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)
So what I'm trying to do is make a code that adds the value of the letters in a name e.g. name: ABCD ---> 1 + 2+ 3+ 4= 10
My code so far is:
def main():
name = input("Please enter your name (all lowercase): ")
print("\nHere is the code: ")
for ch in name:
print(ord(ch)-96,end=" ")
What I want to do is add all the values of the (ord(ch)-96,end=" ")
You could do this:
sum(ord(c) - 96 for c in name)
Relevant documentation
sum
ord
If you don't actually need to print out the value of each character like you currently are, use sum like others have suggested.
However, if you want to keep the loop body which prints out the value of each character as well as summing them all, just create a variable outside the loop and increment it by ord(c)-96 each time:
total = 0
for ch in name:
charValue = ord(ch)-96
print(charValue, end="")
total += charValue
Once the for loop is completed total will hold the sum of all the values of each character.
In [19]: sum(ord(c) - ord('A') + 1 for c in 'ABCD')
Out[19]: 10
One way is to create a mapping of char->value, which you can do using a dict:
>>> from string import ascii_lowercase
>>> lookup = {ch:idx for idx, ch in enumerate(ascii_lowercase, start=1)}
>>> test = 'abcd'
>>> sum(lookup[ch] for ch in test)
10
This saves mucking about with ordinal values and is a bit more explicit...
I'm trying to create a function where you can put in a phrase such as "ana" in the word "banana", and count how many times it finds the phrase in the word. I can't find the error I'm making for some of my test units not to work.
def test(actual, expected):
""" Compare the actual to the expected value,
and print a suitable message.
"""
import sys
linenum = sys._getframe(1).f_lineno # get the caller's line number.
if (expected == actual):
msg = "Test on line {0} passed.".format(linenum)
else:
msg = ("Test on line {0} failed. Expected '{1}', but got '{2}'.".format(linenum, expected, actual))
print(msg)
def count(phrase, word):
count1 = 0
num_phrase = len(phrase)
num_letters = len(word)
for i in range(num_letters):
for x in word[i:i+num_phrase]:
if phrase in word:
count1 += 1
else:
continue
return count1
def test_suite():
test(count('is', 'Mississippi'), 2)
test(count('an', 'banana'), 2)
test(count('ana', 'banana'), 2)
test(count('nana', 'banana'), 1)
test(count('nanan', 'banana'), 0)
test(count('aaa', 'aaaaaa'), 4)
test_suite()
Changing your count function to the following passes the tests:
def count(phrase, word):
count1 = 0
num_phrase = len(phrase)
num_letters = len(word)
for i in range(num_letters):
if word[i:i+num_phrase] == phrase:
count1 += 1
return count1
Use str.count(substring). This will return how many times the substring occurs in the full string (str).
Here is an interactive session showing how it works:
>>> 'Mississippi'.count('is')
2
>>> 'banana'.count('an')
2
>>> 'banana'.count('ana')
1
>>> 'banana'.count('nana')
1
>>> 'banana'.count('nanan')
0
>>> 'aaaaaa'.count('aaa')
2
>>>
As you can see, the function is non-overlapping. If you need overlapping behaviour, look here: string count with overlapping occurrences
You're using the iteration wrong, so:
for i in range(num_letters): #This will go from 1, 2, ---> len(word)
for x in word[i:i+num_phrase]:
#This will give you the letters starting from word[i] to [i_num_phrase]
#but one by one, so : for i in 'dada': will give you 'd' 'a' 'd' 'a'
if phrase in word: #This condition doesnt make sense in your problem,
#if it's true it will hold true trough all the
#iteration and count will be
#len(word) * num_phrase,
#and if it's false it will return 0
count1 += 1
else:
continue
I guess, str.count(substring) is wrong solution, because it doesn't count overlapping substrings and test suite fails.
There is also builtin str.find method, which could be helpful for the task.
Another way :
def count(sequence,item) :
count = 0
for x in sequence :
if x == item :
count = count+1
return count
A basic question rais this times.
when u see a string like "isisisisisi" howmany "isi" do u count?
at first state you see the string "isi s isi s isi" and return 3 as count.
at the second state you see the string "isisisisisi" and counts the "i" tow times per phrase like this "isi isi isi isi isi".
In other word second 'i' is last character of first 'isi' and first character of second 'isi'.
so you have to return 5 as count.
for first state simply can use:
>>> string = "isisisisisi"
>>> string.count("isi")
3
and for second state you have to recognize the "phrase"+"anything"+"phrase" in the search keyword.
the below function can do it:
def find_iterate(Str):
i = 1
cnt = 0
while Str[i-1] == Str[-i] and i < len(Str)/2:
i += 1
cnt += 1
return Str[0:cnt+1]
Now you have many choice to count the search keyword in the string.
for example I do such below:
if __name__ == "__main__":
search_keyword = "isi"
String = "isisisisisi"
itterated_part = find_iterate(search_keyword)
c = 0
while search_keyword in String:
c += String.count(search_keyword)
String = String.replace(search_keyword, itterated_part)
print c
I do not know if a better way be in python.but I tried to do this with help of Regular Expressions but found no way.