Is i = i + n truly the same as i += n?

up vote
6
down vote

favorite

One block of code works but the other does not. Which would make sense except the second block is the same as the first only with an operation written in shorthand. They are practically the same operation.

l = ['table']
i =

Version 1

for n in l:
 i += n
print(i)

Output: ['t', 'a', 'b', 'l', 'e']

Version 2

for n in l:
 i = i + n
print(i)

Output:

TypeError: can only concatenate list (not "str") to list

What is causing this strange error?

edited 40 mins ago

khelwood

28.3k63858

asked 46 mins ago

Luke Bakare

333

New contributor

4

No, not the same for lists. += extends a list. + concatenates two lists into a new list.
â€“Â khelwood
40 mins ago

"What is causing this strange error?" It's not an error. It's the behavior of an overloaded operator.
â€“Â GreenMatt
37 mins ago

2

Ideally, if you're using this idea in code, it's probably safer to use the append() and extend() methods for adding elements and concatenating lists, respectively, to avoid ambiguity like this.
â€“Â Green Cloak Guy
36 mins ago

add a commentÂ |Â

up vote
6
down vote

favorite

l = ['table']
i =

Version 1

for n in l:
 i += n
print(i)

Output: ['t', 'a', 'b', 'l', 'e']

Version 2

for n in l:
 i = i + n
print(i)

Output:

TypeError: can only concatenate list (not "str") to list

What is causing this strange error?

edited 40 mins ago

khelwood

28.3k63858

asked 46 mins ago

Luke Bakare

333

New contributor

4

No, not the same for lists. += extends a list. + concatenates two lists into a new list.
â€“Â khelwood
40 mins ago

"What is causing this strange error?" It's not an error. It's the behavior of an overloaded operator.
â€“Â GreenMatt
37 mins ago

2

Ideally, if you're using this idea in code, it's probably safer to use the append() and extend() methods for adding elements and concatenating lists, respectively, to avoid ambiguity like this.
â€“Â Green Cloak Guy
36 mins ago

add a commentÂ |Â

up vote
6
down vote

favorite

l = ['table']
i =

Version 1

for n in l:
 i += n
print(i)

Output: ['t', 'a', 'b', 'l', 'e']

Version 2

for n in l:
 i = i + n
print(i)

Output:

TypeError: can only concatenate list (not "str") to list

What is causing this strange error?

edited 40 mins ago

khelwood

28.3k63858

asked 46 mins ago

Luke Bakare

333

New contributor

l = ['table']
i =

Version 1

for n in l:
 i += n
print(i)

Output: ['t', 'a', 'b', 'l', 'e']

Version 2

for n in l:
 i = i + n
print(i)

Output:

TypeError: can only concatenate list (not "str") to list

What is causing this strange error?

python python-3.x

edited 40 mins ago

khelwood

28.3k63858

asked 46 mins ago

Luke Bakare

333

New contributor

edited 40 mins ago

khelwood

28.3k63858

asked 46 mins ago

Luke Bakare

333

New contributor

edited 40 mins ago

khelwood

28.3k63858

edited 40 mins ago

khelwood

28.3k63858

edited 40 mins ago

khelwood

28.3k63858

asked 46 mins ago

Luke Bakare

333

New contributor

asked 46 mins ago

Luke Bakare

333

asked 46 mins ago

Luke Bakare

333

New contributor

Luke Bakare is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

4

No, not the same for lists. += extends a list. + concatenates two lists into a new list.
â€“Â khelwood
40 mins ago

"What is causing this strange error?" It's not an error. It's the behavior of an overloaded operator.
â€“Â GreenMatt
37 mins ago

2

Ideally, if you're using this idea in code, it's probably safer to use the append() and extend() methods for adding elements and concatenating lists, respectively, to avoid ambiguity like this.
â€“Â Green Cloak Guy
36 mins ago

add a commentÂ |Â

4

No, not the same for lists. += extends a list. + concatenates two lists into a new list.
â€“Â khelwood
40 mins ago

"What is causing this strange error?" It's not an error. It's the behavior of an overloaded operator.
â€“Â GreenMatt
37 mins ago

2

Ideally, if you're using this idea in code, it's probably safer to use the append() and extend() methods for adding elements and concatenating lists, respectively, to avoid ambiguity like this.
â€“Â Green Cloak Guy
36 mins ago

No, not the same for lists. += extends a list. + concatenates two lists into a new list.
â€“Â khelwood
40 mins ago

"What is causing this strange error?" It's not an error. It's the behavior of an overloaded operator.
â€“Â GreenMatt
37 mins ago

Ideally, if you're using this idea in code, it's probably safer to use the append() and extend() methods for adding elements and concatenating lists, respectively, to avoid ambiguity like this.
â€“Â Green Cloak Guy
36 mins ago

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
9
down vote

accepted

They don't have to be the same.

