Как найти элемент в списке по имени

asked Feb 28, 2009 at 18:06

You can get a list of all matching elements with a list comprehension:

[x for x in myList if x.n == 30]  # list of all elements with .n==30

If you simply want to determine if the list contains any element that matches and do it (relatively) efficiently, you can do

def contains(list, filter):
    for x in list:
        if filter(x):
            return True
    return False

if contains(myList, lambda x: x.n == 3)  # True if any element has .n==3
    # do stuff

Simple, Elegant, and Powerful:

A generator expression in conjuction with a builtin… (python 2.5+)

any(x for x in mylist if x.n == 10)

Uses the Python any() builtin, which is defined as follows:

any(iterable) ->
Return True if any element of the iterable is true. Equivalent to:

def any(iterable):
    for element in iterable:
        if element:
            return True
    return False

Just for completeness, let’s not forget the Simplest Thing That Could Possibly Work:

for i in list:
  if i.n == 5:
     # do something with it
     print "YAY! Found one!"

answered Feb 28, 2009 at 18:20

[x for x in myList if x.n == 30]               # list of all matches
[x.n_squared for x in myList if x.n == 30]     # property of matches
any(x.n == 30 for x in myList)                 # if there is any matches
[i for i,x in enumerate(myList) if x.n == 30]  # indices of all matches

def first(iterable, default=None):
  for item in iterable:
    return item
  return default

first(x for x in myList if x.n == 30)          # the first match, if any

answered Feb 28, 2009 at 18:19

filter(lambda x: x.n == 5, myList)

answered Feb 28, 2009 at 18:22

answered Feb 28, 2009 at 18:23

answered Jun 16, 2020 at 6:01

Johannes WeissJohannes Weiss

52.2k16 gold badges102 silver badges136 bronze badges

Consider using a dictionary:

myDict = {}

for i in range(20):
    myDict[i] = i * i

print(5 in myDict)

answered Mar 1, 2009 at 1:14

answered Apr 18, 2021 at 18:55

asked Mar 3, 2009 at 1:45

From Dive Into Python:

>>> li
['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
>>> li.index("example")
5

If you just want to find out if an element is contained in the list or not:

>>> li
['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
>>> 'example' in li
True
>>> 'damn' in li
False

answered Feb 17, 2011 at 10:00

answered Mar 3, 2009 at 1:51

Here is another way using list comprehension (some people might find it debatable). It is very approachable for simple tests, e.g. comparisons on object attributes (which I need a lot):

el = [x for x in mylist if x.attr == "foo"][0]

Of course this assumes the existence (and, actually, uniqueness) of a suitable element in the list.

answered Sep 24, 2010 at 9:35

assuming you want to find a value in a numpy array,
I guess something like this might work:

Numpy.where(arr=="value")[0]

David ZDavid Z

127k27 gold badges253 silver badges278 bronze badges

I use function for returning index for the matching element (Python 2.6):

def index(l, f):
     return next((i for i in xrange(len(l)) if f(l[i])), None)

Then use it via lambda function for retrieving needed element by any required equation e.g. by using element name.

element = mylist[index(mylist, lambda item: item["name"] == "my name")]

If i need to use it in several places in my code i just define specific find function e.g. for finding element by name:

def find_name(l, name):
     return l[index(l, lambda item: item["name"] == name)]

And then it is quite easy and readable:

element = find_name(mylist,"my name")

answered Oct 20, 2011 at 12:30

Jarret HardieJarret Hardie

94.4k10 gold badges132 silver badges126 bronze badges

how’s this one?

def global_index(lst, test):
    return ( pair[0] for pair in zip(range(len(lst)), lst) if test(pair[1]) )

Usage:

>>> global_index([1, 2, 3, 4, 5, 6], lambda x: x>3)
<generator object <genexpr> at ...>
>>> list(_)
[3, 4, 5]

answered Mar 3, 2009 at 2:06

str() выдаёт вам строку, так что это надо бы удалить, а так просто использовать генератор списка.
baskets = [x for x in names if 'пластик' in x]

Если нужно лишь найти все элементы со словом «пластик», то в принципе подойдёт.
baskets = [x for x in names if x.count('пластик')]
Но учитывая некоторую ошибку в самом начале, из-за которой список переделывается в str для применения lower(), немного изменяем это всё, чтоб алгоритм работал.

names = [x.lower() for x in names]

baskets = [x for x in names if x.count('пластик')]