user2988577 - 13 days ago 5x

Python Question

I'm working with a big dictionary and for some reason I also need to work on small random samples from that dictionary. How can I get this small sample (for example of length 2)?

Here is a toy-model:

`dy={'a':1, 'b':2, 'c':3, 'd':4, 'e':5}`

I need to perform some task on dy which involves all the entries. Let us say, to simplify, I need to sum together all the values:

`s=0`

for key in dy.key:

s=s+dy[key]

Now, I also need to perform the same task on a random sample of dy; for that I need a random sample of the keys of dy. The simple solution I can imagine is

`sam=list(dy.keys())[:1]`

In that way I have a list of two keys of the dictionary which are somehow random. So, going back to may task, the only change I need in the code is:

`s=0`

for key in sam:

s=s+dy[key]

The point is I do not fully understand how dy.keys is constructed and then I can't foresee any future issue

Answer

Given your example of:

```
dy = {'a':1, 'b':2, 'c':3, 'd':4, 'e':5}
```

Then the sum of all the values is more simply put as:

```
s = sum(dy.values())
```

Then if it's not memory prohibitive, you can sample using:

```
import random
values = list(dy.values())
s = sum(random.sample(values, 2))
```

Or, since `random.sample`

can take a `set`

-like object, then:

```
from operator import itemgetter
import random
s = sum(itemgetter(*random.sample(dy.keys(), 2))(dy))
```

Or just use:

```
s = sum(dy[k] for k in random.sample(dy.keys(), 2))
```

An alternative is to use a `heapq`

, eg:

```
import heapq
import random
s = sum(heapq.nlargest(2, dy.values(), key=lambda L: random.random()))
```

Source (Stackoverflow)

Comments