Using the + operator calls the method __add__ while using the += operator calls __iadd__. It is completey up to the object in question what happens when one of these methods is called.

If you use x += y but x does not provide an __iadd__ method, __add__ is used as a fallback, e.g. x = x + y happens.

In the case of lists, using l += iterable actually extends the list l with the elements of iterable. In your case, every character from the string (which is an iterable) is appended during the extend operation.

Demo 1: using __iadd__

>>> l = 
>>> l += 'table'
>>> l
['t', 'a', 'b', 'l', 'e']

Demo 2: using extend does the same

>>> l = 
>>> l.extend('table')
>>> l
['t', 'a', 'b', 'l', 'e']

Demo 3: adding a list and a string raises a TypeError.

>>> l = 
>>> l = l + 'table'
[...]
TypeError: can only concatenate list (not "str") to list

Not using += gives you the TypeError here because only __iadd__ implements the extending behavior.

Demo 4: common pitfall: += does not build a new list (note the identical ids)

>>> l = 
>>> id(l)
140359172117512
>>> l += [1, 2, 3]
>>> id(l)
140359172117512

... but the l = l + iterable syntax does.

>>> l = 
>>> id(l)
140359172142472
>>> l = l + [1, 2, 3]
>>> id(l)
140359172167560

In some cases, this can produce subtle bugs, because += mutates the original list, while
l = l + iterable builds a new list and reassigns the name l.

BONUS

Ned Batchelder's challenge to find this in the docs

edited 14 mins ago

answered 39 mins ago

timgeb

38k104876

1

Thorough explanation. Thank you!
â€“Â Luke Bakare
29 mins ago

add a commentÂ |Â

up vote
0
down vote

If in the second case you cast the n to a list to avoid errors:

for n in l:
 i = i + [n]
print(i)

you get

['table']

so they do different operations.

answered 24 mins ago

Jake

278114

add a commentÂ |Â

Your Answer

StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

Luke Bakare is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f52747784%2fis-i-i-n-truly-the-same-as-i-n%23new-answer', 'question_page');

);

Post as a guest

Name

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
9
down vote

accepted

They don't have to be the same.

Using the + operator calls the method __add__ while using the += operator calls __iadd__. It is completey up to the object in question what happens when one of these methods is called.

If you use x += y but x does not provide an __iadd__ method, __add__ is used as a fallback, e.g. x = x + y happens.

Demo 1: using __iadd__

>>> l = 
>>> l += 'table'
>>> l
['t', 'a', 'b', 'l', 'e']

Demo 2: using extend does the same

>>> l = 
>>> l.extend('table')
>>> l
['t', 'a', 'b', 'l', 'e']

Demo 3: adding a list and a string raises a TypeError.

>>> l = 
>>> l = l + 'table'
[...]
TypeError: can only concatenate list (not "str") to list

Not using += gives you the TypeError here because only __iadd__ implements the extending behavior.

Demo 4: common pitfall: += does not build a new list (note the identical ids)

>>> l = 
>>> id(l)
140359172117512
>>> l += [1, 2, 3]
>>> id(l)
140359172117512

... but the l = l + iterable syntax does.

>>> l = 
>>> id(l)
140359172142472
>>> l = l + [1, 2, 3]
>>> id(l)
140359172167560

In some cases, this can produce subtle bugs, because += mutates the original list, while
l = l + iterable builds a new list and reassigns the name l.

BONUS

Ned Batchelder's challenge to find this in the docs

edited 14 mins ago

answered 39 mins ago

timgeb

38k104876

1

Thorough explanation. Thank you!
â€“Â Luke Bakare
29 mins ago

add a commentÂ |Â

up vote
9
down vote

accepted

They don't have to be the same.

Using the + operator calls the method __add__ while using the += operator calls __iadd__. It is completey up to the object in question what happens when one of these methods is called.

If you use x += y but x does not provide an __iadd__ method, __add__ is used as a fallback, e.g. x = x + y happens.

Demo 1: using __iadd__

>>> l = 
>>> l += 'table'
>>> l
['t', 'a', 'b', 'l', 'e']

Demo 2: using extend does the same

>>> l = 
>>> l.extend('table')
>>> l
['t', 'a', 'b', 'l', 'e']

Demo 3: adding a list and a string raises a TypeError.

>>> l = 
>>> l = l + 'table'
[...]
TypeError: can only concatenate list (not "str") to list

Not using += gives you the TypeError here because only __iadd__ implements the extending behavior.

Demo 4: common pitfall: += does not build a new list (note the identical ids)

>>> l = 
>>> id(l)
140359172117512
>>> l += [1, 2, 3]
>>> id(l)
140359172117512

... but the l = l + iterable syntax does.

>>> l = 
>>> id(l)
140359172142472
>>> l = l + [1, 2, 3]
>>> id(l)
140359172167560

In some cases, this can produce subtle bugs, because += mutates the original list, while
l = l + iterable builds a new list and reassigns the name l.

