I have a dictionary of points, say:

```
>>> points={'a':(3,4), 'b':(1,2), 'c':(5,5), 'd':(3,3)}
```

I want to create a new dictionary with all the points whose x and y value is smaller than 5, i.e. points 'a', 'b' and 'd'.

According to the the book, each dictionary has the `items()`

function, which returns a list of `(key, pair)`

tuple:

```
>>> points.items()
[('a', (3, 4)), ('c', (5, 5)), ('b', (1, 2)), ('d', (3, 3))]
```

So I have written this:

```
>>> for item in [i for i in points.items() if i[1][0]<5 and i[1][1]<5]:
... points_small[item[0]]=item[1]
...
>>> points_small
{'a': (3, 4), 'b': (1, 2), 'd': (3, 3)}
```

Is there a more elegant way? I was expecting Python to have some super-awesome `dictionary.filter(f)`

function...

Nowadays, in Python 2.7 and up, you can use a dict comprehension:

```
{k: v for k, v in points.iteritems() if v[0] < 5 and v[1] < 5}
```

And in Python 3:

```
{k: v for k, v in points.items() if v[0] < 5 and v[1] < 5}
```

```
dict((k, v) for k, v in points.items() if all(x < 5 for x in v))
```

You could choose to call `.iteritems()`

instead of `.items()`

if you're in Python 2 and `points`

may have a **lot** of entries.

`all(x < 5 for x in v)`

may be overkill if you know for sure each point will always be 2D only (in that case you might express the same constraint with an `and`

) but it will work fine;-).

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