Trying to update an array by changing its value in python [duplicate] - python

I have a list:
my_list = [1, 2, 3, 4, 5]
How can I multiply each element in my_list by 5? The output should be:
[5, 10, 15, 20, 25]


You can just use a list comprehension:
my_list = [1, 2, 3, 4, 5]
my_new_list = [i * 5 for i in my_list]
>>> print(my_new_list)
[5, 10, 15, 20, 25]
Note that a list comprehension is generally a more efficient way to do a for loop:
my_new_list = []
for i in my_list:
my_new_list.append(i * 5)
>>> print(my_new_list)
[5, 10, 15, 20, 25]
As an alternative, here is a solution using the popular Pandas package:
import pandas as pd
s = pd.Series(my_list)
>>> s * 5
0 5
1 10
2 15
3 20
4 25
dtype: int64
Or, if you just want the list:
>>> (s * 5).tolist()
[5, 10, 15, 20, 25]
Finally, one could use map, although this is generally frowned upon.
my_new_list = map(lambda x: x * 5, my_list)
Using map, however, is generally less efficient. Per a comment from ShadowRanger on a deleted answer to this question:
The reason "no one" uses it is that, in general, it's a performance
pessimization. The only time it's worth considering map in CPython is
if you're using a built-in function implemented in C as the mapping
function; otherwise, map is going to run equal to or slower than the
more Pythonic listcomp or genexpr (which are also more explicit about
whether they're lazy generators or eager list creators; on Py3, your
code wouldn't work without wrapping the map call in list). If you're
using map with a lambda function, stop, you're doing it wrong.
And another one of his comments posted to this reply:
Please don't teach people to use map with lambda; the instant you
need a lambda, you'd have been better off with a list comprehension
or generator expression. If you're clever, you can make map work
without lambdas a lot, e.g. in this case, map((5).__mul__, my_list), although in this particular case, thanks to some
optimizations in the byte code interpreter for simple int math, [x * 5 for x in my_list] is faster, as well as being more Pythonic and simpler.


A blazingly faster approach is to do the multiplication in a vectorized manner instead of looping over the list. Numpy has already provided a very simply and handy way for this that you can use.
>>> import numpy as np
>>>
>>> my_list = np.array([1, 2, 3, 4, 5])
>>>
>>> my_list * 5
array([ 5, 10, 15, 20, 25])
Note that this doesn't work with Python's native lists. If you multiply a number with a list it will repeat the items of the as the size of that number.
In [15]: my_list *= 1000
In [16]: len(my_list)
Out[16]: 5000
If you want a pure Python-based approach using a list comprehension is basically the most Pythonic way to go.
In [6]: my_list = [1, 2, 3, 4, 5]
In [7]: [5 * i for i in my_list]
Out[7]: [5, 10, 15, 20, 25]
Beside list comprehension, as a pure functional approach, you can also use built-in map() function as following:
In [10]: list(map((5).__mul__, my_list))
Out[10]: [5, 10, 15, 20, 25]
This code passes all the items within the my_list to 5's __mul__ method and returns an iterator-like object (in python-3.x). You can then convert the iterator to list using list() built in function (in Python-2.x you don't need that because map return a list by default).
benchmarks:
In [18]: %timeit [5 * i for i in my_list]
463 ns ± 10.6 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [19]: %timeit list(map((5).__mul__, my_list))
784 ns ± 10.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [20]: %timeit [5 * i for i in my_list * 100000]
20.8 ms ± 115 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [21]: %timeit list(map((5).__mul__, my_list * 100000))
30.6 ms ± 169 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [24]: arr = np.array(my_list * 100000)
In [25]: %timeit arr * 5
899 µs ± 4.98 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)


You can do it in-place like so:
l = [1, 2, 3, 4, 5]
l[:] = [x * 5 for x in l]
This requires no additional imports and is very pythonic.


