I'm trying to count the number of times "bob" has occurred in a given string. this is what I tried:
s = input("give me a string:")
count = 0
for i in s:
if i=="b":
for j in s:
x=0
if j!="b":
x+=1
else:
break
if s[x+1]=="o" and s[x+2]=="b":
count+=1
print(count)
if I give the string "bob", it gives back 2, and if I give something like "jbhxbobalih", it gives back 0. I don't know why this happens. any idea?
The easiest manual count would probably use indeces and slices. The main difference between this and the much simpler s.count("bob") is that it also counts overlapping occurrences:
# s = "aboboba" -> 2
count = 0
for i, c in enumerate(s):
if s[i:i+3] == "bob":
count += 1
You can try checking 3 consecutive characters, if they are 'bob', just add our counter up, and do nothing otherwise.
Your code should be like this:
s = input("give me a string:")
count = 0
for i in range(0, len(s) - 3):
if s[i] == 'b' and s[i + 1] == 'o' and s[i + 2] == 'b':
count += 1
print(count)
100 % working this will work for all string.
import re
def check(string, sub_str):
count = 0
val = re.findall(sub_str, string)
for i in val:
count+=1
print(count)
# driver code
string = "baadbobaaaabobsasddswqbobdwqdwqsbob"
sub_str = "bob"
check(string, sub_str)
This gives the correct output.
I want to count the number of instances where consecutive letters are identical in a given string.
For example, my string input is:
EOOOEOEE
I would only like to find the number of occasions where there is more than one consecutive 'O'.
The Output should be:
1
Since, there is only one set of O's that come consecutively.
This is possible with itertools.groupby:
from itertools import groupby
x = 'EOOOEOEE'
res = sum(len(list(j)) > 1 for i, j in groupby(x) if i == 'O') # 1
You can use a regex:
>>> import re
>>> s = 'EOOOEOEEOO'
>>> sum(1 for x in re.finditer(r'O{2,}', s))
2
Just count with a for-loop:
n = 0
g = 0
s = 'EOOOEOEE'
for c in s:
if c == 'O':
g += 1
elif g:
if g > 1:
n += 1
g = 0
if g:
n += 1
which gives n as 1.
I assume you want to know the number of times that all letters are consecutive in a string and not just for the letter 'O'.
Make a character dictionary that will hold this count as values to keys as the characters in a string. char_dict = {}
The idea is to have two conditions to match
(1) Is the current character same as the previous character
(2) If the first condition is true then is the current pair of consecutive characters part of a larger substring that has the same consecutive characters.
Simply put, take for example ABBBCBB. When we encounter the third B we want to check whether it is part of sub-string that is consecutive and already accounted for. i.e. BBB should give consecutive count to be 1 and not 2. To implement this we use a flag variable that checks this conditions.
If we use only the (1)st condition BBB will count as BB and BB and not as a single BBB.
Rest of the code is pretty straight forward.
char_dict = {}
string = "EOOOEOEEFFFOFEOOO"
prev_char = None
flag=0
for char in list(string):
if char not in char_dict:
#initialize the count by zero
char_dict[char] = 0
if char == prev_char and flag !=0:
char_dict[char] += 1
flag = 0
else:
flag = 1
prev_char = char
print(char_dict)
I have this function to check if a string contains three or more lowercase letters.
def lowerCaseValid(word):
lowCharList = ['abcdefghijklmnopqrstuvwxyz']
i = 0
flag = 0
while i <= len(word):
j = 0
while j <= len(lowCharList):
if lowCharList[j] == word[i]:
flag += 1
j = 0
else:
j += 1
i += 1
if flag >= 3:
return True
In simple terms, I pass in a string (word) and create a list of acceptable characters (lowCharList).
Then, I set up a nested while loop that checks word[i] at every index of lowCharList, until it finds a match.
Then it resets lowCharList counter and adds 1 to flag, and moves on to word[i+1].
If it doesn't find a match by the time it reaches z, then it moves onto word[i+1] anyways.
