Sorting Python list based on the length of the string
Question:
I want to sort a list of strings based on the string length. I tried to use sort as follows, but it doesn’t seem to give me correct result.
xs = ['dddd','a','bb','ccc']
print xs
xs.sort(lambda x,y: len(x) < len(y))
print xs
['dddd', 'a', 'bb', 'ccc']
['dddd', 'a', 'bb', 'ccc']
What might be wrong?
Answers:
When you pass a lambda
to sort
, you need to return an integer, not a boolean. So your code should instead read as follows:
xs.sort(lambda x,y: cmp(len(x), len(y)))
Note that cmp is a builtin function such that cmp(x, y)
returns -1 if x
is less than y
, 0 if x
is equal to y
, and 1 if x
is greater than y
.
Of course, you can instead use the key
parameter:
xs.sort(key=lambda s: len(s))
This tells the sort
method to order based on whatever the key function returns.
EDIT: Thanks to balpha and Ruslan below for pointing out that you can just pass len
directly as the key parameter to the function, thus eliminating the need for a lambda
:
xs.sort(key=len)
And as Ruslan points out below, you can also use the built-in sorted function rather than the list.sort
method, which creates a new list rather than sorting the existing one in-place:
print(sorted(xs, key=len))
The same as in Eli’s answer – just using a shorter form, because you can skip a lambda
part here.
Creating new list:
>>> xs = ['dddd','a','bb','ccc']
>>> sorted(xs, key=len)
['a', 'bb', 'ccc', 'dddd']
In-place sorting:
>>> xs.sort(key=len)
>>> xs
['a', 'bb', 'ccc', 'dddd']
I Would like to add how the pythonic key function works while sorting :
Decorate-Sort-Undecorate Design Pattern :
Python’s support for a key function when sorting is implemented using what is known as the
decorate-sort-undecorate design pattern.
It proceeds in 3 steps:
-
Each element of the list is temporarily replaced with a “decorated” version that includes the result of the key function applied to the element.
-
The list is sorted based upon the natural order of the keys.
-
The decorated elements are replaced by the original elements.
Key parameter to specify a function to be called on each list element prior to making comparisons. docs
Write a function lensort to sort a list of strings based on length.
def lensort(a):
n = len(a)
for i in range(n):
for j in range(i+1,n):
if len(a[i]) > len(a[j]):
temp = a[i]
a[i] = a[j]
a[j] = temp
return a
print lensort(["hello","bye","good"])
def lensort(list_1):
list_2=[];list_3=[]
for i in list_1:
list_2.append([i,len(i)])
list_2.sort(key = lambda x : x[1])
for i in list_2:
list_3.append(i[0])
return list_3
This works for me!
The easiest way to do this is:
list.sort(key = lambda x:len(x))
I can do it using below two methods, using function
def lensort(x):
list1 = []
for i in x:
list1.append([len(i),i])
return sorted(list1)
lista = ['a', 'bb', 'ccc', 'dddd']
a=lensort(lista)
print([l[1] for l in a])
In one Liner using Lambda, as below, a already answered above.
lista = ['a', 'bb', 'ccc', 'dddd']
lista.sort(key = lambda x:len(x))
print(lista)
I want to sort a list of strings based on the string length. I tried to use sort as follows, but it doesn’t seem to give me correct result.
xs = ['dddd','a','bb','ccc']
print xs
xs.sort(lambda x,y: len(x) < len(y))
print xs
['dddd', 'a', 'bb', 'ccc']
['dddd', 'a', 'bb', 'ccc']
What might be wrong?
When you pass a lambda
to sort
, you need to return an integer, not a boolean. So your code should instead read as follows:
xs.sort(lambda x,y: cmp(len(x), len(y)))
Note that cmp is a builtin function such that cmp(x, y)
returns -1 if x
is less than y
, 0 if x
is equal to y
, and 1 if x
is greater than y
.
Of course, you can instead use the key
parameter:
xs.sort(key=lambda s: len(s))
This tells the sort
method to order based on whatever the key function returns.
EDIT: Thanks to balpha and Ruslan below for pointing out that you can just pass len
directly as the key parameter to the function, thus eliminating the need for a lambda
:
xs.sort(key=len)
And as Ruslan points out below, you can also use the built-in sorted function rather than the list.sort
method, which creates a new list rather than sorting the existing one in-place:
print(sorted(xs, key=len))
The same as in Eli’s answer – just using a shorter form, because you can skip a lambda
part here.
Creating new list:
>>> xs = ['dddd','a','bb','ccc']
>>> sorted(xs, key=len)
['a', 'bb', 'ccc', 'dddd']
In-place sorting:
>>> xs.sort(key=len)
>>> xs
['a', 'bb', 'ccc', 'dddd']
I Would like to add how the pythonic key function works while sorting :
Decorate-Sort-Undecorate Design Pattern :
Python’s support for a key function when sorting is implemented using what is known as the
decorate-sort-undecorate design pattern.
It proceeds in 3 steps:
-
Each element of the list is temporarily replaced with a “decorated” version that includes the result of the key function applied to the element.
-
The list is sorted based upon the natural order of the keys.
-
The decorated elements are replaced by the original elements.
Key parameter to specify a function to be called on each list element prior to making comparisons. docs
Write a function lensort to sort a list of strings based on length.
def lensort(a):
n = len(a)
for i in range(n):
for j in range(i+1,n):
if len(a[i]) > len(a[j]):
temp = a[i]
a[i] = a[j]
a[j] = temp
return a
print lensort(["hello","bye","good"])
def lensort(list_1):
list_2=[];list_3=[]
for i in list_1:
list_2.append([i,len(i)])
list_2.sort(key = lambda x : x[1])
for i in list_2:
list_3.append(i[0])
return list_3
This works for me!
The easiest way to do this is:
list.sort(key = lambda x:len(x))
I can do it using below two methods, using function
def lensort(x):
list1 = []
for i in x:
list1.append([len(i),i])
return sorted(list1)
lista = ['a', 'bb', 'ccc', 'dddd']
a=lensort(lista)
print([l[1] for l in a])
In one Liner using Lambda, as below, a already answered above.
lista = ['a', 'bb', 'ccc', 'dddd']
lista.sort(key = lambda x:len(x))
print(lista)