SubodhD - 5 months ago 7

Python Question

I have list like this:

`lst = [1, 5, 0, 0, 4, 0]`

EDIT:

I would like find the location of the first number in list that is not equal to zero and it is

`a = 3 # (given number)`

It should return 4 as index and value too.

I tried with this:

`min(enumerate(lst), key=lambda x: abs(x[1]-a))`

but its shows the 0 element too.

which is the better way for doing this?

Answer

If I understood you correctly, just filter out the zeros after enumerating:

```
In [1]: lst = [1, 5, 0, 0, 4, 0]
In [2]: a = 3
In [3]: min(filter(lambda x: x[1] != 0, enumerate(lst)),
key=lambda x: abs(x[1] - a))
Out[3]: (4, 4)
```

Your own example does return `(4, 4)`

too for the given values though. With `a = 0`

there's a difference.

Source (Stackoverflow)

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