BONUS

Ned Batchelder's challenge to find this in the docs

edited 14 mins ago

answered 39 mins ago

timgeb

38k104876

1

Thorough explanation. Thank you!
â€“Â Luke Bakare
29 mins ago

add a commentÂ |Â

up vote
9
down vote

accepted

They don't have to be the same.

Using the + operator calls the method __add__ while using the += operator calls __iadd__. It is completey up to the object in question what happens when one of these methods is called.

If you use x += y but x does not provide an __iadd__ method, __add__ is used as a fallback, e.g. x = x + y happens.

Demo 1: using __iadd__

>>> l = 
>>> l += 'table'
>>> l
['t', 'a', 'b', 'l', 'e']

Demo 2: using extend does the same

>>> l = 
>>> l.extend('table')
>>> l
['t', 'a', 'b', 'l', 'e']

Demo 3: adding a list and a string raises a TypeError.

>>> l = 
>>> l = l + 'table'
[...]
TypeError: can only concatenate list (not "str") to list

Not using += gives you the TypeError here because only __iadd__ implements the extending behavior.

Demo 4: common pitfall: += does not build a new list (note the identical ids)

>>> l = 
>>> id(l)
140359172117512
>>> l += [1, 2, 3]
>>> id(l)
140359172117512

... but the l = l + iterable syntax does.

>>> l = 
>>> id(l)
140359172142472
>>> l = l + [1, 2, 3]
>>> id(l)
140359172167560

In some cases, this can produce subtle bugs, because += mutates the original list, while
l = l + iterable builds a new list and reassigns the name l.

BONUS

Ned Batchelder's challenge to find this in the docs

edited 14 mins ago

answered 39 mins ago

timgeb

38k104876

They don't have to be the same.

Using the + operator calls the method __add__ while using the += operator calls __iadd__. It is completey up to the object in question what happens when one of these methods is called.

If you use x += y but x does not provide an __iadd__ method, __add__ is used as a fallback, e.g. x = x + y happens.

Demo 1: using __iadd__

>>> l = 
>>> l += 'table'
>>> l
['t', 'a', 'b', 'l', 'e']

Demo 2: using extend does the same

>>> l = 
>>> l.extend('table')
>>> l
['t', 'a', 'b', 'l', 'e']

Demo 3: adding a list and a string raises a TypeError.

>>> l = 
>>> l = l + 'table'
[...]
TypeError: can only concatenate list (not "str") to list

Not using += gives you the TypeError here because only __iadd__ implements the extending behavior.

Demo 4: common pitfall: += does not build a new list (note the identical ids)

>>> l = 
>>> id(l)
140359172117512
>>> l += [1, 2, 3]
>>> id(l)
140359172117512

... but the l = l + iterable syntax does.

>>> l = 
>>> id(l)
140359172142472
>>> l = l + [1, 2, 3]
>>> id(l)
140359172167560

In some cases, this can produce subtle bugs, because += mutates the original list, while
l = l + iterable builds a new list and reassigns the name l.

BONUS

Ned Batchelder's challenge to find this in the docs

edited 14 mins ago

answered 39 mins ago

timgeb

38k104876

edited 14 mins ago

answered 39 mins ago

timgeb

38k104876

answered 39 mins ago

timgeb

38k104876

answered 39 mins ago

timgeb

38k104876

1

Thorough explanation. Thank you!
â€“Â Luke Bakare
29 mins ago

add a commentÂ |Â

1

Thorough explanation. Thank you!
â€“Â Luke Bakare
29 mins ago

Thorough explanation. Thank you!
â€“Â Luke Bakare
29 mins ago

add a commentÂ |Â

up vote
0
down vote

If in the second case you cast the n to a list to avoid errors:

for n in l:
 i = i + [n]
print(i)

you get

['table']

so they do different operations.

answered 24 mins ago

Jake

278114

add a commentÂ |Â

up vote
0
down vote

If in the second case you cast the n to a list to avoid errors:

for n in l:
 i = i + [n]
print(i)

you get

['table']

so they do different operations.

answered 24 mins ago

Jake

278114

add a commentÂ |Â

up vote
0
down vote

If in the second case you cast the n to a list to avoid errors:

for n in l:
 i = i + [n]
print(i)

you get

['table']

so they do different operations.

answered 24 mins ago

Jake

278114

If in the second case you cast the n to a list to avoid errors:

for n in l:
 i = i + [n]
print(i)

you get

['table']

so they do different operations.

answered 24 mins ago

Jake

278114

answered 24 mins ago

Jake

278114

answered 24 mins ago

Jake

278114

answered 24 mins ago

Jake

278114

add a commentÂ |Â

Luke Bakare is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Luke Bakare is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Post as a guest

Name

Search This Blog

Iyfjky