I'm practicing coding on codingbat.com since I'm a complete beginner in python, and here is one of the exercises:
Given a string, return a new string made of every other char starting with the first, so "Hello" yields "Hlo".
Here is my attempt at defining the function string_bits(str):
def string_bits(str):
char = 0
first = str[char]
for char in range(len(str)):
char += 2
every_other = str[char]
return (first + every_other)
Running the code gives an error. What's wrong with my code?
A different approach, with an explanation:
If you need to handle a sentence, where spaces would be included, you can do this using slicing. On a string slicing works as:
[start_of_string:end_of_string:jump_this_many_char_in_string]
So, you want to jump only every second letter, so you do:
[::2]
The first two are empty, because you just want to step every second character.
So, you can do this in one line, like this:
>>> " ".join(i[::2] for i in "Hello World".split())
'Hlo Wrd'
What just happened above, is we take our string, use split to make it a list. The split by default will split on a space, so we will have:
["Hello", "World"]
Then, what we will do from there, is using a comprehension, iterate through each item of the list, which will give us a word at a time, and from there we will perform the desired string manipulation per i[::2].
The comprehension is: (documentation)
i[::2] for i in "Hello World".split()
Finally, we call "".join (doc), which will now change our list back to a string, to finally give us the output:
"Hlo Wrd"
Check out the slicing section from the docs: https://docs.python.org/3/tutorial/introduction.html
The problem is that the char += 2 returns a value greater than len(str) as len(str)-1 (the range) + 2 is longer than the string. You could do:
def string_bits(string):
if len(string) == 2:
return string[0]
result = ''
for char in range(0,len(string),2):#range created value sin increments of two
result += string[char]
return result
A more succinct method would be:
def string_bits(string):
return string[::2]
You should avoid using 'str' as a variable name as it is a reserved word by Python.
Ok, for me:
You should not use str as a variable name as it is a python built-in function (replace str by my_str for example)
For example, 'Hello' length is 5, so 0 <= index <= 4. Here you are trying to access index 3+2=5 (when char = 3) in your for loop.
You can achieve what you want with the following code:
def string_bits(my_str):
result = ""
for char in range(0, len(my_str), 2):
result += my_str[char]
return result
The error you are getting means that you are trying to get the nth letter of a string that has less than n characters.
As another suggestion, strings are Sequence-types in Python, which means they have a lot of built-in functionalities for doing exactly what you're trying to do here. See Built-in Types - Python for more information, but know that sequence types support slicing - that is, selection of elements from the sequence.
So, you could slice your string like this:
def string_bits(input_string):
return input_string[::2]
Meaning "take my input_string from the start (:) to the end (:) and select every second (2) element"
Related
How can I replace even and odd-indexed letters in my strings? I'd like to replace odd-indexed characters with uppercased letters and even-indexed characters with lowercased ones.
x=input("Enter String: ")
How can I modify the inputted string?
This sounds a little like a "do my homework for me" post, but I'll help you out, as I need the training myself.
You can do this by breaking down the problem. (As I am quite new with python syntax, I'm gonna assume that the user has already given an input to string x)
Make a loop, or otherwise iterate through the characters of your string
Make sure you have an index number for each character, which increments for each one
Check if the number is even, by using modulus of 2 (%2). This returns the remainder of a number when divided by 2. In the case of even numbers, that will be 0.
If %2 == 0 set letter to lower case, else set letter to upper case.
append letter to new String, which you defined before the loop. You cannot directly alter a single character in a String, because they are immutable. This means that you cannot change the String itself, but you can assign a new String to the variable.
Done. Print and see if it worked.
Code:
x = "seMi Long StRing WiTH COMPLetely RaNDOM CasINg"
result_string = ""
index = 0;
for c in x:
if(index%2 == 0):
result_string += c.lower()
else:
result_string += c.upper()
index+=1
print(result_string)
s=input()
l=[]
s=s.lower()
l=[i.upper() if s.index(i)%2==0 else i for i in s ]
print("".join(l))
x = 'myname'
for item in range(len(x)):
if item%2==0:
print(x[item].upper())
else:
print(x[item].lower())
this is the for loop i was referring to. but the thing with this line of code is that it is specific to the value you have assigned to the variable x where as the function i provided above can take any string value without us having to repeat the code each time.
def myfunc(string):
result=''
for x in range(len(string)):
if x%2==0:
result=result+string[x].upper()
else:
result=result+string[x].lower()
return result
The above is a function for the question you asked.
A non-function for loop might be easier to grasp right now (like you I am very new to Python as well. So for me it was easier to understand the for loop before I got into functions. Look at my next post for the same.
