I have a string-array, myString = ' 23.654 ' and a character, Char = '.'. Is there some way to use the find function to determine the index of Char in myString?
You can use strfind which will identify the locations where one string exists within another.
index = strfind(myString, '.')
Or you could use == combined with find. The == performs an element-wise equality check between each character in the string and your character of interest. It will yield a logical array that is true where the character occurs, and false otherwise. If you need the actual index, find will return the location of all of the true values.
index = find(myString == '.')
Related
Say the character which we want to check if it appears consecutively in a string s is the dot '.'.
For example, 'test.2.1' does not have consecutive dots, whereas 'test..2.2a...' has consecutive dots. In fact, for the second example, we should not even bother with checking the rest of the string after the first occurence of the consecutive dots.
I have come up with the following simple method:
def consecutive_dots(s):
count = 0
for c in data:
if c == '.':
count += 1
if count == 2:
return True
else:
count = 0
return False
I was wondering if the above is the most 'pythonic' and efficient way to achieve this goal.
You can just use in to check if two consecutive dots (that is, the string "..") appear in the string s
def consecutive_dots(s):
return '..' in s
if you look at the pseudo code, i am trying to make a new string without certain elements.
thesentence = 'i need help!*'
bettersentence = ''.join([char for char in thesentence if char != '!' and '*'])
print(bettersentence)
comparing a character with two strings at the same time doesnt work. But im wondering wether there isnt any easy approach to this?
You can turn your string into a list and then use a for loop to see which characters in the string equal to * or ! and replace them with an empty string
def bettersentence(text):
text = list(text)
for i, character in enumerate(text):
if character == '*' or character == '!':
text[i] = ''
return ('').join(text)
I am having some issues with some code I wrote for this problem:
“Write a function namedd calc that will evaluate a simple arithmetic expression. The input to your program will be a string of the form:
operand1 operator operand2
where operand1 and operand2 are non-negative integers and operator is a single-character operator, which is either +, -, or *. You may assume that there is a space between each operand and the operator. You may further assume that the input is a valid mathemat- ical expression, i.e. your program is not responsible for the case where the user enters gibberish.
Your function will return an integer, such that the returned value is equal to the value produced by applying the given operation to the given operands.
Sample execution:
calc("5 + 10") # 15
“You may not use the split or eval functions in your solution.
Hint: the hard part here is breaking the input string into its three component. You may use the find and rfind functions to find the position of the first and last space, and then use the slice operator (that is, s[startindex:endindex]) to extract the relevant range of characters. Be careful of off-by-one errors in using the slice operator.
Hint: it’s best to test your code as you work. The first step should be to break the input string into its three components. Write a program that does that, have it print out the operator and the two operands on separate lines, and test it until you are convinced that it works. Then, modifying it to perform the desired mathematical operation should be straightforward. Test your program with several different inputs to make sure it works as you expect.”
Here is my code:
def calc(exp):
operand1 = int(exp[:1])
operand2 = int(exp[4:6])
operator = exp[2:3]
if(operator == "+"):
addition = operand1+operand2
return addition
if(operator == "-"):
subtraction = operand1-operand2
return subtraction
if(operator == "*"):
multiplication = operand1*operand2
return multiplication
print(calc("5 + 10"))
print(calc("4 - 8"))
print(calc("4 * 3"))
My code does not fully meet the criteria of this question. It only works for single digit numbers. How can I make my code work for any number?
Like:
“504 + 507”
”5678 + 76890”
and so on?
Thank you. Any help is appreciated.
As the hint says, get the position of the first and last space of the expression, use it to extract the operand and the operators, and then evaluate accordingly.
def calc(exp):
#Get the position for first space with find
low_idx = exp.find(' ')
#Get the position for last space with rfind
high_idx = exp.rfind(' ')
#Extract operators and operand with slice, converting operands to int
operand1 = int(exp[0:low_idx])
operator = exp[low_idx+1:high_idx]
operand2 = int(exp[high_idx:])
result = 0
#Evaluate based on operator
if operator == '+':
result = operand1 + operand2
elif operator == '-':
result = operand1 - operand2
elif operator == '*':
result = operand1 * operand2
return result
print(calc("5 + 10"))
print(calc("4 - 8"))
print(calc("4 * 3"))
print(calc("504 + 507"))
print(calc("5678 + 76890"))
#15
#-4
#12
#1011
#82568
The answer is in the specification:
You may use the find and rfind functions to find the position of the first and last space, and then use the slice operator (that is, s[startindex:endindex]) to extract the relevant range of characters.
find and rfind are methods of string objects.
You could split it into three components using this code: (note: this doesn't use split or eval)
def splitExpression(e):
numbers = ["1","2","3","4","5","6","7","8","9","0"] # list of all numbers
operations = ["+","-","*","/"] # list of all operations
output = [] # output components
currentlyParsing = "number" # the component we're currently parsing
buildstring = "" # temporary variable
for c in e:
if c == " ":
continue # ignore whitespace
if currentlyParsing == "number": # we are currently parsing a number
if c in numbers:
buildstring += c # this is a number, continue
elif c in operations:
output.append(buildstring) # this component has reached it's end
buildstring = c
currentlyParsing = "operation" # we are expecting an operation now
else:
pass # unknown symbol!
elif currentlyParsing == "operation": # we are currently parsing an operation
if c in operations:
buildstring += c # this is an operation, continue
elif c in numbers:
output.append(buildstring) # this component has reached it's end
buildstring = c
currentlyParsing = "number" # we are expecting a number now
else:
pass # unknown symbol!
if buildstring: # anything left in the buffer?
output.append(buildstring)
buildstring = ""
return output
Usage: splitExpression("281*14") returns ["281","*","14"]
This function also accepts spaces between numbers and operations
You can simply take the string and use the split method for the string object, which will return a list of strings based on some separator.
