Check if all values in list are greater than a certain number
Question:
my_list1 = [30,34,56]
my_list2 = [29,500,43]
How to I check if all values in list are >= 30? my_list1 should work and my_list2 should not.
The only thing I could think of doing was:
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:
In hindsight, after dealing with bigger datasets where speed actually matters and utilizing numpy…I would do this:
>>> 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:
len([*filter(lambda x: x >= 30, my_list1)]) > 0
Answers:
Use the all() function with a generator expression:
>>> 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 to 30, otherwise my_list1 would not pass the test either.
If you wanted to do this in a function, you’d use:
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 a value below 30, you return False, and return True if you found no evidence to the contrary.
Similarly, you can use the any() function to test if at least 1 value matches the condition.
There is a builtin function all:
all (x > limit for x in my_list)
Being limit the value greater than which all numbers must be.
You can use 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.
…any reason why you can’t use min()?
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…
You could do the following:
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.
The overall winner between using the np.sum, np.min, and all seems to be np.min in terms of speed for large arrays:
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)
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.
When the criteria is almost immediately met and the all loop exits fast, the all function is winning just slightly over np.min:
>>> %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:
>>> %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
np.sum(my_list<x)
can be very useful is one wants to know how many values are below x.
I write this function
def larger(x, than=0):
if not x or min(x) > than:
return True
return False
Then
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 x in condition.
a = [[a, 2], [b, 3], [c, 4], [d, 5], [a, 1], [b, 6], [e, 7], [h, 8]]
I need this from above one
a = [[a, 3], [b, 9], [c, 4], [d, 5], [e, 7], [h, 8]]
a.append([0, 0])
for i in range(len(a)):
for j in range(i + 1, len(a) - 1):
if a[i][0] == a[j][0]:
a[i][1] += a[j][1]
del a[j]
a.pop()
A solution based on numpy array and all function:
my_list1 = [30, 34, 56]
my_list2 = [29, 500, 43]
#
import numpy as np
all(np.array(my_list1)>=30)
Output: True
all(np.array(my_list2)>=30)
Output: False
my_list1 = [30,34,56]
my_list2 = [29,500,43]
How to I check if all values in list are >= 30? my_list1 should work and my_list2 should not.
The only thing I could think of doing was:
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:
In hindsight, after dealing with bigger datasets where speed actually matters and utilizing numpy…I would do this:
>>> 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:
len([*filter(lambda x: x >= 30, my_list1)]) > 0
Use the all() function with a generator expression:
>>> 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 to 30, otherwise my_list1 would not pass the test either.
If you wanted to do this in a function, you’d use:
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 a value below 30, you return False, and return True if you found no evidence to the contrary.
Similarly, you can use the any() function to test if at least 1 value matches the condition.
There is a builtin function all:
all (x > limit for x in my_list)
Being limit the value greater than which all numbers must be.
You can use 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.
…any reason why you can’t use min()?
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…
You could do the following:
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.
The overall winner between using the np.sum, np.min, and all seems to be np.min in terms of speed for large arrays:
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)
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.
When the criteria is almost immediately met and the all loop exits fast, the all function is winning just slightly over np.min:
>>> %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:
>>> %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
np.sum(my_list<x)
can be very useful is one wants to know how many values are below x.
I write this function
def larger(x, than=0):
if not x or min(x) > than:
return True
return False
Then
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 x in condition.
a = [[a, 2], [b, 3], [c, 4], [d, 5], [a, 1], [b, 6], [e, 7], [h, 8]]
I need this from above one
a = [[a, 3], [b, 9], [c, 4], [d, 5], [e, 7], [h, 8]]
a.append([0, 0])
for i in range(len(a)):
for j in range(i + 1, len(a) - 1):
if a[i][0] == a[j][0]:
a[i][1] += a[j][1]
del a[j]
a.pop()
A solution based on numpy array and all function:
my_list1 = [30, 34, 56]
my_list2 = [29, 500, 43]
#
import numpy as np
all(np.array(my_list1)>=30)
Output: True
all(np.array(my_list2)>=30)
Output: False