Seb - 6 months ago 15

Python Question

I am trying to return a tuple the smallest second index value (y value) from a list of tuples. If there are two tuples with the lowest y value, then select the tuple with the largest x value (i.e first index).

For example, suppose I have the tuple:

`x = [(2, 3), (4, 3), (6, 9)]`

The the value returned should be

`(4, 3)`

`(2, 3)`

`x[0][1]`

`3`

`x[1][1]`

`x[0][0]`

`x[1][0]`

So far I have tried:

`start_point = min(x, key = lambda t: t[1])`

However, this only checks the second index, and does not compare two tuples first index if their second index's are equivalent.

Answer

Include the `x`

value in a tuple returned from the key; this second element in the key will be then used when there is a tie for the `y`

value. To inverse the comparison (from smallest to largest), just negate that value:

```
min(x, key=lambda t: (t[1], -t[0]))
```

After all, `-4`

is smaller than `-2`

.

Demo:

```
>>> x = [(2, 3), (4, 3), (6, 9)]
>>> min(x, key=lambda t: (t[1], -t[0]))
(4, 3)
```

Source (Stackoverflow)

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