I'm trying to write a function that takes a string and prints it normally, and then in reverse, like so:
string = "hello"
mirror(string)
'helloolleh'
This is the code i have so far:
def mirror(x) :
sentence = " "
length = len(x)
lengthstring = str(len(x))
lengthint = int(lengthstring)
sentence = x[lengthint, 0]
print x + sentence
but it keeps saying that len (x) is a tuple and not an int and it can't be part of the string slice?
The error I get with your code is:
File "<stdin>", line 6, in mirror
TypeError: string indices must be integers, not tuple
which says nothing about len(x). In fact, it is referring to the line
sentence = x[lengthint, 0]
in which you are trying to index x using lengthint, 0. Python assumes that your use of the comma indicates you want to index x using the tuple (lengthint, 0), which is not valid.
It seems that you are trying to use string slice syntax to reverse the string. String slice syntax uses :, not ,, like this:
sentence = x[lengthint:0]
But, since lengthint is greater than or equal to zero, that will never produce anything other than an empty result. You need to also tell Python to index backwards, like this:
sentence = x[lengthint:0:-1]
This almost works. It omits the first character of your original string in the reversed copy, because 0 indicates the position past the end position. So you have to use the special value None:
sentence = x[lengthint:None:-1]
After you do that and verify that it works, you can remove the use of lengthstring and lengthint (which I presume were added in an attempt to avoid the original error):
sentence = x[len(x):None:-1]
Finally, Python lets you omit the first two slice parameters in this case, leaving just:
sentence = x[::-1]
If your goal is to make this function work I think your approach is far too complicated. You can simple add the string to its reverse:
>>> def mirror(example):
... return example + example[::-1]
...
>>> mirror('hello')
'helloolleh'
Python supports stride-indexing, meaning that the [::-1] simply reverses the string.
you need to use ':' instead of ',' to slice string, for example
> value = '1234'
> print value[0:2]
12
In order to reverse the string you can just do x[::-1], thus, you can write mirror as below
def mirror(x):
return x + x[::-1]
Refer to https://docs.python.org/2.3/whatsnew/section-slices.html for details about slicing.
For my assignment, I'm required to first come up with a function that reverses a string that is inputted (which I've already done using the code below)
def reverse(s):
if len(s) <= 1:
return s
return reverse(s[1:]) + s[0]
The next part is to construct a new string from a given one by breaking it at a certain index value, and concatenating the reverse of the second part (the suffix) to the beginning of the first part (the prefix)
For example, if the input string is 'laptop' and the chosen index value is, say, 3, the string is broken as 'lap' + 'top'. 'top' would then be reversed to 'pot' and would be concatenated with the prefix (in the order) as 'pot' + 'lap'
The assignment is somewhat confusing and since I'm a novice with little to no experience besides a couple of days working in Python, I'm a little confused as to what to do. I'm pretty sure I have to use the slice and concatenation operators but I'm not sure how I should go about constructing a function that suits the above criteria. Any pointers?
Combining the other two answers, and implementing your reverse function:
def concatreverse(s, i):
"""This function takes in a string, s, which
is split at an index, i, reverses the second of the split,
and concatenates it back on to the first part"""
#Takes your inputs and processes
part1,part2 = s[0:i], s[i:]
#Reverse part2 with the function you already created
#this assumes it is accessible (in the same file, for instance)
rev_part2 = reverse(part2)
#concatenate the result
result = part1 +rev_part2
#Give it back to the caller
return result
As a beginner, it helps to step through line by line, or doing tests using the interpreter to see exactly what's going on :)
Something like:
def concatreverse(s, i):
return s[:i] + reverse(s[i:])
You could do as follows:
s = 'laptop'
i = 3;
# split the string into two parts
part1,part2 = s[0:i], s[i:]
# make new string starting with reversed second part.
s2 = part2[::-1] + part1
print(s2)
# prints: potlap
This is what I have so far:
count=0
mystring=input("enter")
for ch in mystring:
if mystring.lower():
count+=1
print(count)
I figured out how to make a program that displays the number of lower case letters in a string, but it requires that I list each letter individually: if ch=='a' or ch=='b' or ch=='c', etc. I am trying to figure out how to use a command to do so.
This sounds like homework! Anway, this is a fun way of doing it:
#the operator module contains functions that can be used like
#their operator counter parts. The eq function works like the
#'=' operator; it takes two arguments and test them for equality.
from operator import eq
#I want to give a warning about the input function. In python2
#the equivalent function is called raw_input. python2's input
#function is very different, and in this case would require you
#to add quotes around strings. I mention this in case you have
#been manually adding quotes if you are testing in both 2 and 3.
mystring = input('enter')
#So what this line below does is a little different in python 2 vs 3,
#but comes to the same result in each.
