Add string concatenation

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Satwik Kansal 2017-08-30 14:52:11 +05:30
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README.md
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@ -975,6 +975,75 @@ TabError: inconsistent use of tabs and spaces in indentation
> First, tabs are replaced (from left to right) by one to eight spaces such that the total number of characters up to and including the replacement is a multiple of eight <...>
* So the "tab" at the last line of `square` function is replace with 8 spaces and it gets into the loop.
## Class attributes and instance attributes
1.
```py
class A:
x = 1
class B(A):
pass
class C(A):
pass
```
**Ouptut:**
```py
>>> A.x, B.x, C.x
(1, 1, 1)
>>> B.x = [2]
>>> A.x, B.x, C.x
(1, 2, 1)
>>> A.x = 3
>>> A.x, B.x, C.x
(3, 2, 3)
>>> a = A()
>>> a.x, A.x
(3, 3)
>>> a.x += 1
>>> a.x, A.x
(4, 3)
```
2.
```py
class SomeClass:
some_var = 15
some_list = [5]
another_list = [5]
def __init__(self, x):
self.some_var = x + 1
self.some_list = self.some_list + [x]
self.another_list += [x]
```
**Output:**
```py
>>> some_obj = SomeClass(420)
>>> some_obj.some_list
[5, 420]
>>> some_obj.another_list
[5, 420]
>>> another_obj = SomeClass(111)
>>> another_obj.some_list
[5, 111]
>>> another_obj.another_list
[5, 420, 111]
>>> another_obj.another_list is SomeClass.another_list
True
>>> another_obj.another_list is some_obj.another_list
True
```
### Explaination:
* Class variables and variables in class instances are internally handled as dictionaries of a class object. If a variable name is not found in the dictionary of current class, the parent classes are searched for it.
* The `+=` operator modifies the mutable object in-place without creating a new object.
## Catching the Exceptions!
```py
@ -1048,6 +1117,53 @@ SyntaxError: invalid syntax
list.remove(x): x not in list
```
## String concatenation
This is not a WTF, but just a nice thing to be aware of :)
```py
def add_string_with_plus(iters):
s = ""
for i in range(iters):
s += "xyz"
assert len(s) == 3*iters
def add_string_with_format(iters):
fs = "{}"*iters
s = fs.format(*(["xyz"]*iters))
assert len(s) == 3*iters
def add_string_with_join(iters):
l = []
for i in range(iters):
l.append("xyz")
s = "".join(l)
assert len(s) == 3*iters
def convert_list_to_string(l, iters):
s = "".join(l)
assert len(s) == 3*iters
```
**Output:**
```py
>>> timeit(add_string_with_plus(10000))
100 loops, best of 3: 9.73 ms per loop
>>> timeit(add_string_with_format(10000))
100 loops, best of 3: 5.47 ms per loop
>>> timeit(add_string_with_join(10000))
100 loops, best of 3: 10.1 ms per loop
>>> l = ["xyz"]*10000
>>> timeit(convert_list_to_string(l, 10000))
10000 loops, best of 3: 75.3 µs per loop
```
### Explanination
- You can read more about [timeit](https://docs.python.org/3/library/timeit.html) from here. It is generally used to measure the execution time of snippets.
- Don't use `+` for generating long strings — In Python, `str` is immutable, so the left and right strings have to be copied into the new string for every pair of concatenations. If you concatenate four strings of length 10, you'll be copying (10+10) + ((10+10)+10) + (((10+10)+10)+10) = 90 characters instead of just 40 characters. Things get quadratically worse as the number and size of the string increases.
- Therefore, it's advised to use .format or % syntax (however, they are slightly slower than + for short strings).
- Or better, if already you've contents available in the form of an iterable object, then use ''.join(iterable_object) which is much faster.
## Minor Ones
@ -1112,6 +1228,9 @@ tuple()
* The correct statement for expected behavior is `t = ('one',)` or `t = 'one',` (missing comma) otherwise the interpreter considers `t` to be a `str` and iterates over it character by character.
* `()` is a special token and denotes empty `tuple`.
# TODO: Hell of an example!
Trying to comeup with an example that combines multiple examples discussed above, making it difficult for the reader to guess the output correctly :sweat_smile:.
# Contributing