如何在Python列表中查找元素的所有索引
如果我有清单如何在Python列表中查找元素的所有索引,python,list,Python,List,如果我有清单 a=[1,0,0,1,0,1,1,1,0,1,0,0] 我想分别找到0和1的索引,在这种情况下 index_0 = [1,2,4,8,10,11] index_1 = [0,3,5,6,7,9] 有没有一种有效的方法可以做到这一点 index_0 = [i for i, v in enumerate(a) if v == 0] index_1 = [i for i, v in enumerate(a) if v == 1] 或与numpy一起: import numpy as
a=[1,0,0,1,0,1,1,1,0,1,0,0]
我想分别找到0和1的索引,在这种情况下
index_0 = [1,2,4,8,10,11]
index_1 = [0,3,5,6,7,9]
有没有一种有效的方法可以做到这一点
index_0 = [i for i, v in enumerate(a) if v == 0]
index_1 = [i for i, v in enumerate(a) if v == 1]
或与numpy一起:
import numpy as np
a = np.array(a)
index_0 = np.where(a == 0)[0]
index_1 = np.where(a == 1)[0]
或与numpy一起:
import numpy as np
a = np.array(a)
index_0 = np.where(a == 0)[0]
index_1 = np.where(a == 1)[0]
或与numpy一起:
import numpy as np
a = np.array(a)
index_0 = np.where(a == 0)[0]
index_1 = np.where(a == 1)[0]
或与numpy一起:
import numpy as np
a = np.array(a)
index_0 = np.where(a == 0)[0]
index_1 = np.where(a == 1)[0]
另一种方法是:
import os
a = [1,0,0,1,0,1,1,1,0,1,0,0]
index_0 = []
index_1 = []
aux = 0
for i in a:
if i == 0:
index_0.append(aux)
aux += 1
else:
index_1.append(aux)
aux += 1
print index_0
print index_1
另一种方法是:
import os
a = [1,0,0,1,0,1,1,1,0,1,0,0]
index_0 = []
index_1 = []
aux = 0
for i in a:
if i == 0:
index_0.append(aux)
aux += 1
else:
index_1.append(aux)
aux += 1
print index_0
print index_1
另一种方法是:
import os
a = [1,0,0,1,0,1,1,1,0,1,0,0]
index_0 = []
index_1 = []
aux = 0
for i in a:
if i == 0:
index_0.append(aux)
aux += 1
else:
index_1.append(aux)
aux += 1
print index_0
print index_1
另一种方法是:
import os
a = [1,0,0,1,0,1,1,1,0,1,0,0]
index_0 = []
index_1 = []
aux = 0
for i in a:
if i == 0:
index_0.append(aux)
aux += 1
else:
index_1.append(aux)
aux += 1
print index_0
print index_1
使用
itertools.compress
:
>>> a=[1,0,0,1,0,1,1,1,0,1,0,0]
>>> index_1 = [x for x in itertools.compress(range(len(a)),a)]
>>> index_1
[0, 3, 5, 6, 7, 9]
>>> index_0 = [x for x in itertools.compress(range(len(a)),map(lambda x:not x,a))]
>>> index_0
[1, 2, 4, 8, 10, 11]
您可以实现使用一个for循环:以获得更好和更高的效率
>>> a=[1,0,0,1,0,1,1,1,0,1,0,0]
>>> index_0 = []
>>> index_1 = []
>>> for i,x in enumerate(a):
... if x: index_1.append(i)
... else: index_0.append(i)
...
>>> index_0
[1, 2, 4, 8, 10, 11]
>>> index_1
[0, 3, 5, 6, 7, 9]
使用
itertools.compress
:
>>> a=[1,0,0,1,0,1,1,1,0,1,0,0]
>>> index_1 = [x for x in itertools.compress(range(len(a)),a)]
>>> index_1
[0, 3, 5, 6, 7, 9]
>>> index_0 = [x for x in itertools.compress(range(len(a)),map(lambda x:not x,a))]
>>> index_0
[1, 2, 4, 8, 10, 11]
您可以实现使用一个for循环:以获得更好和更高的效率
>>> a=[1,0,0,1,0,1,1,1,0,1,0,0]
>>> index_0 = []
>>> index_1 = []
>>> for i,x in enumerate(a):
... if x: index_1.append(i)
... else: index_0.append(i)
...
>>> index_0
[1, 2, 4, 8, 10, 11]
>>> index_1
[0, 3, 5, 6, 7, 9]
使用
itertools.compress
:
>>> a=[1,0,0,1,0,1,1,1,0,1,0,0]
>>> index_1 = [x for x in itertools.compress(range(len(a)),a)]
>>> index_1
[0, 3, 5, 6, 7, 9]
>>> index_0 = [x for x in itertools.compress(range(len(a)),map(lambda x:not x,a))]
>>> index_0
[1, 2, 4, 8, 10, 11]
您可以实现使用一个for循环:以获得更好和更高的效率
>>> a=[1,0,0,1,0,1,1,1,0,1,0,0]
>>> index_0 = []
>>> index_1 = []
>>> for i,x in enumerate(a):
... if x: index_1.append(i)
... else: index_0.append(i)
...
>>> index_0
[1, 2, 4, 8, 10, 11]
>>> index_1
[0, 3, 5, 6, 7, 9]
使用
itertools.compress
:
>>> a=[1,0,0,1,0,1,1,1,0,1,0,0]
>>> index_1 = [x for x in itertools.compress(range(len(a)),a)]
>>> index_1
[0, 3, 5, 6, 7, 9]
>>> index_0 = [x for x in itertools.compress(range(len(a)),map(lambda x:not x,a))]
>>> index_0
[1, 2, 4, 8, 10, 11]
您可以实现使用一个for循环:以获得更好和更高的效率
>>> a=[1,0,0,1,0,1,1,1,0,1,0,0]
>>> index_0 = []
>>> index_1 = []
>>> for i,x in enumerate(a):
... if x: index_1.append(i)
... else: index_0.append(i)
...
>>> index_0
[1, 2, 4, 8, 10, 11]
>>> index_1
[0, 3, 5, 6, 7, 9]
完美答案。我只是照样打:非常感谢。还有一个快速的问题,如果a的长度很大,哪一个更有效?@user3029108 numpy将随着数组大小的增加而更有效。完美的答案。我只是照样打:非常感谢。还有一个快速的问题,如果a的长度很大,哪一个更有效?@user3029108 numpy将随着数组大小的增加而更有效。完美的答案。我只是照样打:非常感谢。还有一个快速的问题,如果a的长度很大,哪一个更有效?@user3029108 numpy将随着数组大小的增加而更有效。完美的答案。我只是照样打:非常感谢。还有一个快速的问题,如果a的长度较大,哪一个更有效?@user3029108 numpy将随着数组大小的增长而更有效。可能的重复可能的重复可能的重复可能的重复