#First, map is a function that takes a function as its first argument,
#and applies that to each element of the rest of the arguments, which
#are all sequences. Since eq is a function of two arguments, you can
#use map to apply it to the corresponding elements in two sequences.
#in python2, map returns a list of the elements. In python3, map
#returns a map object, which uses a 'lazy' evaluation of the function
#you give on the sequence elements. This means that the function isn't
#actually used until each item of the result is needed. The 'sum' function
#takes a sequence of values and adds them up. The results of eq are all
#True or False, which are really just special names for 1 and 0 respectively.
#Adding them up is the same as adding up a sequence of 1s and 0s.
#so, map is using eq to check each element of two strings (i.e. each letter)
#for equality. mystring.lower() is a copy of mystring with all the letters
#lowercase. sum adds up all the Trues to get the answer you want.
sum(map(eq, mystring, mystring.lower()))
or the one-liner:
#What I am doing here is using a generator expression.
#I think reading it is the best way to understand what is happening.
#For every letter in the input string, check if it is lower, and pass
#that result to sum. sum sees this like any other sequence, but this sequence
#is also 'lazy,' each element is generated as you need it, and it isn't
#stored anywhere. The results are just given to sum.
sum(c.islower() for c in input('enter: '))
You have a typo in your code. Instead of:
if my.string.lower():
It should be:
if ch.islower():
If you have any questions ask below. Good luck!
I'm not sure if this will handle UTF or special characters very nicely but should work for at least ASCII in Python3, using the islower() function.
count=0
mystring=input("enter:")
for ch in mystring:
if ch.islower():
count+=1
print(count)
The correct version of your code would be:
count=0
mystring=input("enter")
for ch in mystring:
if ch.islower():
count += 1
print(count)
The method lower converts a string/char to lowercase. Here you want to know if it IS lowercase (you want a boolean), so you need islower.
Tip: With a bit of wizardry you can even write this:
mystring= input("enter")
count = sum(map(lambda x: x.islower(), mystring))
or
count = sum([x.islower() for x in mystring])
(True is automatically converted to 1 and False to 0)
:)
I think you can use following method:
mystring=input("enter:")
[char.lower() for char in mystring].count( True ) )
I have to make a function called countLetterString(char, str) where
I need to use recursion to find the amount of times the given character appears in the string.
My code so far looks like this.
def countLetterString(char, str):
if not str:
return 0
else:
return 1 + countLetterString(char, str[1:])
All this does is count how many characters are in the string but I can't seem to figure out how to split the string then see whether the character is the character split.
The first step is to break this problem into pieces:
1. How do I determine if a character is in a string?
If you are doing this recursively you need to check if the first character of the string.
2. How do I compare two characters?
Python has a == operator that determines whether or not two things are equivalent
3. What do I do after I know whether or not the first character of the string matches or not?
You need to move on to the remainder of the string, yet somehow maintain a count of the characters you have seen so far. This is normally very easy with a for-loop because you can just declare a variable outside of it, but recursively you have to pass the state of the program to each new function call.
Here is an example where I compute the length of a string recursively:
def length(s):
if not s: # test if there are no more characters in the string
return 0
else: # maintain a count by adding 1 each time you return
# get all but the first character using a slice
return 1 + length( s[1:] )
from this example, see if you can complete your problem. Yours will have a single additional step.
4. When do I stop recursing?
This is always a question when dealing with recursion, when do I need to stop recalling myself. See if you can figure this one out.
EDIT:
not s will test if s is empty, because in Python the empty string "" evaluates to False; and not False == True
First of all, you shouldn't use str as a variable name as it will mask the built-in str type. Use something like s or text instead.
The if str == 0: line will not do what you expect, the correct way to check if a string is empty is with if not str: or if len(str) == 0: (the first method is preferred). See this answer for more info.
So now you have the base case of the recursion figured out, so what is the "step". You will either want to return 1 + countLetterString(...) or 0 + countLetterString(...) where you are calling countLetterString() with one less character. You will use the 1 if the character you remove matches char, or 0 otherwise. For example you could check to see if the first character from s matches char using s[0] == char.
To remove a single character in the string you can use slicing, so for the string s you can get all characters but the first using s[1:], or all characters but the last using s[:-1]. Hope that is enough to get you started!
Reasoning about recursion requires breaking the problem into "regular" and "special" cases. What are the special cases here? Well, if the string is empty, then char certainly isn't in the string. Return 0 in that case.
