## Working with lists

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It is not too uncommon to have a list of data, and then to apply a function to every element, to filter the list, or extract elements that meet some criteria. In this example, we take a string and split it into words. Then, we will examine several ways to apply functions to the words, to filter the list to get data that meets some criteria. Here is the string splitting.

text = '''
As we have seen, handling units with third party functions is fragile, and often requires additional code to wrap the function to handle the units. An alternative approach that avoids the wrapping is to rescale the equations so they are dimensionless. Then, we should be able to use all the standard external functions without modification. We obtain the final solutions by rescaling back to the answers we want.

Before doing the examples, let us consider how the quantities package handles dimensionless numbers.

import quantities as u

a = 5 * u.m
L = 10 * u.m # characteristic length

print a/L
print type(a/L)

'''

words = text.split()
print words

... ... ... ... ... ... ... ... ... ... ... ... >>> >>> >>> ['As', 'we', 'have', 'seen,', 'handling', 'units', 'with', 'third', 'party', 'functions', 'is', 'fragile,', 'and', 'often', 'requires', 'additional', 'code', 'to', 'wrap', 'the', 'function', 'to', 'handle', 'the', 'units.', 'An', 'alternative', 'approach', 'that', 'avoids', 'the', 'wrapping', 'is', 'to', 'rescale', 'the', 'equations', 'so', 'they', 'are', 'dimensionless.', 'Then,', 'we', 'should', 'be', 'able', 'to', 'use', 'all', 'the', 'standard', 'external', 'functions', 'without', 'modification.', 'We', 'obtain', 'the', 'final', 'solutions', 'by', 'rescaling', 'back', 'to', 'the', 'answers', 'we', 'want.', 'Before', 'doing', 'the', 'examples,', 'let', 'us', 'consider', 'how', 'the', 'quantities', 'package', 'handles', 'dimensionless', 'numbers.', 'import', 'quantities', 'as', 'u', 'a', '=', '5', '*', 'u.m', 'L', '=', '10', '*', 'u.m', '#', 'characteristic', 'length', 'print', 'a/L', 'print', 'type(a/L)']


Let us get the length of each word.

print [len(word) for word in words]

# functional approach with a lambda function
print map(lambda word: len(word), words)

# functional approach with a builtin function
print map(len, words)

# functional approach with a user-defined function
def get_length(word):
return len(word)

print map(get_length, words)

[2, 2, 4, 5, 8, 5, 4, 5, 5, 9, 2, 8, 3, 5, 8, 10, 4, 2, 4, 3, 8, 2, 6, 3, 6, 2, 11, 8, 4, 6, 3, 8, 2, 2, 7, 3, 9, 2, 4, 3, 14, 5, 2, 6, 2, 4, 2, 3, 3, 3, 8, 8, 9, 7, 13, 2, 6, 3, 5, 9, 2, 9, 4, 2, 3, 7, 2, 5, 6, 5, 3, 9, 3, 2, 8, 3, 3, 10, 7, 7, 13, 8, 6, 10, 2, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 3, 1, 14, 6, 5, 3, 5, 9]
>>> ... [2, 2, 4, 5, 8, 5, 4, 5, 5, 9, 2, 8, 3, 5, 8, 10, 4, 2, 4, 3, 8, 2, 6, 3, 6, 2, 11, 8, 4, 6, 3, 8, 2, 2, 7, 3, 9, 2, 4, 3, 14, 5, 2, 6, 2, 4, 2, 3, 3, 3, 8, 8, 9, 7, 13, 2, 6, 3, 5, 9, 2, 9, 4, 2, 3, 7, 2, 5, 6, 5, 3, 9, 3, 2, 8, 3, 3, 10, 7, 7, 13, 8, 6, 10, 2, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 3, 1, 14, 6, 5, 3, 5, 9]
>>> [2, 2, 4, 5, 8, 5, 4, 5, 5, 9, 2, 8, 3, 5, 8, 10, 4, 2, 4, 3, 8, 2, 6, 3, 6, 2, 11, 8, 4, 6, 3, 8, 2, 2, 7, 3, 9, 2, 4, 3, 14, 5, 2, 6, 2, 4, 2, 3, 3, 3, 8, 8, 9, 7, 13, 2, 6, 3, 5, 9, 2, 9, 4, 2, 3, 7, 2, 5, 6, 5, 3, 9, 3, 2, 8, 3, 3, 10, 7, 7, 13, 8, 6, 10, 2, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 3, 1, 14, 6, 5, 3, 5, 9]
>>> ... ... ... >>> [2, 2, 4, 5, 8, 5, 4, 5, 5, 9, 2, 8, 3, 5, 8, 10, 4, 2, 4, 3, 8, 2, 6, 3, 6, 2, 11, 8, 4, 6, 3, 8, 2, 2, 7, 3, 9, 2, 4, 3, 14, 5, 2, 6, 2, 4, 2, 3, 3, 3, 8, 8, 9, 7, 13, 2, 6, 3, 5, 9, 2, 9, 4, 2, 3, 7, 2, 5, 6, 5, 3, 9, 3, 2, 8, 3, 3, 10, 7, 7, 13, 8, 6, 10, 2, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 3, 1, 14, 6, 5, 3, 5, 9]


Now let us get all the words that start with the letter “a”. This is sometimes called filtering a list. We use a string function startswith to check for upper and lower-case letters. We will use list comprehension with a condition.

print [word for word in words if word.startswith('a') or word.startswith('A')]

# make word lowercase to simplify the conditional statement
print [word for word in words if word.lower().startswith('a')]

['As', 'and', 'additional', 'An', 'alternative', 'approach', 'avoids', 'are', 'able', 'all', 'answers', 'as', 'a', 'a/L']
['As', 'and', 'additional', 'An', 'alternative', 'approach', 'avoids', 'are', 'able', 'all', 'answers', 'as', 'a', 'a/L']


A slightly harder example is to find all the words that are actually numbers. We could use a regular expression for that, but we will instead use a function we create. We use a function that tries to cast a word as a float. If this fails, we know the word is not a float, so we return False.

def float_p(word):
try:
float(word)
return True
except ValueError:
return False

print [word for word in words if float_p(word)]

# here is a functional approach
print filter(float_p, words)

... ... ... ... ... >>> ['5', '10']
['5', '10']


Finally, we consider filtering the list to find all words that contain certain symbols, say any character in this string “./=*#”. Any of those characters will do, so we search each word for one of them, and return True if it contains it, and False if none are contained.

def punctuation_p(word):
S = './=*#'
for s in S:
if s in word:
return True
return False

print [word for word in words if punctuation_p(word)]
print filter(punctuation_p, words)

... ... ... ... ... >>> ['units.', 'dimensionless.', 'modification.', 'want.', 'numbers.', '=', '*', 'u.m', '=', '*', 'u.m', '#', 'a/L', 'type(a/L)']
['units.', 'dimensionless.', 'modification.', 'want.', 'numbers.', '=', '*', 'u.m', '=', '*', 'u.m', '#', 'a/L', 'type(a/L)']


In this section we examined a few ways to interact with lists using list comprehension and functional programming. These approaches make it possible to work on arbitrary size lists, without needing to know in advance how big the lists are. New lists are automatically generated as results, without the need to preallocate lists, i.e. you do not need to know the size of the output. This can be handy as it avoids needing to write loops in some cases and leads to more compact code.