Since I think you are new with Python, lets do the long way, iterate thru your list using for loop and multiply and append each element to a new list.
using for loop
lst = [5, 20 ,15]
product = []
for i in lst:
product.append(i*5)
print product
using list comprehension, this is also same as using for-loop but more 'pythonic'
lst = [5, 20 ,15]
prod = [i * 5 for i in lst]
print prod


With map (not as good, but another approach to the problem):
list(map(lambda x: x*5,[5, 10, 15, 20, 25]))
also, if you happen to be using numpy or numpy arrays, you could use this:
import numpy as np
list(np.array(x) * 5)


from functools import partial as p
from operator import mul
map(p(mul,5),my_list)
is one way you could do it ... your teacher probably knows a much less complicated way that was probably covered in class


Multiplying each element in my_list by k:
k = 5
my_list = [1,2,3,4]
result = list(map(lambda x: x * k, my_list))
resulting in: [5, 10, 15, 20]


I found it interesting to use list comprehension or map with just one object name x.
Note that whenever x is reassigned, its id(x) changes, i.e. points to a different object.
x = [1, 2, 3]
id(x)
2707834975552
x = [1.5 * x for x in x]
id(x)
2707834976576
x
[1.5, 3.0, 4.5]
list(map(lambda x : 2 * x / 3, x))
[1.0, 2.0, 3.0]
id(x) # not reassigned
2707834976576
x = list(map(lambda x : 2 * x / 3, x))
x
[1.0, 2.0, 3.0]
id(x)
2707834980928


var1 = [2,4,6,8,10,12]
#for Integer multiplier us int(x).__mull__ with map
var2 = list( map(int(2).__mul__,var1 ))
#for float multiplier
var2 = list( map(float(2.5).__mul__,var1 ))


Best way is to use list comprehension:
def map_to_list(my_list, n):
# multiply every value in my_list by n
# Use list comprehension!
my_new_list = [i * n for i in my_list]
return my_new_list
# To test:
print(map_to_list([1,2,3], -1))
Returns:
[-1, -2, -3]

Related

How to check if an array contains all the elements of another array? If not, output the missing elements

I need to check if an array A contains all elements of another array B. If not, output the missing elements. Both A and B are integers, and B is always from 0 to N with an interval of 1.
import numpy as np
A=np.array([1,2,3,6,7,8,9])
B=np.arange(10)
I know that I can use the following to check if there is any missing elements, but it does not give the index of the missing element.
np.all(elem in A for elem in B)
Is there a good way in python to output the indices of the missing elements?

IIUC you can try the following and assuming that B always is an "index" list:
[i for i in B if i not in A]
The output would be : [0, 4, 5]
Best way to do it with numpy
Numpy actually has a function to perform this : numpy.insetdiff1d
np.setdiff1d(B, A)
# Which returns
array([0, 4, 5])

You can use enumerate to get both index and content of a list. The following code would do what you want
idx = [idx for idx, element in enumerate(B) if element not in A]

I am assuming we want to get the elements exclusive to B, when compared to A.
Approach #1
Given the specific of B is always from 0 to N with an interval of 1, we can use a simple mask-based one -
mask = np.ones(len(B), dtype=bool)
mask[A] = False
out = B[mask]
Approach #2
Another one that edits B and would be more memory-efficient -
B[A] = -1
out = B[B>=0]
Approach #3
A more generic case of integers could be handled differently -
def setdiff_for_ints(B, A):
N = max(B.max(), A.max()) - min(min(A.min(),B.min()),0) + 1
mask = np.zeros(N, dtype=bool)
mask[B] = True
mask[A] = False
out = np.flatnonzero(mask)
return out
Sample run -
In [77]: A
Out[77]: array([ 1, 2, 3, 6, 7, 8, -6])
In [78]: B
Out[78]: array([1, 3, 4, 5, 7, 9])
In [79]: setdiff_for_ints(B, A)
Out[79]: array([4, 5, 9])
# Using np.setdiff1d to verify :
In [80]: np.setdiff1d(B, A)
Out[80]: array([4, 5, 9])
Timings -
In [81]: np.random.seed(0)
...: A = np.unique(np.random.randint(-10000,100000,1000000))
...: B = np.unique(np.random.randint(0,100000,1000000))
# #Hugolmn's soln with np.setdiff1d
In [82]: %timeit np.setdiff1d(B, A)
4.78 ms ± 96.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [83]: %timeit setdiff_for_ints(B, A)
599 µs ± 6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Reducing a X,Y,Z list to an X,Y list and retrieve the min & max X of the XY PTS

