在 python 时间跳过列表的特定范围

Question

我有一个数组，我想选择前 2 个或范围，跳过下 2 个，选择下 2 个并继续此操作直到列表结束

list = [2, 4, 6, 7, 9,10, 13, 11, 12,2]
results_wanted = [2,4,9,10,12,2] # note how it skipping 2. 2 is used here as and example

有没有办法在 python 中实现这个？

Answer 1

您可以尝试使用 range(start, end, skip)

进行迭代

并且您可以指定在序列中添加多少个 (add=2) 和跳过多少个 (skip=2)

list1 = [2, 4, 6, 7, 9,10, 13, 11, 12,2, 4, 6, 7, 9,10, 13, 11, 12,2]
list2 = []
add = 2
skip = 2

for i in range(0, len(list1), skip+add):
    list2 += list1[i:i+add]

print(list2)

输出：

[2, 4, 9, 10, 12, 2, 7, 9, 11, 12]

顺便说一句，你应该避免使用 Python 保留字 list 作为变量名。

Answer 2

我没有使用任何 python 预构建技术。我在 if-else 条件下使用传统的 for 循环

我们必须根据特定数字跳过。
这个skip需要根据我定义为skipMode的boolean参数来完成
如果 skipMode 为真，则不会将数字添加到列表中 4 这个skipMode会根据skipNumber

test = [2, 4, 6, 7, 9,10, 13, 11, 12,2]

def skipElementsByPositions(test,skip):
    if(skip> len(test)):
        return -1
    else:
        desireList = []
        skipMode = False
        for i in range(0,len(test)):
            if skipMode==False:
                desireList.append(test[i])
            if (i+1)%skip==0:
                skipMode=not skipMode
        return desireList

print(skipElementsByPositions(test,2)) #2,4,9,10,12,2
print(skipElementsByPositions(test,3)) #2, 4, 6, 13, 11, 12

Answer 3

from itertools import compress, cycle

results = list(compress(lst, cycle([1,1,0,0])))

或

results = [x for i, x in enumerate(lst) if i % 4 < 2]

或者如果您不再需要原始列表，那么可以修改它而不是构建一个新列表（如果您确实想要一个新列表，您仍然可以在副本上使用它）：

del lst[2::4], lst[2::3]

具有一百万个元素的列表的基准：

  8.6 ms ± 0.1 ms _del_slices
 11.6 ms ± 0.1 ms _compress_bools
 13.6 ms ± 0.0 ms _compress_ints
 14.1 ms ± 0.0 ms _copy_del_slices
 22.6 ms ± 0.2 ms _copy_slices
 46.0 ms ± 0.2 ms Black_Raven
 59.5 ms ± 0.4 ms Vikrant_Sharma
 84.8 ms ± 0.1 ms _enumerate_modulo
161.8 ms ± 0.6 ms RCvaram

不包括 numpy 解决方案，因为我必须为此切换到较旧的 Python 版本，并且因为比较没有意义（np.array(lst) 已经花费了约 55 毫秒）。

基准代码(Try it online!):

def _compress_ints(lst):
    return list(compress(lst, cycle([1,1,0,0])))

def _compress_bools(lst):
    return list(compress(
        lst,
        cycle(chain(repeat(True, 2),
                    repeat(False, 2)))
    ))

def _enumerate_modulo(lst):
    return [x for i, x in enumerate(lst) if i % 4 < 2]

def _del_slices(lst):
    del lst[2::4], lst[2::3]
    return lst

def _copy_del_slices(lst):
    results = lst[:]
    del results[2::4], results[2::3]
    return results

def _copy_slices(lst):
    a = lst[::4]
    b = lst[1::4]
    results = [None] * (len(a) + len(b))
    results[::2] = a
    results[1::2] = b
    return results
    
def Black_Raven(list1):
    add = skip = 2
    list2 = []
    for i in range(0, len(list1), skip+add):
        list2 += list1[i:i+add]
    return list2

def Vikrant_Sharma(l):
    n = 2
    return [x for i in range(0, len(l), n + n) for x in l[i: i + n]]

def RCvaram(test):
        skip = 2
        desireList = []
        skipMode = False
        for i in range(0,len(test)):
            if skipMode==False:
                desireList.append(test[i])
            if (i+1)%skip==0:
                skipMode=not skipMode
        return desireList

funcs = [_compress_ints, _compress_bools, _enumerate_modulo, _del_slices, _copy_del_slices, _copy_slices, Black_Raven, Vikrant_Sharma, RCvaram]

from timeit import default_timer as timer
from itertools import compress, cycle, repeat, chain, islice
from random import shuffle
from statistics import mean, stdev
import gc

# Correctness
lst = [2, 4, 6, 7, 9,10, 13, 11, 12,2]
results_wanted = [2,4,9,10,12,2]
for func in funcs:
    assert func(lst[:]) == results_wanted
for n in range(100):
    lst = list(range(n))
    expect = funcs[0](lst[:])
    for func in funcs:
        assert func(lst[:]) == expect, func.__name__

# Speed
times = {func: [] for func in funcs}
def stats(func):
    ts = [t * 1e3 for t in sorted(times[func])[:3]]
    return f'{mean(ts):5.1f} ms ± {stdev(ts):.1f} ms'
original = list(range(1000000))
for _ in range(15):
    shuffle(funcs)
    for func in funcs:
        lst = original.copy()
        gc.collect()
        t0 = timer()
        result = func(lst)
        t = timer() - t0
        del result
        times[func].append(t)
for func in sorted(funcs, key=stats):
    print(stats(func), func.__name__)

Answer 4

取 n 个元素并跳过下一个 n。

l = [2, 4, 6, 7, 9, 10, 13, 11, 12, 2]
n = 2
wanted = [x for i in range(0, len(l), n + n) for x in l[i: i + n]]
### Output : [2, 4, 9, 10, 12, 2]

Answer 5

因为你有一个 numpy 标签，这里是 numpy 方法。

使用面具：

lst = [2, 4, 6, 7, 9,10, 13, 11, 12,2]
a = np.array(lst)

mask = [True, True, False, False]
n = (len(a)+3)//4
a[np.tile(mask, n)[:len(a)]]

或中间重塑为二维数组：

n = (len(a)+3)//4
extra = n*4-len(a)
(np
  .pad(a, (0, extra), mode='constant', constant_values=0)
  .reshape(-1,4)
  [:,:2]
  .ravel()
  )

输出：array([ 2, 4, 9, 10, 12, 2])

在 python 时间跳过列表的特定范围

skipping a certain range of a list at time in python

python

numpy

python-xarray