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dark fader

I have a dictionary of values read from 2 fields in a database: a string field and a numeric field. The string field is unique so that is the key of the dictionary.

I can sort on the keys, but how can I sort based on the values?

Note: I have read this post and probably could change my code to have a list of dictionaries but since I do not really need a list of dictionaries I wanted to know if there a simpler solution.

Answered By: Devin Jeanpierre ( 485)

It is not possible to sort a dict, only to get a representation of a dict that is sorted. Dicts are inherently orderless, but other types, such as lists and tuples, are not. So you need a sorted representation, which will be a list—probably a list of tuples. For instance,

import operator
x = {1: 2, 3: 4, 4:3, 2:1, 0:0}
sorted_x = sorted(x.iteritems(), key=operator.itemgetter(1))

sorted_x will be a list of tuples sorted by the second element in each tuple. dict(sorted_x) == x.

Ray Vega

For example, if passed the following:

a = []

How do I check to see if a is empty?

Answered By: Patrick ( 559)
if not a:
  print "List is empty"

Using the implicit booleanness of the empty list is quite pythonic.

Joan Venge

Does anyone know how to access the index itself for a list like this:

ints = [8,23,45,12,78]

When I loop through it using a for loop, how do I access the loop index, from 1 to 5 in this case?

Answered By: Mike Hordecki ( 506)

Using additional state variable, such as index variable (which you would normally use in languages such as C or PHP), is considered non-pythonic.

The better option is to use the builtin function enumerate:

for idx, val in enumerate(ints):
    print idx, val

Check out PEP 279 for more.

items = []

items.amount()  // should return 3

How I do it right?

Answered By: gnud ( 300)

The len function can be used with a lot of types in python - both built-in types and library types.


I have a list of arbitrary length, and I need to split it up into equal size chunks and operate on it. There are some obvious ways to do this, like keeping a counter and two lists, and when the second list fills up, add it to the first list and empty the second list for the next round of data, but this is potentially extremely expensive.

I was wondering if anyone had a good solution to this for lists of any length, e.g. using generators.

This should work:

l = range(1, 1000)
print chunks(l, 10) -> [ [ 1..10 ], [ 11..20 ], .., [ 991..999 ] ]

I was looking for something useful in itertools but I couldn't find anything obviously useful. Might've missed it, though.

Related question: What is the most “pythonic” way to iterate over a list in chunks?

Answered By: Ned Batchelder ( 247)

Here's a generator that yields the chunks you want:

def chunks(l, n):
    """ Yield successive n-sized chunks from l.
    for i in xrange(0, len(l), n):
        yield l[i:i+n]

import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
 [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
 [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
 [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
 [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
 [70, 71, 72, 73, 74]]

I got a list of dictionaries and want that to be sorted by a value of that dictionary.


[{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]

sorted by name, should become

[{'name':'Bart', 'age':10}, {'name':'Homer', 'age':39}]
Answered By: Mario Fernandez ( 286)

It may look cleaner using a key instead a cmp:

newlist = sorted(list_to_be_sorted, key=lambda k: k['name']) 

or as J.F.Sebastian and others suggested,

from operator import itemgetter
newlist = sorted(list_to_be_sorted, key=itemgetter('name')) 
Ray Vega

Let's say, as an example, I have the following list:

foo = ['a', 'b', 'c', 'd', 'e']

What is the best way to retrieve an item at random from this list?

Answered By: Pēteris Caune ( 368)
foo = ['a', 'b', 'c', 'd', 'e']
from random import choice
print choice(foo)

Possible Duplicates:
Flattening a shallow list in Python
Comprehension for flattening a sequence of sequences?

I wonder whether there is a shortcut to make a simple list out of list of lists in Python.

I can do that in a for loop, but maybe there is some cool "one-liner"? I tried it with reduce, but I get an error.


l = [[1,2,3],[4,5,6], [7], [8,9]]
reduce(lambda x,y: x.extend(y),l)

Error message

Traceback (most recent call last): File "", line 1, in File "", line 1, in AttributeError: 'NoneType' object has no attribute 'extend'

Answered By: Alex Martelli ( 343)

[item for sublist in l for item in sublist] is faster than the shortcuts posted so far.

For evidence, as always, you can use the timeit module in the standard library:

$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in l for item in sublist]'
10000 loops, best of 3: 143 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])'
1000 loops, best of 3: 969 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,l)'
1000 loops, best of 3: 1.1 msec per loop

Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So (for simplicity and without actual loss of generality) say you have L sublists of I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (L**2)/2.

The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.

Sorry for the waffly title - if I could come up with a concise title, I wouldn't have to ask the question.

Suppose I have an immutable list type. It has an operation Foo(x) which returns a new immutable list with the specified argument as an extra element at the end. So to build up a list of strings with values "Hello", "immutable", "world" you could write:

var empty = new ImmutableList<string>();
var list1 = empty.Foo("Hello");
var list2 = list1.Foo("immutable");
var list3 = list2.Foo("word");

(This is C# code, and I'm most interested in a C# suggestion if you feel the language is important. It's not fundamentally a language question, but the idioms of the language may be important.)