I have large data sets of XYZ data I need to extract only the XY as PTS and find the minimum and maximum. The real data sets are floats... ( I still struggle to understand the list comprehension I guess... ) This is part of a bigger problem with Ascii Grid Files, that Im trying to resolve by creating a larger XYZ data set... (Which is a poor approach I know...)
# I have got this... xyz data...
PTSXYZ = [[1,1,3],[4,4,2],[6,4,1],[6,6,5]]
# I want to get this....xy data PTSXY = [[1,1],[4,4],[6,4],[6,6]]
# I have Tried.... this ??
PTSXY = [PTSXYZ[0][i],PTSXYZ[1][i] for i in PTSXYZ] # Addressing Wrong ?
#Then If I want the Minimum & Maximum X Values..
print min(PTSXY[:][0]),max(PTSXY[:][0])

Retrieving the XY slice
For retrieving from the XYZ matrix the XY matrix you can proceed with slicing:
import numpy as np
a = np.array([[1, 1, 3], [4, 4, 2], [6, 4, 1], [6, 6, 5]])
a[:, :2]
With its output being:
array([[1, 1],
[4, 4],
[6, 4],
[6, 6]])
Retrieving the max and min from x
Then, for retrieving the min and max of the X you can use numpy.min and numpy.max:
_max, _min = np.max(a[:, 0]), np.min(a[:, 0])
Obtaining:
6, 1
How numpy efficiency differs from list comprehension
Let's run a test to see how the difference between numpy and list comprehension using numpy:
import numpy as np
a = np.random.randint(1000, size=(1000, 3))
%timeit [[i[0], i[1]] for i in a]
>>> 483 µs ± 10.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit a[:,:2]
>>> 391 ns ± 6.29 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
As expected, numpy is amazingly faster.

For your first question, you can use list comprehension to create a new list based on the first two elements of PTSXYZ:
PTSXY = [i[:2] for i in PTSXYZ]
print PTSXY
# [[1, 1], [4, 4], [6, 4], [6, 6]]
You might also want to consider a generator, which won't store the entire list in memory:
PTSXY = (i[:2] for i in PTSXYZ)
For the second question, you can get the min and max by creating a list of X elements and taking their min and max:
print min(i[0] for i in PTSXY)
# 1
print max(i[0] for i in PTSXY)
# 6

Once you get the XY list as #101 explained, you can get the minimum and maximum in a similar way:
PTSXY = [[i[0], i[1]] for i in PTSXYZ]
minimum = min([i[0] for i in PTSXY])
maximum = max([i[0] for i in PTSXY])

Cast each digit of string to int

So I have a string say "1234567", and my desired endpoint is a list of the form [1, 2, 3, 4, 5, 6, 7]
What I'm currently doing is this
[int(x) for x in "1234567"]
What I'm wondering is if there is a better or more Pythonic way to do this? Possibly using built-ins or standard library functions.

You can use map function:
map(int, "1234567")
or range:
range(1,8)
With range result will be same:
>>> map(int, "1234567")
[1, 2, 3, 4, 5, 6, 7]
>>> range(1,8)
[1, 2, 3, 4, 5, 6, 7]

One way is to use map. map(int, "1234567")

There isn't any 'more pythonic' way to do it. And AFAIK, whether you prefer map or list comprehension is a matter of personal taste, but more people seem to prefer list comprehensions.
For what you are doing though, if this is in performance-sensitive code, take a page from old assembly routines and use a dict instead, it will be faster than int, and not much more complicated:
In [1]: %timeit [int(x) for x in '1234567']
100000 loops, best of 3: 4.69 µs per loop
In [2]: %timeit map(int, '1234567')
100000 loops, best of 3: 4.38 µs per loop
# Create a lookup dict for each digit, instead of using the builtin 'int'
In [5]: idict = dict(('%d'%x, x) for x in range(10))
# And then, for each digit, just look up in the dict.
In [6]: %timeit [idict[x] for x in '1234567']
1000000 loops, best of 3: 1.21 µs per loop

Is it possible to use argsort in descending order?

