I have an array of numbers called data, and I want to slice it to get only the numbers between a lower limit ll and an upper limit ul. The way I've been creating indices like this is:
index1 = where(data>ll)
data2 = data[index1]
index2 = where(data2<ul)
data3 = data2[index2]
Is there a less clumsy and wasteful way to do this, with where or another command?
I'm not quite understand your question, if you just want to retrieve numbers in a range, here is a solution.
If data = [3,5,9,1,34,11,8],
[a for a in data if 1 < a < 10]
will get [3,5,9,8]
where is not any function in python. To get numbers between limits, use comprehension
[i for i in data if ll < i < ul]
or use filter:
filter(lambda x: ll < x < ul, data)
Related
#create a simple list in python
#list comprehension
x = [i for i in range(100)]
print (x)
#using loops
squares = []
for x in range(10):
squares.append(x**2)
print (squares)
multiples = k*[z for z in x] for k in squares
So in the last line of code I am trying to multiply both the lists. the problem is the lists are not of the same side and k*[z for z in x] this part is also incorrect.
For problems with iteration, I suggest anyone to check Loop Like A Native by Ned Batchelder and Looping like a Pro by David Baumgold
Option 1
If you want to multiply them as far as the shortest list goes, zip is your friend:
multiples = [a * b for a, b in zip (x, squares)]
Option 2
If you want a matrix with the product, then you can do it like this
result = [
[a * b for a in x]
for b in squares
]
I don't quite understand what the desired output would be. As the function stands now, you would have a list of lists, where the first element has 100 elements, the second one 400, the third 900, and so on.
One thing that's strange: The expression [z for z in x] defines a list that is identical to x. So, you might just write k*x
If you want to multiply the elements of both lists, you would have to write [[k*z for z in x] for k in squares]. This would lead to a list of 10 lists of 100 elements (or a 10x100-matrix) containing the products of your lists.
If you want to have one list of length 100 in the end that holds some kind of products, you will have to think about how to proceed with the shorter list.
EDIT: Or if you want to multiply them as far as possible until you reach the end of the shorter list, FRANCOIS CYRIL's solution is an elegant way to do so.
You can loop on each array to multiply element by element at the same position in a result array:
i = 0
arrayRes = []
while i < min (len(array1),len(array2)):
arrayRes.append(array1[i]*array2[i])
i+=1
Or do you prefer to multiply them, matrix way?
x = 0
y = 0
arrayRes = []
while x < len(array1):
arrayRes.append([])
while y < len(array2):
arrayRes[x].append(array1[x]*array2[y])
y+=1
x+=1
I have an array A = [1 - 100] and I need to find the sum of all the two digit values in this array. How would I approach this? I have tried :
def solution(A):
A =array[0-100])
while A > 9 & A < 99
total = sum(A)
print "%s" % total
)
Is there a function that given an array consisting of N integers returns the sum of all two digit numbers i.e A = [1,1000,80, -91] the function should return -11(as the two are 80 and -91). not a range, multiple array
You can use a list comprehension and check if the length of the string-format is equal to 2, like so:
sum([x if len(str(x))==2 else 0 for x in xrange(1,101)])
Use the keyword and rather than the bitwise &.
Edit: a fuller answer, as that's not the only thing wrong:
def solution():
A = range(101)
total = sum([a for a in A if 9 < a <= 99])
print total
This uses list comprehension and chained inequalities, so is pretty 'pythonic'.
There is tons of errors in your code, please next time before posting,spend some time try to figure it out yourself and be sure that your code at lest doesn't contain any obvious syntax error.
By array, I assume you're talking about a list. And change it to range(101) for every number from 0 to 100
def solution(A):
return sum([x for x in range(A) if len(str(abs(x))) == 2])
print(solution(101))
As a side note, use and instead of & since that's a bitwise-or sign.
Here are a couple of ways to go about the problem, the first is most similar to the approach you appear to be trying:
def solution1(array):
total = 0
for a in array:
if 9 < a < 100:
total += a
return total
print(solution1(range(101)))
And here's a more compact solution using a comprehension (actually, a generator expression):
def solution2(array):
return sum(a for a in array if 9 < a < 100)
print(solution2(range(101)))
Note that in your original you're confusing loops and conditionals.
