There is no built in reverse function for Python's str object. What is the best way of implementing this method?
If supplying a very concise answer, please elaborate on its efficiency. For example, whether the str object is converted to a different object, etc.
Using slicing:
>>> 'hello world'[::-1]
'dlrow olleh'
Slice notation takes the form [start:stop:step]. In this case, we omit the start and stop positions since we want the whole string. We also use step = -1, which means, "repeatedly step from right to left by 1 character".
#Paolo's s[::-1] is fastest; a slower approach (maybe more readable, but that's debatable) is ''.join(reversed(s)).
What is the best way of implementing a reverse function for strings?
My own experience with this question is academic. However, if you're a pro looking for the quick answer, use a slice that steps by -1:
>>> 'a string'[::-1]
'gnirts a'
or more readably (but slower due to the method name lookups and the fact that join forms a list when given an iterator), str.join:
>>> ''.join(reversed('a string'))
'gnirts a'
or for readability and reusability, put the slice in a function
def reversed_string(a_string):
return a_string[::-1]
and then:
>>> reversed_string('a_string')
'gnirts_a'
Longer explanation
If you're interested in the academic exposition, please keep reading.
There is no built-in reverse function in Python's str object.
Here is a couple of things about Python's strings you should know:
In Python, strings are immutable. Changing a string does not modify the string. It creates a new one.
Strings are sliceable. Slicing a string gives you a new string from one point in the string, backwards or forwards, to another point, by given increments. They take slice notation or a slice object in a subscript:
string[subscript]
The subscript creates a slice by including a colon within the braces:
string[start:stop:step]
To create a slice outside of the braces, you'll need to create a slice object:
slice_obj = slice(start, stop, step)
string[slice_obj]
A readable approach:
While ''.join(reversed('foo')) is readable, it requires calling a string method, str.join, on another called function, which can be rather relatively slow. Let's put this in a function - we'll come back to it:
def reverse_string_readable_answer(string):
return ''.join(reversed(string))
Most performant approach:
Much faster is using a reverse slice:
'foo'[::-1]
But how can we make this more readable and understandable to someone less familiar with slices or the intent of the original author? Let's create a slice object outside of the subscript notation, give it a descriptive name, and pass it to the subscript notation.
start = stop = None
step = -1
reverse_slice = slice(start, stop, step)
'foo'[reverse_slice]
Implement as Function
To actually implement this as a function, I think it is semantically clear enough to simply use a descriptive name:
def reversed_string(a_string):
return a_string[::-1]
And usage is simply:
reversed_string('foo')
What your teacher probably wants:
If you have an instructor, they probably want you to start with an empty string, and build up a new string from the old one. You can do this with pure syntax and literals using a while loop:
def reverse_a_string_slowly(a_string):
new_string = ''
index = len(a_string)
while index:
index -= 1 # index = index - 1
new_string += a_string[index] # new_string = new_string + character
return new_string
This is theoretically bad because, remember, strings are immutable - so every time where it looks like you're appending a character onto your new_string, it's theoretically creating a new string every time! However, CPython knows how to optimize this in certain cases, of which this trivial case is one.
Best Practice
Theoretically better is to collect your substrings in a list, and join them later:
def reverse_a_string_more_slowly(a_string):
new_strings = []
index = len(a_string)
while index:
index -= 1
new_strings.append(a_string[index])
return ''.join(new_strings)
However, as we will see in the timings below for CPython, this actually takes longer, because CPython can optimize the string concatenation.
Timings
Here are the timings:
>>> a_string = 'amanaplanacanalpanama' * 10
>>> min(timeit.repeat(lambda: reverse_string_readable_answer(a_string)))
10.38789987564087
>>> min(timeit.repeat(lambda: reversed_string(a_string)))
0.6622700691223145
>>> min(timeit.repeat(lambda: reverse_a_string_slowly(a_string)))
25.756799936294556
>>> min(timeit.repeat(lambda: reverse_a_string_more_slowly(a_string)))
38.73570013046265
CPython optimizes string concatenation, whereas other implementations may not:
... do not rely on CPython's efficient implementation of in-place string concatenation for statements in the form a += b or a = a + b . This optimization is fragile even in CPython (it only works for some types) and isn't present at all in implementations that don't use refcounting. In performance sensitive parts of the library, the ''.join() form should be used instead. This will ensure that concatenation occurs in linear time across various implementations.