Consider the following code:
avgDists = np.array([1, 8, 6, 9, 4])
ids = avgDists.argsort()[:n]
This gives me indices of the n smallest elements. Is it possible to use this same argsort in descending order to get the indices of n highest elements?

If you negate an array, the lowest elements become the highest elements and vice-versa. Therefore, the indices of the n highest elements are:
(-avgDists).argsort()[:n]
Another way to reason about this, as mentioned in the comments, is to observe that the big elements are coming last in the argsort. So, you can read from the tail of the argsort to find the n highest elements:
avgDists.argsort()[::-1][:n]
Both methods are O(n log n) in time complexity, because the argsort call is the dominant term here. But the second approach has a nice advantage: it replaces an O(n) negation of the array with an O(1) slice. If you're working with small arrays inside loops then you may get some performance gains from avoiding that negation, and if you're working with huge arrays then you can save on memory usage because the negation creates a copy of the entire array.
Note that these methods do not always give equivalent results: if a stable sort implementation is requested to argsort, e.g. by passing the keyword argument kind='mergesort', then the first strategy will preserve the sorting stability, but the second strategy will break stability (i.e. the positions of equal items will get reversed).
Example timings:
Using a small array of 100 floats and a length 30 tail, the view method was about 15% faster
>>> avgDists = np.random.rand(100)
>>> n = 30
>>> timeit (-avgDists).argsort()[:n]
1.93 µs ± 6.68 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
>>> timeit avgDists.argsort()[::-1][:n]
1.64 µs ± 3.39 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
>>> timeit avgDists.argsort()[-n:][::-1]
1.64 µs ± 3.66 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
For larger arrays, the argsort is dominant and there is no significant timing difference
>>> avgDists = np.random.rand(1000)
>>> n = 300
>>> timeit (-avgDists).argsort()[:n]
21.9 µs ± 51.2 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
>>> timeit avgDists.argsort()[::-1][:n]
21.7 µs ± 33.3 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
>>> timeit avgDists.argsort()[-n:][::-1]
21.9 µs ± 37.1 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Please note that the comment from nedim below is incorrect. Whether to truncate before or after reversing makes no difference in efficiency, since both of these operations are only striding a view of the array differently and not actually copying data.

Just like Python, in that [::-1] reverses the array returned by argsort() and [:n] gives that last n elements:
>>> avgDists=np.array([1, 8, 6, 9, 4])
>>> n=3
>>> ids = avgDists.argsort()[::-1][:n]
>>> ids
array([3, 1, 2])
The advantage of this method is that ids is a view of avgDists:
>>> ids.flags
C_CONTIGUOUS : False
F_CONTIGUOUS : False
OWNDATA : False
WRITEABLE : True
ALIGNED : True
UPDATEIFCOPY : False
(The 'OWNDATA' being False indicates this is a view, not a copy)
Another way to do this is something like:
(-avgDists).argsort()[:n]
The problem is that the way this works is to create negative of each element in the array:
>>> (-avgDists)
array([-1, -8, -6, -9, -4])
ANd creates a copy to do so:
>>> (-avgDists_n).flags['OWNDATA']
True
So if you time each, with this very small data set:
>>> import timeit
>>> timeit.timeit('(-avgDists).argsort()[:3]', setup="from __main__ import avgDists")
4.2879798610229045
>>> timeit.timeit('avgDists.argsort()[::-1][:3]', setup="from __main__ import avgDists")
2.8372560259886086
The view method is substantially faster (and uses 1/2 the memory...)