Meaning "if each item is within range of other item with the same index".
price = [1, 2]
budget = [5, 7]
This works:
if price[0] in range(budget[0]) and price[1] in range(budget[1]):
affordable = True
I figure there's some way to just reference the whole array though. Like so: if price in budget:
You could use:
if all(x in range(y) for x,y in zip(price,budget)):
affordable = True
This will create tuples of price[i],budget[i] and then for each of these tuples we check that price[i] is in range(budget[i]). Nevertheless, you can optimize this further to:
if all(0 <= x < y for x,y in zip(price,budget)):
affordable = True
Note that this makes the assumption that prices are all integers. If you however use x in range(y) it will fail if x is not an integer. So 0.7 in range(10) would fail whereas our second approach will succeed (but it depends of course on what you want).
Assuming that both prices and budgets must be non-negative, using in range seems to be over-complicating things. Instead, you could just use the < operator.
Regardless of whether you use < or in range, it seems like the easiest approach would be to zip both lists and apply the condition on the pairs:
if (all([x[0] >= x[1] for x in zip(budget, price)])):
affordable = True
I have a numpy.array called p2De. The first row has multiple elements may larger than 1. I want to set the elements which smaller than 1 to 1. Following is my code, but shows error... why? How to fix it?
bounde=1
p2De[:0]=map(lambda x:bounde if (x < bounde),p2Di[:0])
File "C:\Users\wange\workspace\cathode\src\diffusion.py", line 86
p2De[:0]=map(lambda x:bounde if (x < bounde),p2Di[:0])
^
SyntaxError: invalid syntax
You need to specify an else for your lambda function :
lambda x:bounde if (x < bounde) else #stuff
It should be
lambda x:bounde if (x < bounde) else x
You can also use list comprehension, which is more readable. Also, I would use the max builtin function instead of your lambda:
p2De[:0] = [max(x, bounde) for x in p2Di[:0]]
As others have noted, the syntax problem is in the lambda.
I don't think you want p2De[:0] - that's an empty array. p2De[0] is the 1st row. p2De[0,:] is the same, and makes it clear to the human readers that you have selected the 1st row of a 2d array.
The use of a map or comprehension works, but they don't offer much of an advantage, if any, over a simple loop (since you don't need to replace all of the values):
for i,v in enumerate(p2De[0,:]):
if v<1:
p2De[0,i] = 1
But none of these iterations is good numpy practice. You should try to think in terms of vector operations. A common practice is to use a boolean mask (or indexing) to select the values that should be changed:
I = p2De[0,:]<1 # boolean vector
p2De[0, I] = 1
p2De[0,p2De[0,:]<1]=1 # or one line form
There is also a numpy function that applies limits like this, np.maximum:
p2De[0,:] = np.maximum(p2De[0,:], 1)
np.clip applies both minimum and maximum bounds:
p2De[0,:] = np.clip(p2De[0,:], minbd, maxbd)
np.clip(p2De[0,:], minbd, maxbd, p2De[0,:]) # alt calling method
The Python(3) bosses encourage us to use functions and comprehensions over maps and lambdas. For example if plist was a list like your p2De[0,:] row:
def clip(x):
return 1 if x<1 else x
plist = [clip(x) for x in plist]
plist = [min(x, 1) for x in plist] # using a builtin for this simple case
say if you have a list a, you can do something like this:
a=[2,3,1,7,0,0,8]
bounde=1
b = map(lambda n: n if n > bounde else bounde,a)
print b
I have two arrays with the same shape in the first two dimensions and I'm looking to record the minimum value in each row of the first array. However I would also like to record the elements in the corresponding position in the third dimension of the second array. I can do it like this:
A = np.random.random((5000, 100))
B = np.random.random((5000, 100, 3))
A_mins = np.ndarray((5000, 4))
for i, row in enumerate(A):
current_min = min(row)
A_mins[i, 0] = current_min
A_mins[i, 1:] = B[i, row == current_min]
I'm new to programming (so correct me if I'm wrong) but I understand that with Numpy doing calculations on whole arrays is faster than iterating over them. With this in mind is there a faster way of doing this? I can't see a way to get rid of the row == current_min bit even though the location of the minimum point must have been 'known' to the computer when it was calculating the min().
Any tips/suggestions appreciated! Thanks.
Something along what #lib talked about:
index = np.argmin(A, axis=1)
A_mins[:,0] = A[np.arange(len(A)), index]
A_mins[:,1:] = B[np.arange(len(A)), index]
It is much faster than using a for loop.
For getting the index of the minimum value, use amin instead of min + comparison
The amin function (and many other functions in numpy) also takes the argument axis, that you can use to get the minimum of each row or each column.
See http://docs.scipy.org/doc/numpy/reference/generated/numpy.amin.html