# Find odd numbers in a list and square them using lambda function

Asked by At

Here is the code I have got so far. Now output is `[1, 3, 5, 7, 9]`

``````N = 10
for i in range(1, 10):
arr.append(i)

arr2 = []
f = lambda x: x ** 2
arr2 = filter(lambda x: x % 2 != 0, arr)
map(lambda x: x ** 2, arr2)
print(list(arr2))```
``````

## 4 Answers On Best Solutions

You are discarding the result of `f(i)` as soon as you create it. You need to append it to some list (also, no need to consume the `filter` object into a list):

``````result = []
for i in arr2:
result.append(f(i))
``````

Please note that binding a lambda to an identifier is discouraged in accordance with PEP 8.

The best way to solve this problem is without list comprehensions is a combination of `filter` and `map` like so:

``````arr2 = list(map(lambda x: x ** 2, filter(lambda x: x % 2 != 0, arr)))
`````` On

Here's a very slightly modified version:

``````arr = []
N = 10
for i in range(1, N):
arr.append(i)

arr2 = []
f = lambda x: x ** 2
arr2 = filter(lambda x: x % 2 != 0, arr)
for i in list(arr2):
print(f(i))
``````

`arr2` isn't a list. It's an iterator, which you can only convert to a list once.

Here's a more compact version:

``````N = 10
arr = range(1, N)

square = lambda x: x ** 2
keep_odd = lambda x: x % 2 != 0
arr2 = list(filter(keep_odd, arr))
for i in arr2:
print(square(i))

print(arr2)
``````

It outputs:

``````1
9
25
49
81
[1, 3, 5, 7, 9]
`````` On

You aren't saving the value in the array, you are just printing it.

``````N = 10
for i in range(1, 10):
arr.append(i)

result = []
f = lambda x: x ** 2
arr2 = filter(lambda x: x % 2 != 0, arr)
for i in arr2:
result.append(f(i))
print(result)
`````` On

Your last for-loop applies the lambda function to the elements in your list, but does not save the result. Try:

``````a = [i for i in range(1,10)]
a2 = filter(lambda x: x % 2 != 0, a)
a3 = map(lambda x: x**2, a2)        # This is a generator object
final_list = list(a3)               # This is a list
``````

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