Instead of using np.argsort you could use np.argpartition - if you only need the indices of the lowest/highest n elements.
That doesn't require to sort the whole array but just the part that you need but note that the "order inside your partition" is undefined, so while it gives the correct indices they might not be correctly ordered:
>>> avgDists = [1, 8, 6, 9, 4]
>>> np.array(avgDists).argpartition(2)[:2] # indices of lowest 2 items
array([0, 4], dtype=int64)
>>> np.array(avgDists).argpartition(-2)[-2:] # indices of highest 2 items
array([1, 3], dtype=int64)

As #Kanmani hinted, an easier to interpret implementation may use numpy.flip, as in the following:
import numpy as np
avgDists = np.array([1, 8, 6, 9, 4])
ids = np.flip(np.argsort(avgDists))
print(ids)
By using the visitor pattern rather than member functions, it is easier to read the order of operations.

You can use the flip commands numpy.flipud() or numpy.fliplr() to get the indexes in descending order after sorting using the argsort command. Thats what I usually do.

You could create a copy of the array and then multiply each element with -1.
As an effect the before largest elements would become the smallest.
The indeces of the n smallest elements in the copy are the n greatest elements in the original.

With your example:
avgDists = np.array([1, 8, 6, 9, 4])
Obtain indexes of n maximal values:
ids = np.argpartition(avgDists, -n)[-n:]
Sort them in descending order:
ids = ids[np.argsort(avgDists[ids])[::-1]]
Obtain results (for n=4):
>>> avgDists[ids]
array([9, 8, 6, 4])

An elegant way could be as follows -
ids = np.flip(np.argsort(avgDists))
This will give you indices of elements sorted in descending order.
Now you can use regular slicing...
top_n = ids[:n]

consider order of equal elements
If you run a sorting routine and 2 elements are equal, the order is usually not changed. However, the flip/[::-1] approach changes the order of equal elements.
>>> arr = np.array([3, 5, 4, 7, 3])
>>>
>>> np.argsort(arr)[::-1]
array([3, 1, 2, 4, 0]) # equal elements reorderd
>>> np.argsort(-arr)
array([3, 1, 2, 0, 4]) # equal elements not reorderd (compatible to other sorting)
For compatibility reasons I would hence prefer the argsort of the negative array approach. This is especially relevant, when arr represents some number representation of more complex elements.
Example:
obj = ['street', 'house', 'bridge', 'station', 'rails']
arr = np.array([3, 5, 4, 7, 3]) # cost of obj in coins
Disclaimer: A more common approach is to solve the example above with sorted(list_of_tuples_obj_cost, key=lambda x: x[1])

Another way is to use only a '-' in the argument for argsort as in : "df[np.argsort(-df[:, 0])]", provided df is the dataframe and you want to sort it by the first column (represented by the column number '0'). Change the column-name as appropriate. Of course, the column has to be a numeric one.

How to find the cumulative sum of numbers in a list?

time_interval = [4, 6, 12]
I want to sum up the numbers like [4, 4+6, 4+6+12] in order to get the list t = [4, 10, 22].
I tried the following:
t1 = time_interval[0]
t2 = time_interval[1] + t1
t3 = time_interval[2] + t2
print(t1, t2, t3) # -> 4 10 22

If you're doing much numerical work with arrays like this, I'd suggest numpy, which comes with a cumulative sum function cumsum:
import numpy as np
a = [4,6,12]
np.cumsum(a)
#array([4, 10, 22])
Numpy is often faster than pure python for this kind of thing, see in comparison to #Ashwini's accumu:
In [136]: timeit list(accumu(range(1000)))
10000 loops, best of 3: 161 us per loop
In [137]: timeit list(accumu(xrange(1000)))
10000 loops, best of 3: 147 us per loop
In [138]: timeit np.cumsum(np.arange(1000))
100000 loops, best of 3: 10.1 us per loop
But of course if it's the only place you'll use numpy, it might not be worth having a dependence on it.

