Update the value of element in 2-dimensional array - arrays

I am wondering why two code segments below have different outputs. Could someone explain this?
Code 1
d = 5
matrix_list = [[0] * d] * d
matrix_list[0][3] = 1
for i in matrix_list:
print(i)
Output:
[0, 0, 0, 1, 0]
[0, 0, 0, 1, 0]
[0, 0, 0, 1, 0]
[0, 0, 0, 1, 0]
[0, 0, 0, 1, 0]
Code 2
matrix_list = [[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
matrix_list[0][3] = 1
print()
for i in matrix_list:
print(i)
Output:
[0, 0, 0, 1, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]


This happens, because lists are mutable in python and when you do as [[0] * d] * d you actually use the same list reference for all lists in the list.
list_1 = [1,2]
list_2 = list_1
list_2[0] = 3
print(list_1) -> output: [3,2]
It happens, because list_1 and list_2 are linked to the same object. You can check with:
print(id(list_1))
print(id(list_2))
Or in your case you can try printing ids instead of value:
for i in matrix_list:
print(id(i))
It means the correct way to create list of lists in your case is:
list_of_lists = []
for i in range(5):
list_of_lists.append([0, 0, 0, 0, 0])
It will create a new list object on every iteration.

Related

convert numpy array of size n into a scalar with sub-item index numbers as scalar values

I want to convert a numpy array
a = array([[1, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0],
[0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 1, 0, 0]])
into a scalar with sub-item index numbers as scalar values
desired output
a1 = array([[1],[1],[3],[3],[6],[4]])
I tried this method:
a1 = []
for item in a:
a1.append(np.where(item==1))
and I get this output:
a1 = [(array([0]),),
(array([0]),),
(array([2]),),
(array([2]),),
(array([6]),),
(array([4]),)]
Is there a more pythonic way to achieve it?

After reading the question more carefully, I think this one-liner should
solve your issue in a more pythonic way:
np.where(a==1)[1].reshape((a.shape[0],1))
If you prefer starting indexing with 1 instead of 0 (as is evident in your desired output), then you just have to add a 1 to the line above, i.e.
np.where(a==1)[1].reshape((a.shape[0],1)) + 1
Note that there seems to be an error in your example output above for the last two elements.

Design a specific algorithm for a nxn array implemented on O(nlogn)

The problem:
Suppose that each row of an n×n array A consists of 1’s and 0’s such that, in any row of A, all the 1’s come before any 0’s in that row. Assuming A is already in memory, describe a method running in O(nlogn) time (not O(n2) time!) for counting the number of 1’s in A.
My experience: I have done it for O(n) but I dont know how can I achieve it with O(nlogN)
I would appreciate any help !

Consider that each individual row consists of all 1s followed by all 0s:
1111111000
You can use a binary search to find the transition point (the last 1 in the row). The way this works is to set low and high to the ends and check the middle.
If you are at the transition point, you're done. Otherwise, if you're in the 1s, set low to one after the midpoint. Otherwise, you're in the 0s, so set high to one before the midpoint.
That would go something like (pseudo-code, with some optimisations):
def countOnes(row):
# Special cases first, , empty, all 0s, or all 1s.
if row.length == 0: return 0
if row[0] == "0": return 0
if row[row.length - 1] == 1: return row.length
# At this point, there must be at least one of each value,
# so length >= 2. That means you're guaranteed to find a
# transition point.
lo = 0
hi = row.length - 1
while true:
mid = (lo + hi) / 2
if row[mid] == 1 and row[mid+1] == 0:
return mid + 1
if row[mid] == 1:
lo = mid + 1
else:
hi = mid - 1
Since a binary search for a single row is O(logN) and you need to do that for N rows, the resultant algorithm is O(NlogN).
For a more concrete example, see the following complete Python program, which generates a mostly random matrix then uses the O(N) method and the O(logN) method (the former as confirmation) of counting the ones in each row:
import random
def slow_count(items):
count = 0
for item in items:
if item == 0:
break
count += 1
return count
def fast_count(items):
# Special cases first, no 1s or all 1s.
if len(items) == 0: return 0
if items[0] == 0: return 0
if items[len(items) - 1] == 1: return len(items)
# At this point, there must be at least one of each value,
# so length >= 2. That means you're guaranteed to find a
# transition point.
lo = 0
hi = len(items) - 1
while True:
mid = (lo + hi) // 2
if items[mid] == 1 and items[mid+1] == 0:
return mid + 1
if items[mid] == 1:
lo = mid + 1
else:
hi = mid - 1
# Ensure test data has rows with all zeros and all ones.
N = 20
matrix = [[1] * N, [0] * N]
# Populate other rows randomly.
random.seed()
for _ in range(N - 2):
numOnes = random.randint(0, N)
matrix.append([1] * numOnes + [0] * (N - numOnes))
# Print rows and counts using slow-proven and fast method.
for row in matrix:
print(row, slow_count(row), fast_count(row))
The fast_count function is the equivalent of what I've provided in this answer.
A sample run is:
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] 20 20
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 0 0
[1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 5 5
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0] 15 15
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 10 10
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 1 1
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0] 11 11
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0] 12 12
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0] 11 11
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 1 1
[1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 6 6
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0] 16 16
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0] 14 14
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0] 11 11
[1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 9 9
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0] 13 13
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 1 1
[1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 4 4
[1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 6 6
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0] 19 19