Are there other special cases? Not really! If the string isn't empty, you can break it into its first character (the_string[0]) and all the rest (the_string[1:]). Then you can recursively count the number of character occurrences in the rest, and add 1 if the first character equals the char you're looking for.
I assume this is an assignment, so I won't write the code for you. It's not hard. Note that your if str == 0: won't work: that's testing whether str is the integer 0. if len(str) == 0: is a way that will work, and if str == "": is another. There are shorter ways, but at this point those are probably clearest.
First of all you I would suggest not using char or str. Str is a built function/type and while I don't believe char would give you any problems, it's a reserved word in many other languages. Second you can achieve the same functionality using count, as in :
letterstring="This is a string!"
letterstring.count("i")
which would give you the number of occurrences of i in the given string, in this case 3.
If you need to do it purely for speculation, the thing to remember with recursion is carrying some condition or counter over which each call and placing some kind of conditional within the code that will change it. For example:
def countToZero(count):
print(str(count))
if count > 0:
countToZero(count-1)
Keep it mind this is a very quick example, but as you can see on each call I print the current value and then the function calls itself again while decrementing the count. Once the count is no longer greater than 0 the function will end.
Knowing this you will want to keep track of you count, the index you are comparing in the string, the character you are searching for, and the string itself given your example. Without doing the code for you, I think that should at least give you a start.
You have to decide a base case first. The point where the recursion unwinds and returns.
In this case the the base case would be the point where there are no (further) instances of a particular character, say X, in the string. (if string.find(X) == -1: return count) and the function makes no further calls to itself and returns with the number of instances it found, while trusting its previous caller information.
Recursion means a function calling itself from within, therefore creating a stack(at least in Python) of calls and every call is an individual and has a specified purpose with no knowledge whatsoever of what happened before it was called, unless provided, to which it adds its own result and returns(not strictly speaking). And this information has to be supplied by its invoker, its parent, or can be done using global variables which is not advisable.
So in this case that information is how many instances of that particular character were found by the parent function in the first fraction of the string. The initial function call, made by us, also needs to be supplied that information, since we are the root of all function calls and have no idea(as we haven't treaded the string) of how many Xs are there we can safely tell the initial call that since I haven't gone through the string and haven't found any or zero/0 X therefore here's the string entire string and could you please tread the rest of it and find out how many X are in there. This 0 as a convenience could be the default argument of the function, or you have to supply the 0 every time you make the call.
When will the function call another function?
Recursion is breaking down the task into the most granular level(strictly speaking, maybe) and leave the rest to the (grand)child(ren). The most granular break down of this task would be finding a single instance of X and passing the rest of the string from the point, exclusive(point + 1) at which it occurred to the next call, and adding 1 to the count which its parent function supplied it with.
if not string.find(X) == -1:
string = string[string.find(X) + 1:]
return countLetterString(char, string, count = count + 1)`
Counting X in file through iteration/loop.
It would involve opening the file(TextFILE), then text = read(TextFile)ing it, text is a string. Then looping over each character (for char in text:) , remember granularity, and each time char (equals) == X, increment count by +=1. Before you run the loop specify that you never went through the string and therefore your count for the number X (in text) was = 0. (Sounds familiar?)
return count.
#This function will print the count using recursion.
def countrec(s, c, cnt = 0):
if len(s) == 0:
print(cnt)
return 0
if s[-1] == c:
countrec(s[0:-1], c, cnt+1)
else:
countrec(s[0:-1], c, cnt)
#Function call
countrec('foobar', 'o')
With an extra parameter, the same function can be implemented.
Woking function code:
def countLetterString(char, str, count = 0):
if len(str) == 0:
return count
if str[-1] == char:
return countLetterString(char, str[0:-1], count+1)
else:
return countLetterString(char, str[0:-1], count)
The below function signature accepts 1 more parameter - count.
(P.S : I was presented this question where the function signature was pre-defined; just had to complete the logic.)
Hereby, the code :
def count_occurrences(s, substr, count=0):
''' s - indicates the string,
output : Returns the count of occurrences of substr found in s
'''
len_s = len(s)
len_substr = len(substr)
if len_s == 0:
return count
if len_s < len_substr:
return count
if substr == s[0:len_substr]:
count += 1
count = count_occurrences(s[1:], substr, count) ## RECURSIVE CALL
return count
output behavior :
count_occurences("hishiihisha", "hi", 0) => 3
count_occurences("xxAbx", "xx") => 1 (not mandatory to pass the count , since it's a positional arg.)