In Python 2 you can define your own generator function like this:
def accumu(lis):
total = 0
for x in lis:
total += x
yield total
In [4]: list(accumu([4,6,12]))
Out[4]: [4, 10, 22]
And in Python 3.2+ you can use itertools.accumulate():
In [1]: lis = [4,6,12]
In [2]: from itertools import accumulate
In [3]: list(accumulate(lis))
Out[3]: [4, 10, 22]

I did a bench-mark of the top two answers with Python 3.4 and I found itertools.accumulate is faster than numpy.cumsum under many circumstances, often much faster. However, as you can see from the comments, this may not always be the case, and it's difficult to exhaustively explore all options. (Feel free to add a comment or edit this post if you have further benchmark results of interest.)
Some timings...
For short lists accumulate is about 4 times faster:
from timeit import timeit
def sum1(l):
from itertools import accumulate
return list(accumulate(l))
def sum2(l):
from numpy import cumsum
return list(cumsum(l))
l = [1, 2, 3, 4, 5]
timeit(lambda: sum1(l), number=100000)
# 0.4243644131347537
timeit(lambda: sum2(l), number=100000)
# 1.7077815784141421
For longer lists accumulate is about 3 times faster:
l = [1, 2, 3, 4, 5]*1000
timeit(lambda: sum1(l), number=100000)
# 19.174508565105498
timeit(lambda: sum2(l), number=100000)
# 61.871223849244416
If the numpy array is not cast to list, accumulate is still about 2 times faster:
from timeit import timeit
def sum1(l):
from itertools import accumulate
return list(accumulate(l))
def sum2(l):
from numpy import cumsum
return cumsum(l)
l = [1, 2, 3, 4, 5]*1000
print(timeit(lambda: sum1(l), number=100000))
# 19.18597290944308
print(timeit(lambda: sum2(l), number=100000))
# 37.759664884768426
If you put the imports outside of the two functions and still return a numpy array, accumulate is still nearly 2 times faster:
from timeit import timeit
from itertools import accumulate
from numpy import cumsum
def sum1(l):
return list(accumulate(l))
def sum2(l):
return cumsum(l)
l = [1, 2, 3, 4, 5]*1000
timeit(lambda: sum1(l), number=100000)
# 19.042188624851406
timeit(lambda: sum2(l), number=100000)
# 35.17324400227517

Try the
itertools.accumulate() function.
import itertools
list(itertools.accumulate([1,2,3,4,5]))
# [1, 3, 6, 10, 15]

Behold:
a = [4, 6, 12]
reduce(lambda c, x: c + [c[-1] + x], a, [0])[1:]
Will output (as expected):
[4, 10, 22]

Assignment expressions from PEP 572 (new in Python 3.8) offer yet another way to solve this:
time_interval = [4, 6, 12]
total_time = 0
cum_time = [total_time := total_time + t for t in time_interval]

You can calculate the cumulative sum list in linear time with a simple for loop:
def csum(lst):
s = lst.copy()
for i in range(1, len(s)):
s[i] += s[i-1]
return s
time_interval = [4, 6, 12]
print(csum(time_interval)) # [4, 10, 22]
The standard library's itertools.accumulate may be a faster alternative (since it's implemented in C):
from itertools import accumulate
time_interval = [4, 6, 12]
print(list(accumulate(time_interval))) # [4, 10, 22]

Since python 3.8 it's possible to use Assignment expressions, so things like this became easier to implement
nums = list(range(1, 10))
print(f'array: {nums}')
v = 0
cumsum = [v := v + n for n in nums]
print(f'cumsum: {cumsum}')
produces
array: [1, 2, 3, 4, 5, 6, 7, 8, 9]
cumsum: [1, 3, 6, 10, 15, 21, 28, 36, 45]
The same technique can be applied to find the cum product, mean, etc.
p = 1
cumprod = [p := p * n for n in nums]
print(f'cumprod: {cumprod}')
s = 0
c = 0
cumavg = [(s := s + n) / (c := c + 1) for n in nums]
print(f'cumavg: {cumavg}')
results in
cumprod: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880]
cumavg: [1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0]

First, you want a running list of subsequences:
subseqs = (seq[:i] for i in range(1, len(seq)+1))
Then you just call sum on each subsequence:
sums = [sum(subseq) for subseq in subseqs]
(This isn't the most efficient way to do it, because you're adding all of the prefixes repeatedly. But that probably won't matter for most use cases, and it's easier to understand if you don't have to think of the running totals.)
If you're using Python 3.2 or newer, you can use itertools.accumulate to do it for you:
sums = itertools.accumulate(seq)
And if you're using 3.1 or earlier, you can just copy the "equivalent to" source straight out of the docs (except for changing next(it) to it.next() for 2.5 and earlier).