python: vectorized cumulative counting

I have a numpy array and would like to count the number of occurences for each value, however, in a cumulative way
in = [0, 1, 0, 1, 2, 3, 0, 0, 2, 1, 1, 3, 3, 0, ...]
out = [0, 0, 1, 1, 0, 0, 2, 3, 1, 2, 3, 1, 2, 4, ...]
I'm wondering if it is best to create a (sparse) matrix with ones at col = i and row = in[i]
1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0
0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0
0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0
Then we could compute the cumsums along the rows and extract the numbers from the locations where the cumsums increment.
However, if we cumsum a sparse matrix, doesn't become dense? Is there an efficient way of doing it?

Here's one vectorized approach using sorting -
def cumcount(a):
# Store length of array
n = len(a)
# Get sorted indices (use later on too) and store the sorted array
sidx = a.argsort()
b = a[sidx]
# Mask of shifts/groups
m = b[1:] != b[:-1]
# Get indices of those shifts
idx = np.flatnonzero(m)
# ID array that will store the cumulative nature at the very end
id_arr = np.ones(n,dtype=int)
id_arr[idx[1:]+1] = -np.diff(idx)+1
id_arr[idx[0]+1] = -idx[0]
id_arr[0] = 0
c = id_arr.cumsum()
# Finally re-arrange those cumulative values back to original order
out = np.empty(n, dtype=int)
out[sidx] = c
return out
Sample run -
In [66]: a
Out[66]: array([0, 1, 0, 1, 2, 3, 0, 0, 2, 1, 1, 3, 3, 0])
In [67]: cumcount(a)
Out[67]: array([0, 0, 1, 1, 0, 0, 2, 3, 1, 2, 3, 1, 2, 4])

Numpy Number Patterns

Is there a function in Numpy that allows you to take 4 records at a time and see where they match with a second dataset? Once there is a match move to the next 4 records of the first data set. It wont always be every 4 records, but i am using this as an example.
So if dataset one had - 1,5,7,8,10,12,6,1,3,6,8,9
And the second dataset had - 1,5,7,8,11,15,6,1,3,6,10,6
My result will be: 1,5,7,8, 6,1,3,6
POST EDIT:
My second example datasets:
import numpy as np
a =np.array([15,15,0,0,10,10,0,0,2,1,8,8,42,2,4,4,3,1,1,3,5,6,0,9,47,1,1,7,7,0,0,45,12,17,45])
b = np.array ([6,0,0,15,15,0,0,10,10,0,0,2,1,8,8,42,2,4,4,3,3,4,6,0,9,47,1,1,7,7,0,0,45,12,16,1,9,3,30])
Here's another snapshot of an example:
Thank you in advance for looking at my question!!

Update: for the more difficult and more interesting alignment problem it is probably best not to reinvent the wheel but to rely on python's difflib:
from difflib import SequenceMatcher
import numpy as np
k=4
a = np.array([15,15,0,0,10,10,0,0,2,1,8,8,42,2,4,4,3,1,1,3,5,6,0,9,47,1,1,7,7,0,0,45,12,17,45])
b = np.array ([6,0,0,15,15,0,0,10,10,0,0,2,1,8,8,42,2,4,4,3,3,4,6,0,9,47,1,1,7,7,0,0,45,12,16,1,9,3,30])
sm = SequenceMatcher(a=a, b=b)
matches = sm.get_matching_blocks()
matches = [m for m in matches if m.size >= k]
# [Match(a=0, b=3, size=17), Match(a=21, b=22, size=12)]
consensus = [a[m.a:m.a+m.size] for m in matches]
# [array([15, 15, 0, 0, 10, 10, 0, 0, 2, 1, 8, 8, 42, 2, 4, 4, 3]), array([ 6, 0, 9, 47, 1, 1, 7, 7, 0, 0, 45, 12])]
consfour = [a[m.a:m.a + m.size // k * k] for m in matches]
# [array([15, 15, 0, 0, 10, 10, 0, 0, 2, 1, 8, 8, 42, 2, 4, 4]), array([ 6, 0, 9, 47, 1, 1, 7, 7, 0, 0, 45, 12])]
summary = [np.c_[np.add.outer(np.arange(m.size // k * k), (m.a, m.b)), c]
for m, c in zip(matches, consfour)]
merge = np.concatenate(summary, axis=0)
Below is my original solution assuming already aligned and same-length arrays:
Here is a hybrid solution using numpy to find consecutive matches and cutting them out and then list comp to apply length constraints:
import numpy as np
d1 = np.array([7,1,5,7,8,0,6,9,0,10,12,6,1,3,6,8,9])
d2 = np.array([8,1,5,7,8,0,6,9,0,11,15,6,1,3,6,10,6])
k = 4
# find matches
m = d1 == d2
# find switches between match, no match
sw = np.where(m[:-1] != m[1:])[0] + 1
# split
mnm = np.split(d1, sw)
# select matches
ones_ = mnm[1-m[0]::2]
# apply length constraint
res = [blck[i:i+k] for blck in ones_ for i in range(len(blck)-k+1)]
# [array([1, 5, 7, 8]), array([5, 7, 8, 0]), array([7, 8, 0, 6]), array([8, 0, 6, 9]), array([0, 6, 9, 0]), array([6, 1, 3, 6])]
res_no_ovlp = [blck[k*i:k*i+k] for blck in ones_ for i in range(len(blck)//k)]
# [array([1, 5, 7, 8]), array([0, 6, 9, 0]), array([6, 1, 3, 6])]

