如何将字典中的数字列表平方？

Question

I want to square all the values in a Python dictionary, and am going about it the wrong way. Below is my attempt which obviously fails and yields a

TypeError: unsupported operand type(s)

And the program is

tides = {'G3': [3, 8, 9, 7], 'G2': [2, 7], 'G1': [1, 6, 7]}

for v in tides.values():
    print v**2

The desired output is

{'G3': [9, 64, 81, 49], 'G2': [4, 49], 'G1': [1, 36, 49]}

Answer 1

The dict.values gives you a list of items and each element of that list is a list. So, when you say v ** 2 , you are actually trying to square a list, which is not possible in Python as such. That is why it is failing.

Instead, just recreate the dictionary with dictionary comprehension, like this

>>> tides = {'G3': [3, 8, 9, 7], 'G2': [2, 7], 'G1': [1, 6, 7]}
>>> {tide: [value ** 2 for value in tides[tide]] for tide in tides}
{'G3': [9, 64, 81, 49], 'G2': [4, 49], 'G1': [1, 36, 49]}

Or you can use dict.iteritems like this

>>> {tide: [value ** 2 for value in values] for tide, values in tides.iteritems()}
{'G3': [9, 64, 81, 49], 'G2': [4, 49], 'G1': [1, 36, 49]}

Note: In both the examples I have shown above, I have avoided using dict.values , because that will create a temporary list of all the values from the dictionary. We don't want that.

In the first example, we iterate through the keys in the dictionary and get values corresponding to the key and square each value in it.

In the second example, we take one pair of key and value at a time from the dictionary. So, we don't create the temporary list.

The code within {...} shown above is called dictionary comprehension and within that, whatever is within [...] is called list comprehension .

Answer 2

When you do tides.values() what you get is:

[[3, 8, 9, 7], [2, 7], [1, 6, 7]]

Therefore when you do

for v in tides.values():
    print v ** 2

you're trying to do

[3, 8, 9, 7] ** 2 # and [2, 7] ** 2 and [1, 6, 7] ** 2

which is obviously gibberish. Instead you should try to do:

for key, value in tides.items():
    tides[key] = [num**2 for num in value]

Answer 3

You can use a dictionary comprehension:

tides = {'G3': [3, 8, 9, 7], 'G2': [2, 7], 'G1': [1, 6, 7]}
tides = {k: map(lambda num: num ** 2, v) for k, v in tides.items()}

Result:

{'G3': [9, 64, 81, 49], 'G2': [4, 49], 'G1': [1, 36, 49]}

map(lambda num: num ** 2, v) squares the values of each value v of tides .

If you just want to replace in place, you can do the following:

for v in tides.values():
    for i in range(len(v)):
        v[i] **= 2

Answer 4

tides.values() returns a list of lists. You are trying to square the lists rather than the values in the lists. You can add a second loop to iterate through the list.

for v in tides.values(): for x in v: print x**2

Answer 5

使用字典理解：

print {k: map(lambda x: x**2, [x for x in v]) for k, v in tides.items()}