Python 检查列表中的所有值是否都大于某个数字
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Check if all values in list are greater than a certain number
提问by O.rka
my_list1 = [30,34,56]
my_list2 = [29,500,43]
How to I check if all values in list are >= 30? my_list1should work and my_list2should not.
如何检查列表中的所有值是否 >= 30？my_list1应该工作，my_list2不应该。
The only thing I could think of doing was:
我唯一能想到的就是：
boolean = 0
def func(ls):
for k in ls:
if k >= 30:
boolean = boolean + 1
else:
boolean = 0
if boolean > 0:
print 'Continue'
elif boolean = 0:
pass
Update 2016:
2016 年更新：
In hindsight, after dealing with bigger datasets where speed actually matters and utilizing numpy...I would do this:
事后看来，在处理速度实际上很重要的更大数据集并利用之后numpy......我会这样做：
>>> my_list1 = [30,34,56]
>>> my_list2 = [29,500,43]
>>> import numpy as np
>>> A_1 = np.array(my_list1)
>>> A_2 = np.array(my_list2)
>>> A_1 >= 30
array([ True, True, True], dtype=bool)
>>> A_2 >= 30
array([False, True, True], dtype=bool)
>>> ((A_1 >= 30).sum() == A_1.size).astype(np.int)
1
>>> ((A_2 >= 30).sum() == A_2.size).astype(np.int)
0
You could also do something like:
您还可以执行以下操作：
len([*filter(lambda x: x >= 30, my_list1)]) > 0
采纳答案by Martijn Pieters
Use the all()functionwith a generator expression:
使用带有生成器表达式的all()函数：
>>> my_list1 = [30, 34, 56]
>>> my_list2 = [29, 500, 43]
>>> all(i >= 30 for i in my_list1)
True
>>> all(i >= 30 for i in my_list2)
False
Note that this tests for greater than or equal to30, otherwise my_list1would not pass the test either.
注意这个测试大于等于30，否则my_list1也不会通过测试。
If you wanted to do this in a function, you'd use:
如果您想在函数中执行此操作，您可以使用：
def all_30_or_up(ls):
for i in ls:
if i < 30:
return False
return True
e.g. as soon as you find a value that proves that there is avalue below 30, you return False, and return Trueif you found no evidence to the contrary.
例如，一旦您发现一个值证明存在低于 30的值，您就返回False，True如果您没有发现相反的证据，则返回。
Similarly, you can use the any()functionto test if at least 1value matches the condition.
同样，您可以使用该any()函数来测试是否至少有 1 个值与条件匹配。
回答by Hyperboreus
There is a builtin function all:
有一个内置函数all：
all (x > limit for x in my_list)
Being limit the value greater than which all numbers must be.
限制大于所有数字必须的值。
回答by Simeon Visser
You can use all():
您可以使用all()：
my_list1 = [30,34,56]
my_list2 = [29,500,43]
if all(i >= 30 for i in my_list1):
print 'yes'
if all(i >= 30 for i in my_list2):
print 'no'
Note that this includes all numbers equal to 30 or higher, not strictly above 30.
请注意，这包括所有等于或大于 30 的数字，而不是严格大于 30 的数字。
回答by Roberto
...any reason why you can't use min()?
...有什么理由不能使用min()吗？
def above(my_list, minimum):
if min(my_list) >= minimum:
print "All values are equal or above", minimum
else:
print "Not all values are equal or above", minimum
I don't know if this is exactly what you want, but technically, this is what you asked for...
我不知道这是否正是您想要的，但从技术上讲，这就是您要求的...
回答by Filip Grebowski
You could do the following:
您可以执行以下操作：
def Lists():
my_list1 = [30,34,56]
my_list2 = [29,500,43]
for element in my_list1:
print(element >= 30)
for element in my_list2:
print(element >= 30)
Lists()
This will return the values that are greater than 30 as True, and the values that are smaller as false.
这会将大于 30 的值返回为 True，将小于 30 的值返回为 false。
回答by Chachni
The overall winner between using the np.sum, np.min, and all seems to be np.min in terms of speed for large arrays:
就大型阵列的速度而言，使用 np.sum、np.min 和 all 的总体赢家似乎是 np.min：
N = 1000000
def func_sum(x):
my_list = np.random.randn(N)
return np.sum(my_list < x )==0
def func_min(x):
my_list = np.random.randn(N)
return np.min(my_list) >= x
def func_all(x):
my_list = np.random.randn(N)
return all(i >= x for i in my_list)
(i need to put the np.array definition inside the function, otherwise the np.min function remembers the value and does not do the computation again when testing for speed with timeit)
（我需要将 np.array 定义放在函数中，否则 np.min 函数会记住该值并且在使用 timeit 测试速度时不会再次进行计算）
The performance of "all" depends very much on when the first element that does not satisfy the criteria is found, the np.sum needs to do a bit of operations, the np.min is the lightest in terms of computations in the general case.
“all”的性能很大程度上取决于何时找到第一个不满足条件的元素，np.sum需要做一些运算，一般情况下np.min在计算方面是最轻的.
When the criteria is almost immediately met and the all loop exits fast, the all function is winning just slightly over np.min:
当几乎立即满足条件并且 all 循环快速退出时， all 函数比 np.min 略胜一筹：
>>> %timeit func_sum(10)
10 loops, best of 3: 36.1 ms per loop
>>> %timeit func_min(10)
10 loops, best of 3: 35.1 ms per loop
>>> %timeit func_all(10)
10 loops, best of 3: 35 ms per loop
But when "all" needs to go through all the points, it is definitely much worse, and the np.min wins:
但是当“all”需要把所有的点都过一遍的时候，肯定差很多，np.min胜出：
>>> %timeit func_sum(-10)
10 loops, best of 3: 36.2 ms per loop
>>> %timeit func_min(-10)
10 loops, best of 3: 35.2 ms per loop
>>> %timeit func_all(-10)
10 loops, best of 3: 230 ms per loop
But using
但是使用
np.sum(my_list<x)
can be very useful is one wants to know how many values are below x.
可能非常有用的是，您想知道 x 以下有多少个值。
回答by Shameem
I write this function
我写这个函数
def larger(x, than=0):
if not x or min(x) > than:
return True
return False
Then
然后
print larger([5, 6, 7], than=5) # False
print larger([6, 7, 8], than=5) # True
print larger([], than=5) # True
print larger([6, 7, 8, None], than=5) # False
Empty list on min()will raise ValueError. So I added if not xin condition.
min()上的空列表将引发 ValueError。所以我加if not x了条件。
