Python在列表中查找项目索引的最快方法
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Python Fastest way to find Indexes of item in list
提问by Tyler Cowan
If one was to attempt to find the indexes of an item in a list you could do it a couple different ways here is what I know to be the fastest
如果要尝试在列表中查找项目的索引，您可以通过几种不同的方式来完成，这是我所知道的最快的方法
aList = [123, 'xyz', 'zara','xyz', 'abc'];
indices = [i for i, x in enumerate(aList) if x == "xyz"]
print(indices)
Another way not pythonic and slower
另一种不是pythonic且速度较慢的方法
count = 0
indices = []
aList = [123, 'xyz', 'zara','xyz', 'abc'];
for i in range(0,len(aList):
if 'xyz' == aList[i]:
indices.append(i)
print(indices)
the first method is undoubtedly faster however what if you wanted to go faster is there a way? for the first index using method
第一种方法无疑更快，但是如果你想更快，有什么办法吗？对于第一个索引使用方法
aList = [123, 'xyz', 'zara','xyz', 'abc'];
print "Index for xyz : ", aList.index( 'xyz' )
is very fast but cant handle multiple indexes How might one go about speeding things up?
速度非常快，但无法处理多个索引 如何加快速度？
采纳答案by Garrett R
def find(target, myList):
for i in range(len(myList)):
if myList[i] == target:
yield i
def find_with_list(myList, target):
inds = []
for i in range(len(myList)):
if myList[i] == target:
inds += i,
return inds
In [8]: x = range(50)*200
In [9]: %timeit [i for i,j in enumerate(x) if j == 3]
1000 loops, best of 3: 598 us per loop
In [10]: %timeit list(find(3,x))
1000 loops, best of 3: 607 us per loop
In [11]: %timeit find(3,x)
1000000 loops, best of 3: 375 ns per loop
In [55]: %timeit find_with_list(x,3)
1000 loops, best of 3: 618 us per loop
Assuming you want a list as your output: All options seemed exhibit similar time performance for my test with the list comprehension being the fastest (barely).
假设你想要一个列表作为你的输出：所有选项似乎在我的测试中表现出相似的时间性能，列表理解是最快的（几乎没有）。
And if you're cool with returning a generator, it's way faster than the other approaches. Thought it doesn't account for actually iterating over the indices, nor does it store them, so the inds cannot be iterated over a second time.
如果你对返回发电机很酷，它比其他方法快得多。认为它不考虑实际迭代索引，也不存储它们，因此不能第二次迭代索引。
回答by John
D=dict()
for i, item in enumerate(l):
if item not in D:
D[item] = [i]
else:
D[item].append(i)
Then simply call D[item] to get the indices that match. You'll give up initial calculation time but gain it during call time.
然后只需调用 D[item] 来获取匹配的索引。您将放弃初始计算时间，但会在调用期间获得它。
回答by poke53280
Use list.index(elem, start)! That uses a for loop in C (see it's implementation list_index_implfunction in the source of CPython's listobject.c).
Avoid looping through all the elements in Python, it is slower than in C.
使用list.index(elem, start)！它在 C 中使用 for 循环（请参阅list_index_implCPython 的listobject.c源代码中的实现函数）。避免循环遍历 Python 中的所有元素，它比 C 中慢。
def index_finder(lst, item):
"""A generator function, if you might not need all the indices"""
start = 0
while True:
try:
start = lst.index(item, start)
yield start
start += 1
except ValueError:
break
import array
def index_find_all(lst, item, results=None):
"""If you want all the indices.
Pass results=[] if you explicitly need a list,
or anything that can .append(..)"""
if results is None:
length = len(lst)
results = array.array('B') if length <= 2**8 else array.array('H') if length <= 2**16 else array.array('L') if length <= 2**32 else array.array('Q')
start = 0
while True:
try:
start = lst.index(item, start)
results.append(start)
start += 1
except ValueError:
return results
# Usage example
l = [1, 2, 3, 4, 5, 6, 7, 8] * 32
print(*index_finder(l, 1))
print(*index_find_all(l, 1))
回答by Harish Mashetty
To get the index of the item, you can use the dictionary.
要获取项目的索引，您可以使用字典。
aList = [123, 'xyz', 'zara','xyz', 'abc'];
#The following apporach works only on lists with unique values
aList = list(np.unique(aList));
dict = enumerate(aList);
# get inverse mapping of above dictionary, replace key with values
inv_dict = dict(zip(dict.values(),dict.keys()))
# to get index of item by value, use 'inv_dict' and to get value by index, use 'dict'
valueofItemAtIndex0 = dict[0]; # value = 123
indexofItemWithValue123 = inv_dict[123]; # index = 0
回答by ?ngelo Polotto
I used another way to find the index of a element in a list in Python 3:
我使用另一种方法在 Python 3 的列表中查找元素的索引：
def index_of(elem, a):
a_e = enumerate(a)
a_f = list(filter(lambda x: x[1] == elem, a_e))
if a_f:
return a_f[0][0]
else:
return -1
Some tests:
一些测试：
a=[1,2,3,4,2]
index_of(2,a)
This function always return the first occurrence of the element. If element ins't in the list, return -1. For my goals, that solution worked well.
这个函数总是返回元素的第一次出现。如果元素不在列表中，则返回 -1。对于我的目标，该解决方案运行良好。
回答by Ahmedov
Simply create a dictionary of item->index from the list of items using zip like so:
只需使用 zip 从项目列表中创建一个 item->index 字典，如下所示：
items_as_dict = dict(zip(list_of_items,range(0,len(list_of_items))))
index = items_as_dict(item)