For some reason, why I try my sample input in my main function.
def main():
word = 'corRe!33'
print(lowerCaseValid(word))
I get this error:
in lowerCaseValid
if lowCharList[j] == word[i]:
IndexError: list index out of range
Why is it throwing this error? Thank you.
using python's in operator is easier...
def lowerCaseValid(word):
cnt = 0
lowCharList = ['abcdefghijklmnopqrstuvwxyz']
chars = lowCharList.pop ()
for ch in word:
if ch in chars:
cnt += 1
return cnt >= 3
or with using sets just 2 lines of code
def lowerCaseValid(word):
lowCharList = ['abcdefghijklmnopqrstuvwxyz']
return len(set(lowCharList.pop()) & set(word)) >=3
or one liner with map and lambda
def lowerCaseValid(word):
return len(list(map(lambda x: x.islower, list(word)))) >=3
Another alternative approach using a list comprehension and string.ascii_lowercase instead of redefining the lowercase letters:
from string import ascii_lowercase
def lowerCaseValid(word):
return sum[x in ascii_lowercase for x in word] >= 3
How this works is that the list comprehension goes through each letter in word. The x in ascii_lowercase will return a boolean value of either True (1) or False (0), then sum up the Trues
Change
lowCharList = ['abcdefghijklmnopqrstuvwxyz']
to
lowCharList = list('abcdefghijklmnopqrstuvwxyz')
I believe this should help, since you have a list containing only 1 item, whereas this way, you create a list with 24 items (all different letters).
As heemayl pointed out in the comments, lowCharList is 1 element long! Now you have 2 options: make lowCharList an actual list (lowCharList = list ("abcd...")), or keep lowCharList a string, which will work just fine (remove the brackets in the definition.
Might I suggest another method: checking if str.islower count adds up to >=3. So:
lower_count = 0
for letter in word:
if letter.islower(): lower_count += 1
if lower_count >= 3: return True
Also, as suggested in the comments:
return len ([letter for letter in word if letter.islower()]) >= 3
would work (better than my answer, which is just the expanded form of it)
I want to calculate the number of integers in the string "abajaao1grg100rgegege".
I tried using isnumeric() but it considers '100' as three different integers and shows the output 4. I want my program to consider 100 as a single integer.
Here is my attempt:
T = int(input())
for x in range(T):
S = input()
m = 0
for k in S:
if (k.isnumeric()):
m += 1
print(m)
I'd use a very basic regex (\d+) then count the number of matches:
import re
string = 'abajaao1grg100rgegege'
print(len(re.findall(r'(\d+)', string)))
# 2
Regex is the go-to tool for this sort of problem, as the other answers have noted. However, here is a solution that uses looping constructs and no regex:
result = sum(y.isdigit() and not x.isdigit() for x,y in zip(myString[1:], myString))
In addition, here is an easy to understand, iterative solution, that also doesn't use regex and is much more clear than the other one, but also more verbose:
def getNumbers(string):
result = 0
for i in range(len(string)):
if string[i].isdigit() and (i==0 or not string[i-1].isdigit()):
result += 1
return result
You can use the regex library to solve this issue.
import re
st = "abajaao1grg100rgegege"
res = re.findall(r'\d+', st)
>>> ['1', '100']
You can check how many numbers you have on that list that the findall returned.
print (len(res))
>>> 2
In order to read more on python regex and the patterns, enter here
Not very Pythonic but for beginners more understandable:
Loop over characters in string and in every iteration remember in the was_digit (logical variable) if the current character is digit - for the next iteration.
Increase the counter only if the previous character was not a digit:
string = 'abajaao1grg100rgegege'
counter = 0 # Reset the counter
was_digit = False # Was previous character a digit?
for ch in string:
if ch.isdigit():
if not was_digit: # previous character was not a digit ...
counter += 1 # ... so it is start of the new number - count it!
was_digit = True # for the next iteration
else:
was_digit = False # for the next iteration
print(counter) # Will print 2
random="1qq11q1qq121a21ws1ssq1";
counter=0
i=0
length=len(random)
while(i<length):
if (random[i].isnumeric()):
z=i+1
counter+=1
while(z<length):
if (random[z].isnumeric()):
z=z+1
continue
else:
break
i=z
else:
i+=1
print ("No of integers",counter)
I am working through the book "Introduction to Computation and Programming Using Python" by Dr. Guttag. I am working on the finger exercises for Chapter 3. I am stuck. It is section 3.2, page 25. The exercise is: Let s be a string that contains a sequence of decimal numbers separated by commas, e.g., s = '1.23,2.4,3.123'. Write a program that prints the sume of the numbers in s.
