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 going through popular interview questions, and came up with the following solution to "Compute all permutations of a string."
def perm(s):
if len(s) == 0:
return ['']
else:
l = []
prev = perm(s[1:])
for old_str in prev:
for i in range(len(old_str)):
new_str = old_str[0:i] + s[0] + old_str[i:]
l.append(new_str)
return l
However, this solution returns [] on all input. If I make the string in the base case any string other than the empty string, the computation runs as expected. For example, with 'hi' in the base case, perm('abc') returns
['abchi', 'bachi', 'bcahi', 'bchai', 'acbhi', 'cabhi', 'cbahi', 'cbhai', 'achbi', 'cahbi', 'chabi', 'chbai', 'abhci', 'bahci', 'bhaci', 'bhcai', 'ahbci', 'habci', 'hbaci', 'hbcai', 'ahcbi', 'hacbi', 'hcabi', 'hcbai']
As far as I can tell the algorithm for this code is correct. I am unsure of why the empty string isn't behaving as I anticipate, while other strings do. I have referenced this thread for better solutions, but I am still puzzled by why this one doesn't work.
You should probably add a special case for len(s)==1.
As written, the recursion will go down to the base case with an empty string, and then prev will be []. The for loop that follows this will never run, because prev is empty, so the final result will be empty too.
I'm having problems with a homework question.
"Write a function, to_str(a), that takes an array, a, converts each of
its elements to a string (using str(a[i])) and appends all these
strings together."
This is what I have
def to_str(a):
for i in a: a.append([i])
return str(a[i])
I have no idea how to use str(a[i]), I was wondering if someone can point me to the right direction
From the docs:
str(object) -> string
Return a nice string representation of the object. If the argument is
a string, the return value is the same object.
So str(a[i]) will return a string representation of a[i], i.e. convert a[i] to a string.
You will then need to concatenates the strings for all values of i.
As for your code, I have the following comments:
i is an element of a, not an index, as you might be thinking;
you are appending elements of a to a (endlessly, I'm afraid);
a[i] can cause an exception, because, like I said, i is an element, not an index;
you need to return a concatenation of strings, not a string from one element.
Also, if using str(a[i]) is not strictly mandatory, I'd suggest to skip it as unpythonic. You don't need indexes at all for this. Examples:
''.join(str(element) for element in a)
or
''.join(map(str, a))
will return what you need. In both cases str is applied to all elements of a.
The simplest-to-understand ("beginner") way without using indexes will be
s = ''
for element in a:
s += str(element)
return s
It's a bit less efficient, though it does effectively the same thing.
Converting each element into a string is easiest to use list comprehension:
[ str(i) for i in a ]
# equivalent to
[ str(a[i]) for i in range(len(a)) ]
# equivalent to
map(str, a) # most concise, use if you want to feel incredibly clever...
So you can write the function:
def to_str2(a):
''.join([str(i) for i in a]) # concatenates the list as a list of strings
.
Your code nearly does this:
def to_str(a):
new_a = [] # rather than use the same a!
for i in a:
new_a.append(str(i)) #convert i to string before appending
return new_a
The code meeting all the task criteria is rather something like:
def to_str(a):
return reduce(lambda x, y: x + str(y), a, '')
Which does things exactly in the mentioned way: first converts them to strings, then adds to the string made from already processed elements (which at the beginning is just emty string).
EDIT: The clearer (and supported by Python 3) way is to use explicit looping through elements. It does exactly the same, clarifying at the same time how reduce() works:
def to_str(a):
result = '' # third argument of reduce()
for item in a:
result += str(item) # does what reduce() lambda argument was doing
return result
the simplest way is:
[str(i) for i in a]
if you want a function:
def to_str(a):
return [str(i) for i in a]
I'm stumped on how to construct a function that works on lists within lists from inside out (I guess that's how you could poorly describe it).
I'm trying to dynamically turn a list like
res = SomeDjangoQuerySet
x = ['neighborhood', ['city', ['metro', 'metro']]]
into:
getattr(getattr(getattr(getattr(res, 'neighborhood'), 'city'), 'metro'), 'metro')
AKA:
getattr(getattr(getattr(getattr(res, x[0]), x[1][0]), x[1][1][0]), x[1][1][1])
Basically, the first value will always be a string, the second value will either be a string or a list. Each list will follow this pattern (string, string OR list). The depth of lists within lists is indeterminate. The innermost first value of the getattr() will be an outside variable ('res' in this case). Any advice?
This sounds like recursion and iteration might be useful. Does this do what you want?
def flatten(data):
res = []
if hasattr(data, '__iter__'):
for el in data:
res.extend(flatten(el))
else:
res.append(data)
return res
reduce(getattr, flatten(x), res)
I ended up putting in some time and learning about recursion and found this to be the simplest solution (although, credit to David Zwicker who also provided a working solution).
def recursion(a, b):
if type(b) is list:
return recursion(getattr(a, b[0]), b[1])
else:
return getattr(a, b)
recursion(res, x)
def nestattr(x, y):
if isinstance(y, str):
return getattr(x, y)
elif isinstance(y, list):
return nestattr(getattr(x, y[0]), y[1])
nestattr(res, x)
So you start off with the first string in the list, and you have the getattr of (1) the query with (2) that string. Then you recurse using the rest of that list, and if it's a string, you just do the getattr on (1) the result of the previous getattr with (2) this string. Otherwise, if it's still a list, you repeat. I think this is what you're looking for? Correct me if I'm wrong.