If You want a pythonic way without numpy working in 2.7 this would be my way of doing it
l = [1,2,3,4]
_d={-1:0}
cumsum=[_d.setdefault(idx, _d[idx-1]+item) for idx,item in enumerate(l)]
now let's try it and test it against all other implementations
import timeit, sys
L=list(range(10000))
if sys.version_info >= (3, 0):
reduce = functools.reduce
xrange = range
def sum1(l):
cumsum=[]
total = 0
for v in l:
total += v
cumsum.append(total)
return cumsum
def sum2(l):
import numpy as np
return list(np.cumsum(l))
def sum3(l):
return [sum(l[:i+1]) for i in xrange(len(l))]
def sum4(l):
return reduce(lambda c, x: c + [c[-1] + x], l, [0])[1:]
def this_implementation(l):
_d={-1:0}
return [_d.setdefault(idx, _d[idx-1]+item) for idx,item in enumerate(l)]
# sanity check
sum1(L)==sum2(L)==sum3(L)==sum4(L)==this_implementation(L)
>>> True
# PERFORMANCE TEST
timeit.timeit('sum1(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.001018061637878418
timeit.timeit('sum2(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.000829620361328125
timeit.timeit('sum3(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.4606760001182556
timeit.timeit('sum4(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.18932826995849608
timeit.timeit('this_implementation(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.002348129749298096

There could be many answers for this depending on the length of the list and the performance. One very simple way which I can think without thinking of the performance is this:
a = [1, 2, 3, 4]
a = [sum(a[0:x]) for x in range(1, len(a)+1)]
print(a)
[1, 3, 6, 10]
This is by using list comprehension and this may work fairly well it is just that here I am adding over the subarray many times, you could possibly improvise on this and make it simple!
Cheers to your endeavor!

values = [4, 6, 12]
total = 0
sums = []
for v in values:
total = total + v
sums.append(total)
print 'Values: ', values
print 'Sums: ', sums
Running this code gives
Values: [4, 6, 12]
Sums: [4, 10, 22]

Try this:
result = []
acc = 0
for i in time_interval:
acc += i
result.append(acc)

l = [1,-1,3]
cum_list = l
def sum_list(input_list):
index = 1
for i in input_list[1:]:
cum_list[index] = i + input_list[index-1]
index = index + 1
return cum_list
print(sum_list(l))

In Python3, To find the cumulative sum of a list where the ith element
is the sum of the first i+1 elements from the original list, you may do:
a = [4 , 6 , 12]
b = []
for i in range(0,len(a)):
b.append(sum(a[:i+1]))
print(b)
OR you may use list comprehension:
b = [sum(a[:x+1]) for x in range(0,len(a))]
Output
[4,10,22]

lst = [4, 6, 12]
[sum(lst[:i+1]) for i in xrange(len(lst))]
If you are looking for a more efficient solution (bigger lists?) a generator could be a good call (or just use numpy if you really care about performance).
def gen(lst):
acu = 0
for num in lst:
yield num + acu
acu += num
print list(gen([4, 6, 12]))

In [42]: a = [4, 6, 12]
In [43]: [sum(a[:i+1]) for i in xrange(len(a))]
Out[43]: [4, 10, 22]
This is slighlty faster than the generator method above by #Ashwini for small lists
In [48]: %timeit list(accumu([4,6,12]))
100000 loops, best of 3: 2.63 us per loop
In [49]: %timeit [sum(a[:i+1]) for i in xrange(len(a))]
100000 loops, best of 3: 2.46 us per loop
For larger lists, the generator is the way to go for sure. . .
In [50]: a = range(1000)
In [51]: %timeit [sum(a[:i+1]) for i in xrange(len(a))]
100 loops, best of 3: 6.04 ms per loop
In [52]: %timeit list(accumu(a))
10000 loops, best of 3: 162 us per loop

Somewhat hacky, but seems to work:
def cumulative_sum(l):
y = [0]
def inc(n):
y[0] += n
return y[0]
return [inc(x) for x in l]
I did think that the inner function would be able to modify the y declared in the outer lexical scope, but that didn't work, so we play some nasty hacks with structure modification instead. It is probably more elegant to use a generator.