Quick Answer (TL;DR)
Example
### example01 -------------------
mystring = 'coup_ate_grouping'
backwards = mystring[::-1]
print(backwards)
### ... or even ...
mystring = 'coup_ate_grouping'[::-1]
print(mystring)
### result01 -------------------
'''
gnipuorg_eta_puoc
'''
Detailed Answer
Background
This answer is provided to address the following concern from #odigity:
Wow. I was horrified at first by the solution Paolo proposed, but that
took a back seat to the horror I felt upon reading the first
comment: "That's very pythonic. Good job!" I'm so disturbed that such
a bright community thinks using such cryptic methods for something so
basic is a good idea. Why isn't it just s.reverse()?
Problem
Context
Python 2.x
Python 3.x
Scenario:
Developer wants to transform a string
Transformation is to reverse order of all the characters
Solution
example01 produces the desired result, using extended slice notation.
Pitfalls
Developer might expect something like string.reverse()
The native idiomatic (aka "pythonic") solution may not be readable to newer developers
Developer may be tempted to implement his or her own version of string.reverse() to avoid slice notation.
The output of slice notation may be counter-intuitive in some cases:
see e.g., example02
print 'coup_ate_grouping'[-4:] ## => 'ping'
compared to
print 'coup_ate_grouping'[-4:-1] ## => 'pin'
compared to
print 'coup_ate_grouping'[-1] ## => 'g'
the different outcomes of indexing on [-1] may throw some developers off
Rationale
Python has a special circumstance to be aware of: a string is an iterable type.
One rationale for excluding a string.reverse() method is to give python developers incentive to leverage the power of this special circumstance.
In simplified terms, this simply means each individual character in a string can be easily operated on as a part of a sequential arrangement of elements, just like arrays in other programming languages.
To understand how this works, reviewing example02 can provide a good overview.
Example02
### example02 -------------------
## start (with positive integers)
print 'coup_ate_grouping'[0] ## => 'c'
print 'coup_ate_grouping'[1] ## => 'o'
print 'coup_ate_grouping'[2] ## => 'u'
## start (with negative integers)
print 'coup_ate_grouping'[-1] ## => 'g'
print 'coup_ate_grouping'[-2] ## => 'n'
print 'coup_ate_grouping'[-3] ## => 'i'
## start:end
print 'coup_ate_grouping'[0:4] ## => 'coup'
print 'coup_ate_grouping'[4:8] ## => '_ate'
print 'coup_ate_grouping'[8:12] ## => '_gro'
## start:end
print 'coup_ate_grouping'[-4:] ## => 'ping' (counter-intuitive)
print 'coup_ate_grouping'[-4:-1] ## => 'pin'
print 'coup_ate_grouping'[-4:-2] ## => 'pi'
print 'coup_ate_grouping'[-4:-3] ## => 'p'
print 'coup_ate_grouping'[-4:-4] ## => ''
print 'coup_ate_grouping'[0:-1] ## => 'coup_ate_groupin'
print 'coup_ate_grouping'[0:] ## => 'coup_ate_grouping' (counter-intuitive)
## start:end:step (or start:end:stride)
print 'coup_ate_grouping'[-1::1] ## => 'g'
print 'coup_ate_grouping'[-1::-1] ## => 'gnipuorg_eta_puoc'
## combinations
print 'coup_ate_grouping'[-1::-1][-4:] ## => 'puoc'
Conclusion
The cognitive load associated with understanding how slice notation works in python may indeed be too much for some adopters and developers who do not wish to invest much time in learning the language.
Nevertheless, once the basic principles are understood, the power of this approach over fixed string manipulation methods can be quite favorable.
For those who think otherwise, there are alternate approaches, such as lambda functions, iterators, or simple one-off function declarations.
If desired, a developer can implement her own string.reverse() method, however it is good to understand the rationale behind this aspect of python.
See also
alternate simple approach
alternate simple approach
alternate explanation of slice notation
This answer is a bit longer and contains 3 sections: Benchmarks of existing solutions, why most solutions here are wrong, my solution.
