对于所有发生的情况，还有一个解决方案(抱歉，如果重复):

values = [1,2,3,1,2,4,5,6,3,2,1]
map(lambda val: (val, [i for i in xrange(len(values)) if values[i] == val]), values)

对于所有发生的情况，还有一个解决方案(抱歉，如果重复):

values = [1,2,3,1,2,4,5,6,3,2,1]
map(lambda val: (val, [i for i in xrange(len(values)) if values[i] == val]), values)

这里是使用np的时间性能比较。Where vs list_comprehension。好像是np。哪里的平均速度更快。

# np.where
start_times = []
end_times = []
for i in range(10000):
    start = time.time()
    start_times.append(start)
    temp_list = np.array([1,2,3,3,5])
    ixs = np.where(temp_list==3)[0].tolist()
    end = time.time()
    end_times.append(end)
print("Took on average {} seconds".format(
    np.mean(end_times)-np.mean(start_times)))

Took on average 3.81469726562e-06 seconds

# list_comprehension
start_times = []
end_times = []
for i in range(10000):
    start = time.time()
    start_times.append(start)
    temp_list = np.array([1,2,3,3,5])
    ixs = [i for i in range(len(temp_list)) if temp_list[i]==3]
    end = time.time()
    end_times.append(end)
print("Took on average {} seconds".format(
    np.mean(end_times)-np.mean(start_times)))

Took on average 4.05311584473e-06 seconds

如果你需要搜索所有元素在某些索引之间的位置，你可以声明它们:

[i for i,x in enumerate([1,2,3,2]) if x==2 & 2<= i <=3] # -> [3]

occurrences = lambda s, lst: (i for i,e in enumerate(lst) if e == s)
list(occurrences(1, [1,2,3,1])) # = [0, 3]

more_itertools。Locate查找满足条件的所有项的索引。

from more_itertools import locate


list(locate([0, 1, 1, 0, 1, 0, 0]))
# [1, 2, 4]

list(locate(['a', 'b', 'c', 'b'], lambda x: x == 'b'))
# [1, 3]

More_itertools是一个第三方库> PIP install More_itertools。