For example:
stringList = "504 + 507".split(" ")
stringList will now be a list such as ["504", "+", "507"] due to the separator " " which is a whitespace. Then just use stringList[1] with your conditionals to solve the problem. Additionally, you can use int(stringList[0]) and int(stringList[2]) to convert the strings to int objects.
EDIT:
Now I realized that your problem said to use find() instead of split(). Simply use the logic above but instead find(" ") the first whitespace. You will then need to find the second whitespace by slicing past the first whitespace using the two additional arguments available for find().
You need to split the string out instead of hard coding the positions of the indexes.
When coding you want to try to make your code as dynamic as possible, that generally means not hard coding stuff that could be a variable or in this case could be grabbed from the spaces.
Also in the if statements I modified them to elif as it is all one contained statement and thus should be grouped.
def calc(exp):
vals = exp.split(' ')
operand1 = int(vals[0])
operand2 = int(vals[2])
operator = vals[1]
if operator == '+':
return operand1+operand2
elif operator == '-':
return operand1-operand2
else:
return operand1*operand2
My question is more or less similar to:
Is there a way to substring a string in Python?
but it's more specifically oriented.
How can I get a par of a string which is located between two known words in the initial string.
Example:
mySrting = "this is the initial string"
Substring = "initial"
knowing that "the" and "string" are the two known words in the string that can be used to get the substring.
Thank you!
You can start with simple string manipulation here. str.index is your best friend there, as it will tell you the position of a substring within a string; and you can also start searching somewhere later in the string:
>>> myString = "this is the initial string"
>>> myString.index('the')
8
>>> myString.index('string', 8)
20
Looking at the slice [8:20], we already get close to what we want:
>>> myString[8:20]
'the initial '
Of course, since we found the beginning position of 'the', we need to account for its length. And finally, we might want to strip whitespace:
>>> myString[8 + 3:20]
' initial '
>>> myString[8 + 3:20].strip()
'initial'
Combined, you would do this:
startIndex = myString.index('the')
substring = myString[startIndex + 3 : myString.index('string', startIndex)].strip()
If you want to look for matches multiple times, then you just need to repeat doing this while looking only at the rest of the string. Since str.index will only ever find the first match, you can use this to scan the string very efficiently:
searchString = 'this is the initial string but I added the relevant string pair a few more times into the search string.'
startWord = 'the'
endWord = 'string'
results = []
index = 0
while True:
try:
startIndex = searchString.index(startWord, index)
endIndex = searchString.index(endWord, startIndex)
results.append(searchString[startIndex + len(startWord):endIndex].strip())
# move the index to the end
index = endIndex + len(endWord)
except ValueError:
# str.index raises a ValueError if there is no match; in that
# case we know that we’re done looking at the string, so we can
# break out of the loop
break
print(results)
# ['initial', 'relevant', 'search']
You can also try something like this:
mystring = "this is the initial string"
mystring = mystring.strip().split(" ")
for i in range(1,len(mystring)-1):
if(mystring[i-1] == "the" and mystring[i+1] == "string"):
print(mystring[i])
I suggest using a combination of list, split and join methods.
This should help if you are looking for more than 1 word in the substring.
Turn the string into array:
words = list(string.split())
Get the index of your opening and closing markers then return the substring:
open = words.index('the')
close = words.index('string')
substring = ''.join(words[open+1:close])
You may want to improve a bit with the checking for the validity before proceeding.
If your problem gets more complex, i.e multiple occurrences of the pair values, I suggest using regular expression.
import re
substring = ''.join(re.findall(r'the (.+?) string', string))
The re should store substrings separately if you view them in list.
I am using the spaces between the description to rule out the spaces between words, you can modify to your needs as well.
Is there a function in Octave that returns the position of the first occurrence of a string in a cell array?
I found findstr but this returns a vector, which I do not want. I want what index does but it only works for strings.
If there is no such function, are there any tips on how to go about it?
As findstr is being deprecated, a combination of find and strcmpi may prove useful. strcmpi compares strings by ignoring the case of the letters which may be useful for your purposes. If this is not what you want, use the function without the trailing i, so strcmp. The input into strcmpi or strcmp are the string to search for str and for your case the additional input parameter is a cell array A of strings to search in. The output of strcmpi or strcmp will give you a vector of logical values where each location k tells you whether the string k in the cell array A matched with str. You would then use find to find all locations of where the string matched, but you can further restrain it by specifying the maximum number of locations n as well as where to constrain your search - specifically if you want to look at the first or last n locations where the string matched.
If the desired string is in str and your cell array is stored in A, simply do:
index = find(strcmpi(str, A)), 1, 'first');
To reiterate, find will find all locations where the string matched, while the second and third parameters tell you to only return the first index of the result. Specifically, this will return the first occurrence of the desired searched string, or the empty array if it can't be found.
Example Run
octave:8> A = {'hello', 'hello', 'how', 'how', 'are', 'you'};
octave:9> str = 'hello';
octave:10> index = find(strcmpi(str, A), 1, 'first')
index = 1
octave:11> str = 'goodbye';
octave:12> index = find(strcmpi(str, A), 1, 'first')
index = [](1x0)