使用Lambda函数进行排序

Question

regular unsorted dictionary

>>> d = {'banana': 3, 'apple': 4, 'pear': 1, 'orange': 2}

dictionary sorted by key

>>> OrderedDict(sorted(d.items(), key=lambda t: t[0]))

dictionary sorted by value

>>> OrderedDict(sorted(d.items(), key=lambda t: t[1]))

How lambda function is working in the above lines. what is t[0], t[1]?

Answer 1

Working of lambda function in here is like, Sort the list of tuple with the second item. sorted(d.items(), key=lambda t: t[1])

d.items() will be like this.

[('orange', 2), ('pear', 1), ('banana', 3), ('apple', 4)]

So lambda t:t[1] it return a second elemnt in a given iterator. So it will sort based on the second element. Same as for the lambda t:t[0] it's sort with first element.

Example to understand the lambda function.

In [39]: func = lambda t:t[1]

In [40]: [func(i) for i in d.items()]
Out[40]: [2, 1, 3, 4]

Answer 2

lambda is anonymous function. It is convenient to use when specifying sort order, but there is nothing magic about lambda , if you wanted too, you could define the function explicitly:

def sorter(x):                                                                                               
    return x[1]                                                                                              

print sorted(d.items(), key=sorter) 

t is the an element in d.items() , so when you choose t[0] you are sorting by first element and when you choose t[1] you are sorting by second element. If you were to choose t[2] you would get IndexError .

Answer 3

First some background: The item() function returns a list of tuples in the dictionary.

This line:

>>> OrderedDict(sorted(d.items(), key=lambda t: t[0]))

In this line, the sorted function iterates through each item (a tuple), and as per the passed lambda uses the first element of the tuple (the key) to sort ( t[0] ). The second example follows the same idea, except with the second element of the tuple (the value). As such, each line respectively results in sorting by key and value.