Jerry Luke - 7 days ago 4x

Python Question

Is there a more idiomatic way to accomplish the following in Python3?

`if i%1 == 0 and i%2 == 0 and i%3 == 0 and i%4 == 0 and i%5 == 0 and i%6 == 0 and i%7 == 0 and i%8 == 0 and i%9 == 0 and i%10 == 0 and i%11 == 0 and i%12 == 0 and i%13 == 0 and i%14 == 0 and i%15 == 0 and i%16 == 0 and i%17 == 0 and i%18 == 0 and i%19 == 0 and i%20 == 0:`

I'm trying to find the smallest positive number that is evenly divisible by all of the numbers from 1 to 20. I'm not looking for a new solution. I'm looking for a neater way to express what I am doing above.

Answer

```
if all(i % j == 0 for j in range(1, 21)): # python2 -> xrange(2, 21)
# do whatever
```

If all `i % j == 0`

, it will return *True* otherwise it will *short circuit* and return *False* if there is any remainder for `i % j`

. Also, checking `if i % 1`

is redundant so you can start at 2.

Or conversly, check if there is *not* any `i % j`

with a remainder.

```
if not any(i % j for j in range(2, 21)):
```

Or if you prefer *functional*:

```
if not any(map(i.__mod__, range(2, 21)))
```

Source (Stackoverflow)

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