You can use matrix masking like,
import numpy as np
from scipy.sparse import dia_matrix
a = np.array([1,5,7,8,10,12,6,1,3,6,8,9])
b = np.array([1,5,7,8,11,15,6,1,3,6,10,6])
mask = dia_matrix((np.ones((1, a.size)).repeat(4, axis=0), np.arange(4)),
shape=(a.size, b.size), dtype=np.int)
print(mask.toarray())
matches = a[mask.T.dot(mask.dot(a == b) == 4).astype(np.bool)]
print(matches)
This will output,
array([[1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]])
[1 5 7 8 6 1 3 6]
You can think about how the matrix multiplication works to get this result.
Scaling
For scaling, I tested with 1e3, 1e5, and 1e7 elements and got,
1e3 - 0.019184964010491967
1e5 - 0.4330314120161347
1e7 - 144.54082221200224
See the gist. Not sure why such a hard jump at 1e7 elements.

This is an exercise is list comprehension. We have the data
data = [1,5,7,8,10,12,6,1,3,6,8,9]
search_data = [1,5,7,8,11,15,6,1,3,6,10,6]
First we can chunk the original data into blocks of length n
n = 4
chunks = [data[i:i + n] for i in range(len(data) - n + 1)]
search_chunks = [search_data[i:i + n] for i in range(len(search_data) - n + 1)]
Now we must select chunks from the first list that appear in the second list
hits = [c for c in chunks if c in search_chunks]
print hits
# [[1, 5, 7, 8], [6, 1, 3, 6]]
This may not be the optimal solution for long lists. It may improve performance to consider sets, if there are likely to repeated chunks
chunks = set(tuple(data[i:i + n]) for i in range(len(data) - n + 1))
search_chunks = set(tuple(search_data[i:i + n]) for i in range(len(search_data) - n + 1))
This can be quite competitive with above numpy solution, e.g.
import numpy as np
import time
# Generate data
len_ = 10000
max_ = 10
data = map(int, np.random.rand(len_) * max_)
search_data = map(int, np.random.rand(len_) * max_)
# Time list comprehension
start = time.time()
n = 4
chunks = set(tuple(data[i:i + n]) for i in range(len(data) - n + 1))
search_chunks = set(tuple(search_data[i:i + n]) for i in range(len(search_data) - n + 1))
hits = [c for c in chunks if c in search_chunks]
print time.time() - start
# Time numpy
a = np.array(data)
b = np.array(search_data)
mask = 1 * (np.abs(np.arange(a.size).reshape((-1, 1)) - np.arange(a.size) - 0.5) < 2)
start = time.time()
matches = a[mask.T.dot(mask.dot(a == b) == 4).astype(np.bool)]
print time.time() - start
It's typically faster here, but it depends on number of repeated chunks etc.

Pandas Series of lists from list of lists

Background
Given a list of lists (or numpy array of array):
array([[0, 0, 0, ..., 0, 0, 0],
[0, 0, 0, ..., 0, 0, 0],
[0, 0, 0, ..., 0, 0, 0],
...,
[0, 0, 0, ..., 0, 0, 0],
[0, 0, 0, ..., 0, 0, 0],
[0, 0, 0, ..., 0, 0, 0]])
Question
How does one create a Pandas Series with an array in each row?
0 [0, 0, 0, ..., 0, 0, 0]
1 [0, 0, 0, ..., 0, 0, 0]
...
999 [0, 0, 0, ..., 0, 0, 0]
1000 [0, 0, 0, ..., 0, 0, 0]
Attempt
df['my_array'] = pd.Series(my_array_of_arrays)
Traceback (most recent call last)
...
Exception: Data must be 1-dimensional

Cast to list so it stores each element as an array in the series:
In [112]:
a = np.array([[0,1,2],[2,3,5]])
Out[112]:
array([[0, 1, 2],
[2, 3, 5]])
In [114]:
pd.Series(list(a))
Out[114]:
0 [0, 1, 2]
1 [2, 3, 5]
dtype: object
You can see that the element is still an array:
In [116]:
pd.Series(list(a))[0]
Out[116]:
array([0, 1, 2])

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