The important thing is that the existing lists are not altered by Foo - so empty.Count would still return 0.

Another (more idiomatic) way of getting to the end result would be:

var list = new ImmutableList<string>().Foo("Hello")

My question is: what's the best name for Foo?

EDIT 3: As I reveal later on, the name of the type might not actually be ImmutableList<T>, which makes the position clear. Imagine instead that it's TestSuite and that it's immutable because the whole of the framework it's a part of is immutable...

(End of edit 3)

Options I've come up with so far:

  • Add: common in .NET, but implies mutation of the original list
  • Cons: I believe this is the normal name in functional languages, but meaningless to those without experience in such languages
  • Plus: my favourite so far, it doesn't imply mutation to me. Apparently this is also used in Haskell but with slightly different expectations (a Haskell programmer might expect it to add two lists together rather than adding a single value to the other list).
  • With: consistent with some other immutable conventions, but doesn't have quite the same "additionness" to it IMO.
  • And: not very descriptive.
  • Operator overload for + : I really don't like this much; I generally think operators should only be applied to lower level types. I'm willing to be persuaded though!

The criteria I'm using for choosing are:

  • Gives the correct impression of the result of the method call (i.e. that it's the original list with an extra element)
  • Makes it as clear as possible that it doesn't mutate the existing list
  • Sounds reasonable when chained together as in the second example above

Please ask for more details if I'm not making myself clear enough...

EDIT 1: Here's my reasoning for preferring Plus to Add. Consider these two lines of code:


In my view (and this is a personal thing) the latter is clearly buggy - it's like writing "x + 5;" as a statement on its own. The first line looks like it's okay, until you remember that it's immutable. In fact, the way that the plus operator on its own doesn't mutate its operands is another reason why Plus is my favourite. Without the slight ickiness of operator overloading, it still gives the same connotations, which include (for me) not mutating the operands (or method target in this case).

EDIT 2: Reasons for not liking Add.

Various answers are effectively: "Go with Add. That's what DateTime does, and String has Replace methods etc which don't make the immutability obvious." I agree - there's precedence here. However, I've seen plenty of people call DateTime.Add or String.Replace and expect mutation. There are loads of newsgroup questions (and probably SO ones if I dig around) which are answered by "You're ignoring the return value of String.Replace; strings are immutable, a new string gets returned."

Now, I should reveal a subtlety to the question - the type might not actually be an immutable list, but a different immutable type. In particular, I'm working on a benchmarking framework where you add tests to a suite, and that creates a new suite. It might be obvious that:

var list = new ImmutableList<string>();

isn't going to accomplish anything, but it becomes a lot murkier when you change it to:

var suite = new TestSuite<string, int>();
suite.Add(x => x.Length);

That looks like it should be okay. Whereas this, to me, makes the mistake clearer:

var suite = new TestSuite<string, int>();
suite.Plus(x => x.Length);

That's just begging to be:

var suite = new TestSuite<string, int>().Plus(x => x.Length);

Ideally, I would like my users not to have to be told that the test suite is immutable. I want them to fall into the pit of success. This may not be possible, but I'd like to try.

I apologise for over-simplifying the original question by talking only about an immutable list type. Not all collections are quite as self-descriptive as ImmutableList<T> :)

Answered By: Ken ( 110)

I'd go with Cons, for one simple reason: it means exactly what you want it to.

  1. I'm a huge fan of saying exactly what I mean, especially in source code. A newbie will have to look up the definition of Cons only once, but then read and use that a thousand times. I find that, in the long term, it's nicer to work with systems that make the common case easier, even if the up-front cost is a little bit higher.

  2. The fact that it would be "meaningless" to people with no FP experience is actually a big advantage. As you pointed out, all of the other words you found already have some meaning, and that meaning is either slightly different or ambiguous. A new concept should have a new word (or in this case, an old one). I'd rather somebody have to look up the definition of Cons, than to assume incorrectly he knows what Add does.

  3. Other operations borrowed from functional languages often keep their original names, with no apparent catastrophes. I haven't seen any push to come up with synonyms for "map" and "reduce" that sound more familiar to non-FPers, nor do I see any benefit from doing so.

(Full disclosure: I'm a Lisp programmer, so I already know what Cons means.)

SaaS Developer

I'm using C# on Framework 3.5. I'm looking to quickly sort a Generic List<>. For the sake of this example lets say I have a List of a Person type with a property of lastname. How would I sort this List using a lambda expression?

List<Person> people = PopulateList();
people.OrderBy(???? => ?????)
Answered By: Marc Gravell ( 271)

If you mean an in-place sort (i.e. the list is updated):

people.Sort((x, y) => string.Compare(x.LastName, y.LastName));

If you mean a new list:

var newList = people.OrderBy(x=>x.LastName).ToList(); // ToList optional