CODEWITHSUNDEEP
pythonlistlist-comprehension
user3050527
user3050527

Reputation: 931

trying to do this simple function with list comprehension

So I'm making a function that basically checks if the string past the 0th element is an exclamation mark and the first element is the letter 'h'.

so example:

>>>f('h!')
True
>>>f('h!!!!!')
True
>>>f('!!!!!')
False
>>>f('!!!h')
False

i wanted to do this using list comprehension but i end up making the output a list with the first element being the truth value. My function is the following:

def f(s):
  return s[0] == 'h' and [i == '!' for i in s[1:]]

this returns
  >>>f('h!!')
  [True]

What's wrong with the way i've done it? if i remove the brackets after the 'and', it raises a syntax error.

Upvotes: 1

Views: 54

Answers (3)

John La Rooy
John La Rooy

Reputation: 304147

This is probably the fastest way

def f(s):
  return s == "h" + "!" * (len(s) - 1)

Upvotes: 1

Christian Tapia
Christian Tapia

Reputation: 34146

Try printing the list comprehension part:

print [i == '!' for i in s[1:]]

you will get a list of booleans like

>>> [True, True, True]

So if you want to apply the AND operator (returns True if all the elements are True) to that list (so you get the correct result), you can use the all() built-in function:

return s[0] == 'h' and all([i == '!' for i in s[1:]])

According to Python docs:

all(iterable): Return True if all elements of the iterable are true (or if the iterable is empty).

Edit:

As @thefourtheye commented, to avoid the list creation, you can pass a generator expression to all():

return s[0] == 'h' and all(i == '!' for i in s[1:])

Upvotes: 3

sashkello
sashkello

Reputation: 17871

Change to this:

s[0] == 'h' and all(i == '!' for i in s[1:])

all returns True if all the elements of the generator expression inside are returning True.

Upvotes: 1

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