The previous example was:
total = 0
for c in '123456789':
total += int(c)
print total.
I've tried and tried but keep getting various errors. Here's my latest attempt.
total = 0
s = '1.23,2.4,3.123'
print s
float(s)
for c in s:
total += c
print c
print total
print 'The total should be ', 1.23+2.4+3.123
I get ValueError: invalid literal for float(): 1.23,2.4,3.123.
Floating point values cannot have a comma. You are passing 1.23,2.4,3.123 as it is to float function, which is not valid. First split the string based on comma,
s = "1.23,2.4,3.123"
print s.split(",") # ['1.23', '2.4', '3.123']
Then convert each and and every element of that list to float and add them together to get the result. To feel the power of Python, this particular problem can be solved in the following ways.
You can find the total, like this
s = "1.23,2.4,3.123"
total = sum(map(float, s.split(",")))
If the number of elements is going to be too large, you can use a generator expression, like this
total = sum(float(item) for item in s.split(","))
All these versions will produce the same result as
total, s = 0, "1.23,2.4,3.123"
for current_number in s.split(","):
total += float(current_number)
Since you are starting with Python, you could try this simple approach:
Use the split(c) function, where c is a delimiter. With this you will have a list numbers (in the code below). Then you can iterate over each element of that list, casting each number to a float (because elements of numbers are strings) and sum them:
numbers = s.split(',')
sum = 0
for e in numbers:
sum += float(e)
print sum
Output:
6.753
From the book Introduction to Computation and Programming using Python at page 25.
"Let s be a string that contains a sequence of decimal numbers separated by commas, e.g., s
= '1.23,2.4,3.123'. Write a program that prints the sum of the numbers in s."
If we use only what has been taught so far, then this code is one approach:
tmp = ''
num = 0
print('Enter a string of decimal numbers separated by comma:')
s = input('Enter the string: ')
for ch in s:
if ch != ',':
tmp = tmp + ch
elif ch == ',':
num = num + float(tmp)
tmp = ''
# Also include last float number in sum and show result
print('The sum of all numbers is:', num + float(tmp))
total = 0
s = '1.23,2.4,3.123'
for c in s.split(','):
total = total + float(c)
print(total)
Works Like A Charm
Only used what i have learned yet
s = raw_input('Enter a string that contains a sequence of decimal ' +
'numbers separated by commas, e.g. 1.23,2.4,3.123: ')
s = "," + s+ ","
total =0
for i in range(0,len(s)):
if s[i] == ",":
for j in range(1,(len(s)-i)):
if s[i+j] == ","
total = total + float(s[(i+1):(i+j)])
break
print total
This is what I came up with:
s = raw_input('Enter a sequence of decimal numbers separated by commas: ')
aux = ''
total = 0
for c in s:
aux = aux + c
if c == ',':
total = total + float(aux[0:len(aux)-1])
aux = ''
total = total + float(aux) ##Uses last value stored in aux
print 'The sum of the numbers entered is ', total
I think they've revised this textbook since this question was asked (and some of the other's have answered.) I have the second edition of the text and the split example is not on page 25. There's nothing prior to this lesson that shows you how to use split.
I wound up finding a different way of doing it using regular expressions. Here's my code:
# Intro to Python
# Chapter 3.2
# Finger Exercises
# Write a program that totals a sequence of decimal numbers
import re
total = 0 # initialize the running total
for s in re.findall(r'\d+\.\d+','1.23, 2.2, 5.4, 11.32, 18.1,22.1,19.0'):
total = total + float(s)
print(total)
I've never considered myself dense when it comes to learning new things, but I'm having a hard time with (most of) the finger exercises in this book so far.
