list[-a:])
con = False
else:
a = a+1
Test Cases
Input: abbbaaccada
Output: ccada
Input: bbccdddcb
Output: (Empty string)
str = input("Enter string: ")
def my_string(string):
if not string:
return ""
if len(string) == 1:
return string
if string[0] == string[1] == string[2]:
return my_string(string[3:])
return string[0] + my_string(string[1:])
print (my_string(str))
I am new to python. and I am trying to remove characters with 3 or more consecutive appearance in a string. In this I could only able to get output of only 1 iteration. e.g. i/p- hhhelllo o/p-eo but for i/p- abbbaaccada o/p is aaaccada but it should be ccada.. please help..
I have done this till 3 repetition but how to generalize it for more than 3 repetition.??
Your problem presents the opportunity to show how else in for loops can be useful. Take a look:
def remover(my_str):
temp = set(my_str)
while True:
for c in temp:
if 3*c in my_str:
my_str = my_str.replace(3*c, '')
break
else:
break
return my_str
test1 = 'abbbaaccada'
print(remover(test1)) # -> ccada
test2 = 'i/p- hhhelllo'
print(remover(test2)) # -> i/p- eo
If you insist on having recursive calls, you can modify the above as follows:
def remover(my_str):
temp = set(my_str)
new_str = my_str
for c in temp:
if 3*c in new_str:
new_str = new_str.replace(3*c, '')
if my_str == new_str:
return new_str
else:
return remover(new_str)
I have added a solution which will work for 3 or more repetition as the above solution didn't work for me. It is a recursive solution.
import re
def format_string(u_str):
f_str = remove_string(u_str)
if f_str == u_str:
return f_str
else:
return format_string(f_str)
def remove_string(u_str):
index = 0 # This will maintain the index while traversing the entire string
while index < len(u_str):
r = re.search(u_str[index]*4 + '*', u_str)
if r:
start, end = r.span() # start and end index of substring matching 3 or more repetition
u_str = u_str[:start] + u_str[end:] # removing the found substring
index = end
else:
index += 1
return u_str
test1 = 'abbbaaccada'
print('output:' + format_string(test1))
test2 = 'bbccdddcb'
print('output:' + format_string(test2))
I have a string like '....(((...((...' for which I have to generate another string 'ss(4)h5(3)ss(3)h2(2)ss(3)'.
'.' corresponds to 'ss' and the number of continous '.' is in the bracket.
'(' corresponds to 'h5' and the number of continuos '(' is in the bracket.
Currently I'm able to get the output 'ss(4)h5(3)ss(3)' and my code ignores the last two character sequences.
This is what I have done so far
def main():
stringInput = raw_input("Enter the string:")
ssCount = 0
h5Count = 0
finalString = ""
ssString = ""
h5String = ""
ssCont = True
h5Cont = True
for i in range(0, len(stringInput), 1):
if stringInput[i] == ".":
h5Cont = False
if ssCont:
ssCount = ssCount + 1
ssString = "ss(" + str(ssCount) + ")"
ssCont = True
else:
finalString = finalString + ssString
ssCont = True
ssCount = 1
elif stringInput[i] == "(":
ssCont = False
if h5Cont:
h5Count = h5Count + 1
h5String = "h5(" + str(h5Count) + ")"
h5Cont = True
else:
finalString = finalString + h5String
h5Cont = True
h5Count = 1
print finalString
main()
How to modify the code to get the desired output?
I don’t know about modifying your existing code, but to me this can be done very succinctly and pythonically using itertools.groupby. Note that I’m not sure if the 'h2' in your expected output is a typo or if it should be 'h5', which I’m assuming.
from itertools import chain, groupby
string = '....(((...((...'
def character_count(S, labels): # this allows you to customize the labels you want to use
for K, G in groupby(S):
yield labels[K], '(', str(sum(1 for c in G)), ')' # sum() counts the number of items in the iterator G
output = ''.join(chain.from_iterable(character_count(string, {'.': 'ss', '(': 'h5'}))) # joins the components into a single string
print(output)
# >>> ss(4)h5(3)ss(3)h5(2)ss(3)
#Kelvin 's answer is great, however if you want to define a function yourself, you could do it like this:
def h5ss(x):
names = {".": "ss", "(": "h5"}
count = 0
current = None
out = ""
for i in x:
if i == current:
count += 1
else:
if current is not None:
out += "{}({})".format(names[current], count)
count = 1
current = i
if current is not None:
out += "{}({})".format(names[current], count)
return out