Without having to use Numpy, you can loop directly over the array and accumulate the sum along the way. For example:
a=range(10)
i=1
while((i>0) & (i<10)):
a[i]=a[i-1]+a[i]
i=i+1
print a
Results in:
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45]

A pure python oneliner for cumulative sum:
cumsum = lambda X: X[:1] + cumsum([X[0]+X[1]] + X[2:]) if X[1:] else X
This is a recursive version inspired by recursive cumulative sums. Some explanations:
The first term X[:1] is a list containing the previous element and is almost the same as [X[0]] (which would complain for empty lists).
The recursive cumsum call in the second term processes the current element [1] and remaining list whose length will be reduced by one.
if X[1:] is shorter for if len(X)>1.
Test:
cumsum([4,6,12])
#[4, 10, 22]
cumsum([])
#[]
And simular for cumulative product:
cumprod = lambda X: X[:1] + cumprod([X[0]*X[1]] + X[2:]) if X[1:] else X
Test:
cumprod([4,6,12])
#[4, 24, 288]

Here's another fun solution. This takes advantage of the locals() dict of a comprehension, i.e. local variables generated inside the list comprehension scope:
>>> [locals().setdefault(i, (elem + locals().get(i-1, 0))) for i, elem
in enumerate(time_interval)]
[4, 10, 22]
Here's what the locals() looks for each iteration:
>>> [[locals().setdefault(i, (elem + locals().get(i-1, 0))), locals().copy()][1]
for i, elem in enumerate(time_interval)]
[{'.0': <enumerate at 0x21f21f7fc80>, 'i': 0, 'elem': 4, 0: 4},
{'.0': <enumerate at 0x21f21f7fc80>, 'i': 1, 'elem': 6, 0: 4, 1: 10},
{'.0': <enumerate at 0x21f21f7fc80>, 'i': 2, 'elem': 12, 0: 4, 1: 10, 2: 22}]
Performance is not terrible for small lists:
>>> %timeit list(accumulate([4, 6, 12]))
387 ns ± 7.53 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
>>> %timeit np.cumsum([4, 6, 12])
5.31 µs ± 67.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
>>> %timeit [locals().setdefault(i, (e + locals().get(i-1,0))) for i,e in enumerate(time_interval)]
1.57 µs ± 12 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
And obviously falls flat for larger lists.
>>> l = list(range(1_000_000))
>>> %timeit list(accumulate(l))
95.1 ms ± 5.22 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
>>> %timeit np.cumsum(l)
79.3 ms ± 1.07 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
>>> %timeit np.cumsum(l).tolist()
120 ms ± 1.23 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
>>> %timeit [locals().setdefault(i, (e + locals().get(i-1, 0))) for i, e in enumerate(l)]
660 ms ± 5.14 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Even though the method is ugly and not practical, it sure is fun.

I think the below code is the easiest:
a=[1,1,2,1,2]
b=[a[0]]+[sum(a[0:i]) for i in range(2,len(a)+1)]

def cumulative_sum(list):
l = []
for i in range(len(list)):
new_l = sum(list[:i+1])
l.append(new_l)
return l
time_interval = [4, 6, 12]
print(cumulative_sum(time_interval)
Maybe a more beginner-friendly solution.

So you need to make a list of cumulative sums. You can do it by using for loop and .append() method
time_interval = [4, 6, 12]
cumulative_sum = []
new_sum = 0
for i in time_interval:
new_sum += i
cumulative_sum.append(new_sum)
print(cumulative_sum)
or, using numpy module
import numpy
time_interval = [4, 6, 12]
c_sum = numpy.cumsum(time_interval)
print(c_sum.tolist())

This would be Haskell-style:
def wrand(vtlg):
def helpf(lalt,lneu):
if not lalt==[]:
return helpf(lalt[1::],[lalt[0]+lneu[0]]+lneu)
else:
lneu.reverse()
return lneu[1:]
return helpf(vtlg,[0])

Categories