s = input('Enter a sequence of decimal numbers separated by commas: ')
x = ''
sum = 0.0
for c in s:
if c != ',':
x = x + c
else:
sum = sum + float(x)
x = ''
sum = sum + float(x)
print(sum)
This is using just the ideas already covered in the book at this point. Basically it goes through each character in the original string, s, using string addition to add each one to the next to build a new string, x, until it encounters a comma, at which point it changes what it has as x to a float and adds it to the sum variable, which started at zero. It then resets x back to an empty string and repeats until all the characters in s have been covered
Here's a solution without using split:
s='1.23,2.4,3.123,5.45343'
pos=[0]
total=0
for i in range(0,len(s)):
if s[i]==',':
pos.append(len(s[0:i]))
pos.append(len(s))
for j in range(len(pos)-1):
if j==0:
num=float(s[pos[j]:pos[j+1]])
total=total+num
else:
num=float(s[pos[j]+1:pos[j+1]])
total=total+num
print total
My way works:
s = '1.23, 211.3'
total = 0
for x in s:
for i in x:
if i != ',' and i != ' ' and i != '.':
total = total + int(i)
print total
My answer is here:
s = '1.23,2.4,3.123'
sum = 0
is_int_part = True
n = 0
for c in s:
if c == '.':
is_int_part = False
elif c == ',':
if is_int_part == True:
total += sum
else:
total += sum/10.0**n
sum = 0
is_int_part = True
n = 0
else:
sum *= 10
sum += int(c)
if is_int_part == False:
n += 1
if is_int_part == True:
total += sum
else:
total += sum/10.0**n
print total
I have managed to answer the question with the knowledge gained up until 3.2 the section for loop
s = '1.0, 1.1, 1.2'
print 'List of decimal number'
print s
total = 0.0
for c in s:
if c == ',':
total += float(s[0:(s.index(','))])
d = int(s.index(','))+1
s = s[(d+1) : len(s)]
s = float(s)
total += s
print '1.0 + 1.1 + 1.2 = ', total
This is the answer to the question i feel that the split function is not good for beginner like you and me.
Considering the fact that you might not yet be exposed to more complex functions, simply try these out.
total = 0
for c in "1.23","2.4",3.123":
total += float(c)
print total
My answer:
s = '2.1,2.0'
countI = 0
countF = 0
totalS = 0
for num in s:
if num == ',' or (countF + 1 == len(s)):
totalS += float(s[countI:countF])
if countF < len(s):
countI = countF + 1
countF += 1
print(totalS) # 4.1
This only works if the numbers are floats
Here is my answer. It is similar to the one by user5716300 above, but since I am also a beginner I explicitly created a separate variable s1 for the split string:
s = "1.23,2.4,3.123"
s1 = s.split(",") #this creates a list of strings
count = 0.0
for i in s1:
count = count + float(i)
print(count)
If we are just sticking with the content for that chapter, I came up with this: (though using that sum method mentioned by theFourthEye is also pretty slick):
s = '1.23,3.4,4.5'
result = s.split(',')
result = list(map(float, result))
n = 0
add = 0
for a in result:
add = add + result[n]
n = n + 1
print(add)
I just wanna to post my answer because I am reading this book now.
s = '1.23,2.4,3.123'
ans = 0.0
i = 0
j = 0
for c in s:
if c == ',':
ans += float(s[i:j])
i = j + 1
j += 1
ans += float(s[i:j])
print(str(ans))
Using knowledge from the book:
s = '4.58,2.399,3.1456,7.655,9.343'
total = 0
index = 0
for string in s:
index += 1
if string == ',':
temp = float(s[:index-1])
s = s[index:]
index = 0
total += temp
temp = 0
print(total)
Here I used string slicing, and by slicing the original string every time our 'string' variable is equal to ','. Also using an index variable to keep track of the number that is before the comma. After slicing the string, the number that gets input into tmp is cleared with the comma in front of it, the string becoming another string without that number.
Because of this, the index variable needs to be reset every time this happens.
Here's mine using the exact string in the question and only what has been taught so far.
total = 0
temp_num = ''
for char in '1.23,2.4,3.123':
if char == ',':
total += float(temp_num)
temp_num = ''
else:
temp_num += char
total += float(temp_num) #to catch the last number that has no comma after it
print(total)
I know this isn't covered in the book up to this point but I happened to learn the use of the eval() function on my own prior to getting to this question and used it to solve.
total = 0
s = "1.23,2.4,3.123"
x = eval(s)
y = sum(x)
print(y)
I think this is the easiest way to answer the question. It uses the split command, which is not introduced in the book at this moment but a very useful command.
s = input('Insert string of decimals, e,g, 1.4,5.55,12.651:')
sList = s.split(',') #create a list of these values
print(sList) #to check if list is correctly created
total = 0 #for creating the variable
for each in sList:
total = total + float(each)
print(total)
total =0
s = {1.23,2.4,3.123}
for c in s:
total = total+float(c)
print(total)