The existing answers are only correct if Unicode Modifiers / grapheme clusters are ignored. I'll deal with that later, but first have a look at the speed of some reversal algorithms:
NOTE: I've what I called list_comprehension should be called slicing
list_comprehension : min: 0.6μs, mean: 0.6μs, max: 2.2μs
reverse_func : min: 1.9μs, mean: 2.0μs, max: 7.9μs
reverse_reduce : min: 5.7μs, mean: 5.9μs, max: 10.2μs
reverse_loop : min: 3.0μs, mean: 3.1μs, max: 6.8μs
list_comprehension : min: 4.2μs, mean: 4.5μs, max: 31.7μs
reverse_func : min: 75.4μs, mean: 76.6μs, max: 109.5μs
reverse_reduce : min: 749.2μs, mean: 882.4μs, max: 2310.4μs
reverse_loop : min: 469.7μs, mean: 577.2μs, max: 1227.6μs
You can see that the time for the list comprehension (reversed = string[::-1]) is in all cases by far the lowest (even after fixing my typo).
String Reversal
If you really want to reverse a string in the common sense, it is WAY more complicated. For example, take the following string (brown finger pointing left, yellow finger pointing up). Those are two graphemes, but 3 unicode code points. The additional one is a skin modifier.
example = "👈🏾👆"
But if you reverse it with any of the given methods, you get brown finger pointing up, yellow finger pointing left. The reason for this is that the "brown" color modifier is still in the middle and gets applied to whatever is before it. So we have
U: finger pointing up
M: brown modifier
L: finger pointing left
and
original: LMU 👈🏾👆
reversed: UML (above solutions) ☝🏾👈
reversed: ULM (correct reversal) 👆👈🏾
Unicode Grapheme Clusters are a bit more complicated than just modifier code points. Luckily, there is a library for handling graphemes:
>>> import grapheme
>>> g = grapheme.graphemes("👈🏾👆")
>>> list(g)
['👈🏾', '👆']
and hence the correct answer would be
def reverse_graphemes(string):
g = list(grapheme.graphemes(string))
return ''.join(g[::-1])
which also is by far the slowest:
list_comprehension : min: 0.5μs, mean: 0.5μs, max: 2.1μs
reverse_func : min: 68.9μs, mean: 70.3μs, max: 111.4μs
reverse_reduce : min: 742.7μs, mean: 810.1μs, max: 1821.9μs
reverse_loop : min: 513.7μs, mean: 552.6μs, max: 1125.8μs
reverse_graphemes : min: 3882.4μs, mean: 4130.9μs, max: 6416.2μs
The Code
#!/usr/bin/env python3
import numpy as np
import random
import timeit
from functools import reduce
random.seed(0)
def main():
longstring = ''.join(random.choices("ABCDEFGHIJKLM", k=2000))
functions = [(list_comprehension, 'list_comprehension', longstring),
(reverse_func, 'reverse_func', longstring),
(reverse_reduce, 'reverse_reduce', longstring),
(reverse_loop, 'reverse_loop', longstring)
]
duration_list = {}
for func, name, params in functions:
durations = timeit.repeat(lambda: func(params), repeat=100, number=3)
duration_list[name] = list(np.array(durations) * 1000)
print('{func:<20}: '
'min: {min:5.1f}μs, mean: {mean:5.1f}μs, max: {max:6.1f}μs'
.format(func=name,
min=min(durations) * 10**6,
mean=np.mean(durations) * 10**6,
max=max(durations) * 10**6,
))
create_boxplot('Reversing a string of length {}'.format(len(longstring)),
duration_list)
def list_comprehension(string):
return string[::-1]
def reverse_func(string):
return ''.join(reversed(string))
def reverse_reduce(string):
return reduce(lambda x, y: y + x, string)
def reverse_loop(string):
reversed_str = ""
for i in string:
reversed_str = i + reversed_str
return reversed_str
def create_boxplot(title, duration_list, showfliers=False):
import seaborn as sns
import matplotlib.pyplot as plt
import operator
plt.figure(num=None, figsize=(8, 4), dpi=300,
facecolor='w', edgecolor='k')
sns.set(style="whitegrid")
sorted_keys, sorted_vals = zip(*sorted(duration_list.items(),
key=operator.itemgetter(1)))
flierprops = dict(markerfacecolor='0.75', markersize=1,
linestyle='none')
ax = sns.boxplot(data=sorted_vals, width=.3, orient='h',
flierprops=flierprops,
showfliers=showfliers)
