Check if all elements in a list are identical

Question

I need a function which takes in a list and outputs True if all elements in the input list evaluate as equal to each other using the standard equality operator and False otherwise  I feel it would be best to iterate through the list comparing adjacent elements and then AND all the resulting Boolean values  But I m not sure what s the most Pythonic way to do that

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A set comparison work   len set the list      1   Using set removes all duplicate elements

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You can use  nunique   to find number of unique items in a list   def identical elements list       series   pd Series list      if series nunique      1  identical   True     else   identical   False     return identical    identical elements   a    a    Out 427   True  identical elements   a    b    Out 428   False

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Can use map and lambda  lst    1 1 1 1 1 1 1 1 1   print all map lambda x  x    lst 0   lst 1

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The simplest and most elegant way is as follows   all x  myList 0  for x in myList     Yes  this even works with the empty list  This is because this is one of the few cases where python has lazy semantics    Regarding performance  this will fail at the earliest possible time  so it is asymptotically optimal

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Or use diff method of numpy   import numpy as np def allthesame l       return np unique l  shape 0  lt  1   And to call   print allthesame  1 1 1      Output   True

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If you re interested in something a little more readable  but of course not as efficient   you could try   def compare lists list1  list2       if len list1     len list2     Weed out unequal length lists          return False     for item in list1          if item not in list2              return False     return True  a list 1     apple    orange    grape    pear   a list 2     pear    orange    grape    apple    b list 1     apple    orange    grape    pear   b list 2     apple    orange    banana    pear    c list 1     apple    orange    grape   c list 2     grape    orange    print compare lists a list 1  a list 2    Returns True print compare lists b list 1  b list 2    Returns False print compare lists c list 1  c list 2    Returns False

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For what it s worth  this came up on the python-ideas mailing list recently   It turns out that there is an itertools recipe for doing this already 1  def all equal iterable        Returns True if all the elements are equal to each other      g   groupby iterable      return next g  True  and not next g  False    Supposedly it performs very nicely and has a few nice properties    Short-circuits   It will stop consuming items from the iterable as soon as it finds the first non-equal item  Doesn t require items to be hashable  It is lazy and only requires O 1  additional memory to do the check    1In other words  I can t take the credit for coming up with the solution -- nor can I take credit for even finding it

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The simple solution is to apply set on list if all elements are identical len will be 1 else greater than 1 lst    1 1 1 1 1 1 1 1 1  len lst   len list set lst     print len lst   1   lst    1 2 1 1 1 1 1 1 1  len lst   len list set lst    print len lst   2

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I d do   not any  x i     x i 1  for i in range 0  len x -1      as any stops searching the iterable as soon as it finds a True condition

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There is also a pure Python recursive option  def checkEqual lst       if len lst   2           return lst 0   lst 1      else          return lst 0   lst 1  and checkEqual lst 1     However for some reason it is in some cases two orders of magnitude slower than other options  Coming from C language mentality  I expected this to be faster  but it is not  The other disadvantage is that there is recursion limit in Python which needs to be adjusted in this case  For example using this

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Check if all elements equal to the first   np allclose array  array 0

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def allTheSame i       j   itertools groupby i      for k in j  break     for k in j  return False     return True   Works in Python 2 4  which doesn t have  all

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This is another option  faster than len set x    1 for long lists  uses short circuit   def constantList x       return x and  x 0   len x     x

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Or use diff method of numpy   import numpy as np def allthesame l       return np all np diff l   0    And to call   print allthesame  1 1 1      Output   True

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You can convert the list to a set  A set cannot have duplicates  So if all the elements in the original list are identical  the set will have just one element  if len set input list      1        input list has all identical elements

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gt  gt  gt  a    1  2  3  4  5  6   gt  gt  gt  z     a x   a x 1   for x in range 0  len a -1    gt  gt  gt  z   1  2    2  3    3  4    4  5    5  6     Replacing it with the test  gt  gt  gt  z     a x     a x 1   for x in range 0  len a -1    gt  gt  gt  z  False  False  False  False  False   gt  gt  gt  if False in z   Print  All elements are not equal

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You can do   reduce and    x  yourList 0  for x in yourList   True    It is fairly annoying that python makes you import the operators like operator and   As of python3  you will need to also import functools reduce    You should not use this method because it will not break if it finds non-equal values  but will continue examining the entire list  It is just included here as an answer for completeness

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This is a simple way of doing it   result   mylist and all mylist 0     elem for elem in mylist    This is slightly more complicated  it incurs function call overhead  but the semantics are more clearly spelled out   def all identical seq       if not seq            empty list is False          return False     first   seq 0      return all first    elem for elem in seq

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Regarding using reduce   with lambda  Here is a working code that I personally think is way nicer than some of the other answers   reduce lambda x  y   x 1   y  y    2  2  2    True  2     Returns a tuple where the first value is the boolean if all items are same or not

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Use itertools groupby  see the itertools recipes   from itertools import groupby def all equal iterable       g   groupby iterable      return next g  True  and not next g  False   or without groupby  def all equal iterator       iterator   iter iterator      try          first   next iterator      except StopIteration          return True     return all first    rest for rest in iterator    There are a number of alternative one-liners you might consider   Converting the input to a set and checking that it only has one or zero  in case the input is empty  items def all equal2 iterator       return len set iterator    lt   1   Comparing against the input list without the first item def all equal3 lst       return lst  -1     lst 1     Counting how many times the first item appears in the list def all equal ivo lst       return not lst or lst count lst 0      len lst    Comparing against a list of the first element repeated def all equal 6502 lst       return not lst or  lst 0   len lst     lst    But they have some downsides  namely   all equal and all equal2 can use any iterators  but the others must take a sequence input  typically concrete containers like a list or tuple  all equal and all equal3 stop as soon as a difference is found  what is called  quot short circuit quot    whereas all the alternatives require iterating over the entire list  even if you can tell that the answer is False just by looking at the first two elements  In all equal2 the content must be hashable  A list of lists will raise a TypeError for example  all equal2  in the worst case  and all equal 6502 create a copy of the list  meaning you need to use double the memory   On Python 3 9  using perfplot  we get these timings  lower Runtime  s  is better

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Doubt this is the  most Pythonic   but something like    gt  gt  gt  falseList    1 2 3 4   gt  gt  gt  trueList    1  1  1   gt  gt  gt    gt  gt  gt  def testList list         for item in list 1            if item    list 0             return False       return True       gt  gt  gt  testList falseList  False  gt  gt  gt  testList trueList  True   would do the trick

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A solution faster than using set   that works on sequences  not iterables  is to simply count the first element  This assumes the list is non-empty  but that s trivial to check  and decide yourself what the outcome should be on an empty list   x count x 0      len x    some simple benchmarks    gt  gt  gt  timeit timeit  len set s1   lt  1    s1  1  5000   number 10000  1 4383411407470703  gt  gt  gt  timeit timeit  len set s1   lt  1    s1  1  4999  2    number 10000  1 4765670299530029  gt  gt  gt  timeit timeit  s1 count s1 0    len s1     s1  1  5000   number 10000  0 26274609565734863  gt  gt  gt  timeit timeit  s1 count s1 0    len s1     s1  1  4999  2    number 10000  0 25654196739196777

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lambda lst  reduce lambda a b  b b  a 0  and a 1    lst   lst 0   True   1    The next one will short short circuit   all itertools imap lambda i yourlist i   yourlist i 1   xrange len yourlist -1

[python] Check if all elements in a list are identical

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