在Python 3中对字典列表进行排序

Question

I am trying to sort a list of dictionaries. My goal is to sort dictionaries with multiple (possibly the same) keys in the same way, even if the dictionaries are in a different order or if the keys are in the dictionary in a different order.

In Python 2, I have used the following:

a = [{1: 2, 7: 8}, {7: 8, 3: 4}, {5: 6}]
b = [{3: 4, 7: 8}, {7: 8, 1: 2}, {5: 6}]
a.sort()
b.sort()
a
Out[20]: [{5: 6}, {1: 2, 7: 8}, {3: 4, 7: 8}]
b
Out[21]: [{5: 6}, {1: 2, 7: 8}, {3: 4, 7: 8}]

This succeeds in my goal of creating two sorted dictionaries that look exactly the same.

I am trying to do the same thing in Python 3, where .sort() does not work for a list of dictionaries.

I have tried different ways.

1.

sorted(a, key=lambda d: max(d.keys()))

This does not work:

a = [{1: 2, 7: 8}, {3: 4, 7: 8}, {5: 6}]
b = [{3: 4, 7: 8}, {1: 2, 7: 8}, {5: 6}]
a2 = sorted(a, key=lambda d: max(d.keys()))
b2 = sorted(b, key=lambda d: max(d.keys()))
a2
Out[1]: [{5: 6}, {1: 2, 7: 8}, {7: 8, 3: 4}]
b2
Out[2]: [{5: 6}, {3: 4, 7: 8}, {7: 8, 1: 2}]

2.

a2 = sorted([list(zip(x.keys(),x.values())) for x in a])
a3 = [{k: v for (k,v) in x} for x in a2]

This does not work:

a = [{1: 2, 7: 8}, {7: 8, 3: 4}, {5: 6}]
b = [{3: 4, 7: 8}, {7: 8, 1: 2}, {5: 6}]
a2 = sorted([list(zip(x.keys(),x.values())) for x in a])
a3 = [{k: v for (k,v) in x} for x in a2]
b2 = sorted([list(zip(x.keys(),x.values())) for x in b])
b3 = [{k: v for (k,v) in x} for x in b2]
a3
Out[1]: [{1: 2, 7: 8}, {5: 6}, {7: 8, 3: 4}]
b3
Out[2]: [{3: 4, 7: 8}, {5: 6}, {7: 8, 1: 2}]

Does anyone have an idea how I can get the Python 2 result in Python 3??

Answer 1

Sorting on all keys in the dictionaries can be done with:

a.sort(key=lambda d: d.keys())
b.sort(key=lambda d: d.keys())

To get the result you want we need to sort the keys on ascending order as follows:

a.sort(key=lambda d: sorted(list(d.keys()), reverse=True))
b.sort(key=lambda d: sorted(list(d.keys()), reverse=True))

This gives the following result:

>>> a
[{5: 6}, {1: 2, 7: 8}, {3: 4, 7: 8}]
>>> b
[{5: 6}, {1: 2, 7: 8}, {3: 4, 7: 8}]

Edit: In order to sort based on the values as well (asked in the comments) the following might work:

a.sort(key=lambda d: sorted(list(d.keys()) + sorted(list(d.values())), reverse=True))
b.sort(key=lambda d: sorted(list(d.keys()) + sorted(list(d.values())), reverse=True))

Answer 2

You need to use all items of each dict while sorting, and d.items() iterates items in arbitrary order, so they need to be sorted. And this is your sorting key:

a = [{1: 2, 7: 8}, {7: 8, 3: 4}, {5: 6}]
b = [{3: 4, 7: 8}, {7: 8, 1: 2}, {5: 6}]
sorted(a, key=lambda d: sorted(d.items()))
Out: [{1: 2, 7: 8}, {3: 4, 7: 8}, {5: 6}]

sorted(b, key=lambda d: sorted(d.items()))
Out: [{1: 2, 7: 8}, {3: 4, 7: 8}, {5: 6}]

x = [{1: 2},{1: 3}]
y = [{1: 3},{1: 2}]
sorted(x, key=lambda d: sorted(d.items()))
Out: [{1: 2}, {1: 3}]

sorted(y, key=lambda d: sorted(d.items()))
Out: [{1: 2}, {1: 3}]

Note that sorted(a, key=...) creates a new list, while a.sort(key=...) makes sorting in place.