ax.set(xlabel="Time in ms", ylabel="")
plt.yticks(plt.yticks()[0], sorted_keys)
ax.set_title(title)
plt.tight_layout()
plt.savefig("output-string.png")
if __name__ == '__main__':
main()
1. using slice notation
def rev_string(s):
return s[::-1]
2. using reversed() function
def rev_string(s):
return ''.join(reversed(s))
3. using recursion
def rev_string(s):
if len(s) == 1:
return s
return s[-1] + rev_string(s[:-1])
A lesser perplexing way to look at it would be:
string = 'happy'
print(string)
'happy'
string_reversed = string[-1::-1]
print(string_reversed)
'yppah'
In English [-1::-1] reads as:
"Starting at -1, go all the way, taking steps of -1"
Reverse a string in python without using reversed() or [::-1]
def reverse(test):
n = len(test)
x=""
for i in range(n-1,-1,-1):
x += test[i]
return x
This is also an interesting way:
def reverse_words_1(s):
rev = ''
for i in range(len(s)):
j = ~i # equivalent to j = -(i + 1)
rev += s[j]
return rev
or similar:
def reverse_words_2(s):
rev = ''
for i in reversed(range(len(s)):
rev += s[i]
return rev
Another more 'exotic' way using bytearray which supports .reverse()
b = bytearray('Reverse this!', 'UTF-8')
b.reverse()
b.decode('UTF-8')`
will produce:
'!siht esreveR'
def reverse(input):
return reduce(lambda x,y : y+x, input)
Here is a no fancy one:
def reverse(text):
r_text = ''
index = len(text) - 1
while index >= 0:
r_text += text[index] #string canbe concatenated
index -= 1
return r_text
print reverse("hello, world!")
There are multiple ways to reverse a string in Python
Slicing Method
string = "python"
rev_string = string[::-1]
print(rev_string)
using reversed function
string = "python"
rev= reversed(string)
rev_string = "".join(rev)
print(rev_string)
Using Recursion
string = "python"
def reverse(string):
if len(string)==0:
return string
else:
return reverse(string[1:])+string[0]
print(reverse(string))
Using for Loop
string = "python"
rev_string =""
for s in string:
rev_string = s+ rev_string
print(rev_string)
Using while Loop
string = "python"
rev_str =""
length = len(string)-1
while length >=0:
rev_str += string[length]
length -= 1
print(rev_str)
original = "string"
rev_index = original[::-1]
rev_func = list(reversed(list(original))) #nsfw
print(original)
print(rev_index)
print(''.join(rev_func))
To solve this in programing way for interview
def reverse_a_string(string: str) -> str:
"""
This method is used to reverse a string.
Args:
string: a string to reverse
Returns: a reversed string
"""
if type(string) != str:
raise TypeError("{0} This not a string, Please provide a string!".format(type(string)))
string_place_holder = ""
start = 0
end = len(string) - 1
if end >= 1:
while start <= end:
string_place_holder = string_place_holder + string[end]
end -= 1
return string_place_holder
else:
return string
a = "hello world"
rev = reverse_a_string(a)
print(rev)
Output:
dlrow olleh
Recursive method:
def reverse(s): return s[0] if len(s)==1 else s[len(s)-1] + reverse(s[0:len(s)-1])
example:
print(reverse("Hello!")) #!olleH
def reverse_string(string):
length = len(string)
temp = ''
for i in range(length):
temp += string[length - i - 1]
return temp
print(reverse_string('foo')) #prints "oof"
This works by looping through a string and assigning its values in reverse order to another string.
a=input()
print(a[::-1])
The above code recieves the input from the user and prints an output that is equal to the reverse of the input by adding [::-1].
OUTPUT:
>>> Happy
>>> yppaH
But when it comes to the case of sentences, view the code output below:
>>> Have a happy day
>>> yad yppah a evaH
But if you want only the characters of the string to be reversed and not the sequence of string, try this:
a=input().split() #Splits the input on the basis of space (" ")
for b in a: #declares that var (b) is any value in the list (a)
print(b[::-1], end=" ") #End declares to print the character in its quotes (" ") without a new line.
