'if' not followed by conditional statement

Question

I'm going through Zed's "Learn Python The Hard Way" and I'm on ex49. I'm quite confused by the following code he gives:

def peek(word_list):
    if word_list: # this gives me trouble
        word = word_list[0]
        return word[0]
    else:
        return None

The condition of the if statement is giving me trouble, as commented. I'm not sure what this means as word_list is an object, not a conditional statement. How can word_list, just by itself, follow if?

Answer 1

The if statement applies the built-in bool() function to the expression which follows. In your case, the code-block inside the if statement only runs if bool(word_list) is True.

Different objects in Python evaluate to either True or False in a Boolean context. These objects are considered to be 'Truthy' or 'Falsy'. For example:

In [180]: bool('abc')
Out[180]: True

In [181]: bool('')
Out[181]: False

In [182]: bool([1, 2, 4])
Out[182]: True

In [183]: bool([])
Out[183]: False

In [184]: bool(None)
Out[184]: False

The above are examples of the fact that:

• strings of length >= 1 are Truthy.
• empty strings are Falsy.
• lists of length >= 1 are Truthy.
• empty lists are Falsy.
• None is Falsy.

So: if word_list will evaluate to True if it is a non-empty list. However, if it is an empty list or None it will evaluate to False.

Answer 2

He is checking if word_list is empty or not. If a list is empty and it is used in a conditional statement, it is evaluated to False. Otherwise, it is evaluated to True.

word_list = ['some value']
if word_list:
    # list is not empty do some stuff
    print "I WILL PRINT"


word_list = []
if word_list:
    # list is empty
    print "I WILL NOT PRINT"

In the above code, only the first snippet will print.

See the following reference: https://docs.python.org/2/library/stdtypes.html#truth-value-testing