Samuel Li - 1 year ago 46

Python Question

I'm having a bit of trouble understanding how Python treats and evaluates lambdas at runtime.

Consider the following code (Python 3.5.2):

`x = 0`

for iteration in range(3):

x = x + 1

print(x)

As expected, this prints 3. Here is my understanding of the way x changes over 3 iterations:

**Initial Value:**`x`

**Iteration 1:**`x+1`

**Iteration 2:**`(x+1) + 1`

**Iteration 3:**`((x+1) + 1) + 1`

Consider the following code:

`add3 = lambda x: x`

for iteration in range(3):

add3 = lambda x: add3(x) + 1

print(add3(0))

Here is my understanding of the way add3

**Initial Value:**`lambda x: x`

**Iteration 1:**`lambda x: (lambda x: x)(x) + 1`

**Iteration 2:**`lambda x: (lambda x: (lambda x: x)(x) + 1)(x) + 1`

**Iteration 3:**`lambda x: (lambda x: (lambda x: (lambda x: x)(x) + 1)(x) + 1)(x) + 1`

Instead, calling add3 causes the maximum recursion depth to be exceeded.

My first thought was that Python is dynamically looking up the function body from its name at call time, rather than storing the function's code as part of the function. However, even the following code does not work:

`functionList = [lambda x: x] #Store each iteration separately`

for i in range(3):

oldFunction = functionList[-1]

newFunction = lambda x: oldFunction(x) + 1 #Should be a completely new lambda object

functionList.append(newFunction)

print(functionList[-1](0)) #Causes infinite recursion

Even with no named functions whatsoever, and following the suggestion here (although I may have misunderstood his answer), it still fails:

`functionList = [lambda x: x]`

for i in range(3):

functionList.append(lambda x, i=i: functionList[-1](x) + 1)

print(functionList[-1](0)) #Causes infinite recursion

The four lambdas contained in functionList are completely separate objects in memory:

`>>> print(functionList)`

[<function <lambda> at 0x00000266D41A12F0>, <function <lambda> at 0x00000266D41D7E18>, <function <lambda> at 0x00000266D41D7730>, <function <lambda> at 0x00000266D41D7840>]

Could someone please explain this behavior?

Answer Source

This behavior has nothing to do with 'iterational' lambda generation. When you say `add3 = lambda x: add3(x) + 1`

, the `add3`

object is *replaced* with a lambda *calling itself recursively with no termination condition*.

So when you call `add3(0)`

, it becomes:

```
add3(0) = add3(0) + 1 = (add3(0) + 1) + 1 = ((add3(0) + 1) + 1) + 1
```

And this goes on forever (well, until the maximum recursion depth is exceeded).

As for other examples, the second function in your list already fails with `RecursionError: maximum recursion depth exceeded`

.

I've got this code for you:

```
import copy
flist=[lambda x: x]
flist.append(lambda x: copy.deepcopy(flist[-1])(x) + 1)
>>> flist
[<function <lambda> at 0x101d45268>, <function <lambda> at 0x101bf1a60>]
```

So we made sure that we call a *copy* of a function. `flist[-1](0)`

results in a `RecursionError`

, and the error is raised in the `deepcopy`

module. So, this means that `copy.deepcopy(flist[-1])(x)`

means 'copy the last element *currently* in the list and run the copy'.

Here it is: the last element of the list calls itself over and over again.