In the above code in line 2 in I said that ** variable b is any value in the list (a)** I said var a to be a list because when you use split in an input the variable of the input becomes a list. Also remember that split can't be used in the case of int(input())
OUTPUT:
>>> Have a happy day
>>> evaH a yppah yad
If we don't add end(" ") in the above code then it will print like the following:
>>> Have a happy day
>>> evaH
>>> a
>>> yppah
>>> yad
Below is an example to understand end():
CODE:
for i in range(1,6):
print(i) #Without end()
OUTPUT:
>>> 1
>>> 2
>>> 3
>>> 4
>>> 5
Now code with end():
for i in range(1,6):
print(i, end=" || ")
OUTPUT:
>>> 1 || 2 || 3 || 4 || 5 ||
Here is how we can reverse a string using for loop:
string = "hello,world"
for i in range(-1,-len(string)-1,-1):
print (string[i], end=(" "))
Just as a different solution(because it's asked in interviews):
def reverse_checker(string):
ns = ""
for h in range(1,len(string)+1):
ns += string[-h]
print(ns)
if ns == string:
return True
else:
return False
I am new and trying to find a way to insert a number of L's at the beginning and end of a string. So if I have a string which says
"where did I put my cupcake this morning"
And I want to insert 1 L at the start and 2 L's at the end, so it looks like: "Lwhere did I put my cupcake this morningLL" How do I do this. thank you
Strings are immutable so you can't insert characters into an existing string. You have to create a new string. You can use string concatenation to do what you want:
yourstring = "L" + yourstring + "LL"
Note that you can also create a string with n Ls by using multiplication:
m = 1
n = 2
yourstring = ("L" * m) + yourstring + ("L" * n)
For completeness along with the other answers:
yourstring = "L%sLL" % yourstring
Or, more forward compatible with Python 3.x:
yourstring = "L{0}LL".format(yourstring)
You can also use join:
yourstring = ''.join(('L','yourstring','LL'))
Result:
>>> yourstring
'LyourstringLL'
If you want to insert other string somewhere else in existing string, you may use selection method below.
Calling character on second position:
>>> s = "0123456789"
>>> s[2]
'2'
Calling range with start and end position:
>>> s[4:6]
'45'
Calling part of a string before that position:
>>> s[:6]
'012345'
Calling part of a string after that position:
>>> s[4:]
'456789'
Inserting your string in 5th position.
>>> s = s[:5] + "L" + s[5:]
>>> s
'01234L56789'
Also s is equivalent to s[:].
With your question you can use all your string, i.e.
>>> s = "L" + s + "LL"
or if "L" is a some other string (for example I call it as l), then you may use that code:
>>> s = l + s + (l * 2)
Adding to C2H5OH's answer, in Python 3.6+ you can use format strings to make it a bit cleaner:
s = "something about cupcakes"
print(f"L{s}LL")
Let's say we have a string called yourstring:
for x in range(0, [howmanytimes you want it at the beginning]):
yourstring = "L" + yourstring
for x in range(0, [howmanytimes you want it at the end]):
yourstring += "L"
you can use f strings for this
foo = "where did I put my cupcake this morning"
bar = 'L'
foobar = f'{bar*10}{foo}'
print(foobar)
you can replace 10 by how many times you want to put L's in your string
for end also you can do the same
foo = "where did I put my cupcake this morning"
bar = 'L'
foobar = f'{bar*10}{foo}{bar*10}'
print(foobar)
I want to reduce the number of "the" words from a by 2. But this code doesn't seem to run. I cannot understand why the multiplication operator works but the subtraction operator doesn't.
b = "the"
a = b * 5
print a
a -= (b * 2)
print a
output
the
the the the the the
Traceback (most recent call last):
a -= (b * 2)
TypeError: unsupported operand type(s) for -=: 'str' and 'str'
How can I reduce the number of "the" 's in a by 2. If this cannot be done like this then is there a easier method perform this?
b = "the"
a = b * 5
print a
a = a[:-2*len(b)]
print a
# returns: thethethe
I am not substracting (you cannot really do it with strings), I am removing twice the length of b from the end of a, ignoring its real value.
