检查列表单调性的高效和pythonic 方法是什么?左边是单调增加还是减少值?
示例:
[0, 1, 2, 3, 3, 4] # This is a monotonically increasing list
[4.3, 4.2, 4.2, -2] # This is a monotonically decreasing list
[2, 3, 1] # This is neither
答案 0 :(得分:127)
def strictly_increasing(L):
return all(x<y for x, y in zip(L, L[1:]))
def strictly_decreasing(L):
return all(x>y for x, y in zip(L, L[1:]))
def non_increasing(L):
return all(x>=y for x, y in zip(L, L[1:]))
def non_decreasing(L):
return all(x<=y for x, y in zip(L, L[1:]))
def monotonic(L):
return non_increasing(L) or non_decreasing(L)
答案 1 :(得分:32)
如果你有大量的数字列表,最好使用numpy,如果你是:
import numpy as np
def monotonic(x):
dx = np.diff(x)
return np.all(dx <= 0) or np.all(dx >= 0)
应该这样做。
答案 2 :(得分:25)
import itertools
import operator
def monotone_increasing(lst):
pairs = zip(lst, lst[1:])
return all(itertools.starmap(operator.le, pairs))
def monotone_decreasing(lst):
pairs = zip(lst, lst[1:])
return all(itertools.starmap(operator.ge, pairs))
def monotone(lst):
return monotone_increasing(lst) or monotone_decreasing(lst)
这种方法在列表的长度上是O(N)
。
答案 3 :(得分:15)
@ 6502有完美的列表代码,我只想添加适用于所有序列的通用版本:
def pairwise(seq):
items = iter(seq)
last = next(items)
for item in items:
yield last, item
last = item
def strictly_increasing(L):
return all(x<y for x, y in pairwise(L))
def strictly_decreasing(L):
return all(x>y for x, y in pairwise(L))
def non_increasing(L):
return all(x>=y for x, y in pairwise(L))
def non_decreasing(L):
return all(x<=y for x, y in pairwise(L))
答案 4 :(得分:4)
import operator, itertools
def is_monotone(lst):
op = operator.le # pick 'op' based upon trend between
if not op(lst[0], lst[-1]): # first and last element in the 'lst'
op = operator.ge
return all(op(x,y) for x, y in itertools.izip(lst, lst[1:]))
答案 5 :(得分:2)
以下是使用reduce
复杂度O(n)
的功能性解决方案:
is_increasing = lambda L: reduce(lambda a,b: b if a < b else 9999 , L)!=9999
is_decreasing = lambda L: reduce(lambda a,b: b if a > b else -9999 , L)!=-9999
将9999
替换为您的值的上限,将-9999
替换为下限。例如,如果您要测试数字列表,则可以使用10
和-1
。
我针对@6502's answer及其速度更快地测试了其效果。
案例真实:[1,2,3,4,5,6,7,8,9]
# my solution ..
$ python -m timeit "inc = lambda L: reduce(lambda a,b: b if a < b else 9999 , L)!=9999; inc([1,2,3,4,5,6,7,8,9])"
1000000 loops, best of 3: 1.9 usec per loop
# while the other solution:
$ python -m timeit "inc = lambda L: all(x<y for x, y in zip(L, L[1:]));inc([1,2,3,4,5,6,7,8,9])"
100000 loops, best of 3: 2.77 usec per loop
第二个元素的案例错误:[4,2,3,4,5,6,7,8,7]
:
# my solution ..
$ python -m timeit "inc = lambda L: reduce(lambda a,b: b if a < b else 9999 , L)!=9999; inc([4,2,3,4,5,6,7,8,7])"
1000000 loops, best of 3: 1.87 usec per loop
# while the other solution:
$ python -m timeit "inc = lambda L: all(x<y for x, y in zip(L, L[1:]));inc([4,2,3,4,5,6,7,8,7])"
100000 loops, best of 3: 2.15 usec per loop
答案 6 :(得分:1)
L = [1,2,3]
L == sorted(L)
L == sorted(L, reverse=True)
答案 7 :(得分:1)
我在不同的条件下计算了这个问题的所有答案,并发现:
以下是试用它的代码:
import timeit
setup = '''
import random
from itertools import izip, starmap, islice
import operator
def is_increasing_normal(lst):
for i in range(0, len(lst) - 1):
if lst[i] >= lst[i + 1]:
return False
return True
def is_increasing_zip(lst):
return all(x < y for x, y in izip(lst, islice(lst, 1, None)))
def is_increasing_sorted(lst):
return lst == sorted(lst)
def is_increasing_starmap(lst):
pairs = izip(lst, islice(lst, 1, None))
return all(starmap(operator.le, pairs))
if {list_method} in (1, 2):
lst = list(range({n}))
if {list_method} == 2:
for _ in range(int({n} * 0.0001)):
lst.insert(random.randrange(0, len(lst)), -random.randrange(1,100))
if {list_method} == 3:
lst = [int(1000*random.random()) for i in xrange({n})]
'''
n = 100000
iterations = 10000
list_method = 1
timeit.timeit('is_increasing_normal(lst)', setup=setup.format(n=n, list_method=list_method), number=iterations)
timeit.timeit('is_increasing_zip(lst)', setup=setup.format(n=n, list_method=list_method), number=iterations)
timeit.timeit('is_increasing_sorted(lst)', setup=setup.format(n=n, list_method=list_method), number=iterations)
timeit.timeit('is_increasing_starmap(lst)', setup=setup.format(n=n, list_method=list_method), number=iterations)
如果列表已经单调增加(list_method == 1
),最慢到最慢的是:
如果列表大部分单调增加(list_method == 2
),最慢到最慢的是:
(starmap或zip是否最快取决于执行情况,我无法识别模式。星图通常更快)
如果列表是完全随机的(list_method == 3
),最慢到最慢的是:
答案 8 :(得分:1)
@ 6502为此提供了精美的python代码。这是一种迭代器更简单且没有潜在昂贵的临时片的替代解决方案:
def strictly_increasing(L):
return all(L[i] < L[i+1] for i in range(len(L)-1))
def strictly_decreasing(L):
return all(L[i] > L[i+1] for i in range(len(L)-1))
def non_increasing(L):
return all(L[i] >= L[i+1] for i in range(len(L)-1))
def non_decreasing(L):
return all(L[i] <= L[i+1] for i in range(len(L)-1))
def monotonic(L):
return non_increasing(L) or non_decreasing(L)
答案 9 :(得分:0)
使用Pandas可以实现,您可以通过pip install pandas
安装。
import pandas as pd
以下命令可处理整数或浮点数的列表。
pd.Series(mylist).is_monotonic_increasing
myseries = pd.Series(mylist)
myseries.is_unique and myseries.is_monotonic_increasing
使用未公开的私有方法的替代方法:
pd.Index(mylist)._is_strictly_monotonic_increasing
pd.Series(mylist).is_monotonic_decreasing
myseries = pd.Series(mylist)
myseries.is_unique and myseries.is_monotonic_decreasing
使用未公开的私有方法的替代方法:
pd.Index(mylist)._is_strictly_monotonic_decreasing
答案 10 :(得分:0)
这是一个接受物化和非物化序列的变体。它会自动确定其是否为monotonic
,如果是,则确定其方向(即increasing
或decreasing
)和strict
的方向。提供内联注释以帮助读者。最后提供的测试用例也是如此。
def isMonotonic(seq):
"""
seq.............: - A Python sequence, materialized or not.
Returns.........:
(True,0,True): - Mono Const, Strict: Seq empty or 1-item.
(True,0,False): - Mono Const, Not-Strict: All 2+ Seq items same.
(True,+1,True): - Mono Incr, Strict.
(True,+1,False): - Mono Incr, Not-Strict.
(True,-1,True): - Mono Decr, Strict.
(True,-1,False): - Mono Decr, Not-Strict.
(False,None,None) - Not Monotonic.
"""
items = iter(seq) # Ensure iterator (i.e. that next(...) works).
prev_value = next(items, None) # Fetch 1st item, or None if empty.
if prev_value == None: return (True,0,True) # seq was empty.
