Getting indices of True values in a boolean list

Question

I have a piece of my code where I m supposed to create a switchboard  I want to return a list of all the switches that are on  Here  on  will equal True and  off  equal False  So now I just want to return a list of all the True values and their position  This is all I have but it only return the position of the first occurrence of True  this is just a portion of my code    self states    False  False  False  False  True  True  False  True  False  False  False  False  False  False  False  False   def which switch self       x    self states index i  for i in self states if i    True    This only returns  4

User · Answer

You can use filter for it   filter lambda x  self states x   range len self states      The range here enumerates elements of your list and since we want only those where self states is True  we are applying a filter based on this condition   For Python   3 0   list filter lambda x  self states x   range len self states

User · Answer

If you have numpy available    gt  gt  gt  import numpy as np  gt  gt  gt  states    False  False  False  False  True  True  False  True  False  False  False  False  False  False  False  False   gt  gt  gt  np where states  0  array  4  5  7

User · Answer

Simply do this   def which index self       return           i for i in range len self states           if self states i     True

User · Answer

TL  DR  use np where as it is the fastest option  Your options are np where  itertools compress  and list comprehension    See the detailed comparison below  where it can be seen np where outperforms both itertools compress and also list comprehension     gt  gt  gt  from itertools import compress  gt  gt  gt  import numpy as np  gt  gt  gt  t    False  False  False  False  True  True  False  True  False  False  False  False  False  False  False  False    gt  gt  gt  t   1000 t    Method 1  Using list comprehension    gt  gt  gt   timeit  i for i  x in enumerate t  if x  457   s    1 5   s per loop  mean    std  dev  of 7 runs  1000 loops each     Method 2  Using itertools compress    gt  gt  gt   timeit list compress range len t    t   210   s    704 ns per loop  mean    std  dev  of 7 runs  1000 loops each     Method 3  the fastest method   Using numpy where    gt  gt  gt   timeit np where t  179   s    593 ns per loop  mean    std  dev  of 7 runs  10000 loops each

User · Answer

Using element-wise multiplication and a set    gt  gt  gt  states    False  False  False  False  True  True  False  True  False  False  False  False  False  False  False  False   gt  gt  gt  set multiply states range 1 len states  1  -1  difference  -1     Output   4  5  7

User · Answer

Use enumerate  list index returns the index of first match found    gt  gt  gt  t    False  False  False  False  True  True  False  True  False  False  False  False  False  False  False  False   gt  gt  gt   i for i  x in enumerate t  if x   4  5  7    For huge lists  it d be better to use itertools compress    gt  gt  gt  from itertools import compress  gt  gt  gt  list compress xrange len t    t    4  5  7   gt  gt  gt  t   t 1000  gt  gt  gt   timeit  i for i  x in enumerate t  if x  100 loops  best of 3  2 55 ms per loop  gt  gt  gt   timeit list compress xrange len t    t   1000 loops  best of 3  696   s per loop

User · Answer

Use dictionary comprehension way    x    k v for k v in enumerate states  if v    True    Input   states    False  False  False  False  True  True  False  True  False  False  False  False  False  False  False  False    Output    4  True  5  True  7  True

[python] Getting indices of True values in a boolean list

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