To reduce the number of "the" by 2 in you word, try with the replace method :
b = "the"
a = b * 5
print a
>>> "thethethethethe"
a = a.replace(b, "", 2) # or a.replace(b*2, "", 1) if you want to remove "thethe" from the string
print a
>>> "thethethe"
If you wanted to remove the "the" by starting from the end, use rsplit()
b = "the"
a = "theAtheBthethe"
a = "".join(a.rsplit("the", 2)) # or "".join(a.rsplit("thethe", 1)) if you want to remove "theth" of the string
print a
>>> "theAtheB"
As described here, the * operator is supported by string (and unicode, list, tuple, bytearray, buffer, xrange types), b * 5 returns 5 copies of b concatenated.
Depends if you want to chop them off the start or the end, you can use array subsets:
>>> a[2*len("the"):]
'thethethe'
>>> a[:-(2*len("the"))]
'thethethe'
There is no support for subtraction operator in case of strings, but you can simply add one:
>>> class MyStr(str):
def __init__(self, val):
return str.__init__(self, val)
def __sub__(self, other):
if self.count(other) > 0:
return self.replace(other, '', 1)
else:
return self
and this will work in the following way:
>>> a = MyStr('thethethethethe')
>>> b = a - 'the'
>>> a
'thethethethethe'
>>> b
'thethethethe'
>>> b = a - 2 * 'the'
>>> b
'thethethe'
Regarding a - 2 * 'the' operation you should be aware that this is not " remove twice 'the' string from a ", but " remove result of (2 times 'the') from a " (first multiply "the" by 2 and then subtract from a).
Is this what you expected?
a = a.rpartition(b * 2)[0]
should do this, cutting from the right side. If you don't have any examples of 'thethe' in a, it will return the empty string ''. It won't work if you have multiple 'the's that are separated by other characters. For that, you could use a.rpartition(b)[0] twice. If you want to cut from the left instead, use a.partition(b * 2)[2].
Why doesn't subtracting work? Using addition and multiplication is a convenience feature for working with strings. The semantics of subtracting (or dividing) strs is not defined for Python, so you can't use it that way.
Plus operator work because "+" concatenate whilst minus don't operate on strings. You can try something using regular expression, like:
import re
s="the"*5
expr="the"
print s
# s -= 2
print "".join(re.findall(expr,s)[:-2])
# s -=3
print "".join(re.findall(expr,s)[:-3])
I'm creating a function to create all 26 combinations of words with a fixed suffix. The script works except for the JOIN in the second-to-last line.
def create_word(suffix):
e=[]
letters="abcefghijklmnopqrstuvwxyz"
t=list(letters)
for i in t:
e.append(i)
e.append(suffix)
' '.join(e)
print e
Currently, it is printing ['a', 'suffix', 'b', 'suffix, ...etc]. And I want it to print out as one long string: 'aSuffixbSuffixcSuffix...etc.' Why isn't the join working in this? How can I fix this?
In addition, how would I separate the characters once I have the string? For example to translate "take the last character of the suffix and add a space to it every time ('aSuffixbSuffixcSuffix' --> 'aSuffix bSuffix cSuffix')". Or, more generally, to replace the x-nth character, where x is any integer (e.g., to replace the 3rd, 6th, 9th, etc. character some something I choose).
str.join returns the new value, not transform the existing one. Here's one way to accomplish it.
result = ' '.join(e)
print result
But if you're feeling clever, you can streamline a lot of the setup.
import string
def create_word(suffix):
return ' '.join(i + suffix for i in string.ascii_lowercase)
join doesn't change its arguments - it just returns a new string:
result = ' '.join(e)
return result
If you really want the output you specified (all of the results concatenated together):
>>> import string
>>> string.ascii_lowercase
'abcdefghijklmnopqrstuvwxyz'
>>> letters = string.ascii_lowercase
>>> suffix = 'Suffix'
>>> ''.join('%s%s' % (l, suffix) for l in letters)
'aSuffixbSuffixcSuffixdSuffixeSuffixfSuffixgSuffixhSuffixiSuffixjSuffixkSuffixlSuffixmSuffixnSuffixoSuffixpSuffixqSuffixrSuffixsSuffixtSuffixuSuffixvSuffixwSuffixxSuffixySuffixzSuffix'
Beside the problem already mentioned by rekursive, you should have a look at list comprehension:
def create_word(suffix):
return ''.join(
[i+suffix for i in "abcefghijklmnopqrstuvwxyz"]
)
print create_word('suffix')