# ============================================================
# The next for/loop scans until it finds first value-change.
# ============================================================
# Ex: [3,3,3,78,...] --or- [-5,-5,-5,-102,...]
# ============================================================
# -- If that 'change-value' represents an Increase or Decrease,
# then we know to look for Monotonically Increasing or
# Decreasing, respectively.
# -- If no value-change is found end-to-end (e.g. [3,3,3,...3]),
# then it's Monotonically Constant, Non-Strict.
# -- Finally, if the sequence was exhausted above, which means
# it had exactly one-element, then it Monotonically Constant,
# Strict.
# ============================================================
isSequenceExhausted = True
curr_value = prev_value
for item in items:
isSequenceExhausted = False # Tiny inefficiency.
if item == prev_value: continue
curr_value = item
break
else:
return (True,0,True) if isSequenceExhausted else (True,0,False)
# ============================================================
# ============================================================
# If we tricked down to here, then none of the above
# checked-cases applied (i.e. didn't short-circuit and
# 'return'); so we continue with the final step of
# iterating through the remaining sequence items to
# determine Monotonicity, direction and strictness.
# ============================================================
strict = True
if curr_value > prev_value: # Scan for Increasing Monotonicity.
for item in items:
if item < curr_value: return (False,None,None)
if item == curr_value: strict = False # Tiny inefficiency.
curr_value = item
return (True,+1,strict)
else: # Scan for Decreasing Monotonicity.
for item in items:
if item > curr_value: return (False,None,None)
if item == curr_value: strict = False # Tiny inefficiency.
curr_value = item
return (True,-1,strict)
# ============================================================
# Test cases ...
assert isMonotonic([1,2,3,4]) == (True,+1,True)
assert isMonotonic([4,3,2,1]) == (True,-1,True)
assert isMonotonic([-1,-2,-3,-4]) == (True,-1,True)
assert isMonotonic([]) == (True,0,True)
assert isMonotonic([20]) == (True,0,True)
assert isMonotonic([-20]) == (True,0,True)
assert isMonotonic([1,1]) == (True,0,False)
assert isMonotonic([1,-1]) == (True,-1,True)
assert isMonotonic([1,-1,-1]) == (True,-1,False)
assert isMonotonic([1,3,3]) == (True,+1,False)
assert isMonotonic([1,2,1]) == (False,None,None)
assert isMonotonic([0,0,0,0]) == (True,0,False)
我想这可能更多Pythonic
,但这很棘手,因为它避免了创建中间集合(例如list
,genexps
等);以及采用fall/trickle-through
和short-circuit
方法来过滤各种情况:边缘序列(如空序列或单项序列;或具有所有相同项目的序列);确定增加或减少的单调性,严格性等。希望对您有所帮助。
答案 11 :(得分:0)
def IsMonotonic(data):
''' Returns true if data is monotonic.'''
data = np.array(data)
# Greater-Equal
if (data[-1] > data[0]):
return np.all(data[1:] >= data[:-1])
# Less-Equal
else:
return np.all(data[1:] <= data[:-1])
我的提议(用 numpy)作为这里几个想法的总结。用途
np.array
以创建每个列表比较的布尔值,np.all
用于检查所有结果是否都为 True
>=, <=
而不是计算 np.diff
,答案 12 :(得分:0)
这里有两种方法可以仅使用 range
或列表推导式来确定列表是单调递增还是单调递减。使用 range 稍微更有效,因为它可以短路,而列表推导式必须迭代整个列表。享受。
a = [1,2,3,4,5]
b = [0,1,6,1,0]
c = [9,8,7,6,5]
def monotonic_increase(x):
if len(x) <= 1: return False
for i in range(1, len(x)):
if x[i-1] >= x[i]:
return False
return True
def monotonic_decrease(x):
if len(x) <= 1: return False
for i in range(1, len(x)):
if x[i-1] <= x[i]:
return False
return True
monotonic_increase = lambda x: len(x) > 1 and all(x[i-1] < x[i] for i in range(1, len(x)))
monotonic_decrease = lambda x: len(x) > 1 and all(x[i-1] > x[i] for i in range(1, len(x)))
print(monotonic_increase(a))
print(monotonic_decrease(c))
print(monotonic_decrease([]))
print(monotonic_increase(c))
print(monotonic_decrease(a))
print(monotonic_increase(b))
print(monotonic_decrease(b))
答案 13 :(得分:-2)
>>> l = [0,1,2,3,3,4]
>>> l == sorted(l) or l == sorted(l, reverse=True)