I am using python 3.2.0 and numpy. I would like to check if one of the arrays is in between two other specified arrays. I would like it if you suggest a function or few of them together. Any help is appreciated , as it is a school project and I need to submit it quickly.
If you mean how the last item of arr1 is less than all items of input_arr, and the first item of arr2 is greater than all those in input_arr, you can do this, with "biggest" of arr1 and "smallest" of arr2:
biggest = arr1[len(arr1)-1]
smallest = arr2[0]
between=True
for item in input_arr:
if not (biggest<item and smallest>item):
between=False
break
Alternatively, you could change the < and/or the > to <= or >= if you allow equals to (so [1,3,4],[4,6,8],[8,17,18] is True)
This assumes that the lists are consecutive. If they're not, you'll have to loop through arr1 to find the largest number and arr2 to find the smallest first.
biggest=0
for item in arr1:
if item>biggest:
biggest=item
smallest=arr2[0]
for item in arr2:
if item<smallest:
smallest = item
Use this as a skeleton guide, don't just copy and paste. If you don't understand it and therefore can't construct your own version, you probably need to do some sort of online course (eg. Codecademy). In the mean time, copy the 2nd bit first if you need it, then the first.
If you had 3 arrays,
#lower bound
In[1]: small = np.zeros((3,3))
#Array we are testing
In[2]: test = np.ones((3,3))
#Upper bound
In[3]: large = np.ones((3,3))*2
then you could do a logical_and and test the entire array with the Boolean sum against the size
In[4]: np.logical_and(small<=testm,testm<=large).sum() == l.size
out[4]: True
Related
I am working on a dataset as a part of my course practice and am stuck in a particular step. I have tried that using R, but I wish to do the same in python. I am comparatively new to python and so require help.
The data set consists of a column with name 'Seq' with seq(5000+) records. I have another column of name 'MainSeq' that contains the substring seq values in it. I need to check the presence of seq on MainSeq based on the start position given and then print 7 letters before and after each letter of the seq. i.e.
I have a a value in col 'MainSeq' as 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'.
Col 'Seq' contains value JKLMNO
Start Position of J= 10 and O= 15
I need to create a new column such that it takes 7 letters before and after the start letter from J till O i.e. having a total length of 15
CDEFGHI**J**KLMNOPQ
DEFGHIJ**K**LMNOPQR
EFGHIJK**L**MNOPQRS
FGHIJKL**M**NOPQRST
GHIJKLM**N**OPQRSTU
HIJKLMN**O**PQRSTUV
I know to apply the logic on a specific seq. But since I have around 5000+ seq records, I need to figure out a way to apply the same on all the seq records.
seq = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
i = seq.index('J')
j = seq.index('O')
value = 7
for mid in range(i, 1+j):
print(seq[mid-value:mid+value+1])
I'm not sure this will do exactly what you want, you've not really supplied a lot of data to test with, but it might work or at least give you a start.
import pandas as pd
df = pd.DataFrame({'MainSeq':['ABCDEFGHIJKLMNOPQRSTUVWZYZ','ABCDEFGHIJKLMNOPQRSTUVWZYZ'], 'Seq':'JKLMNO'})
def get_sequences(seq, letters, value):
sequences = [seq[seq.index(letter)-value:seq.index(letter)+value+1] for letter in letters]
return sequences
df['new_seq'] = df.apply(lambda row : get_sequences(row['MainSeq'], row['Seq'], 7), axis = 1)
df = df.explode('new_seq')
print(df)
I have the following pieces of code doing the sorting of a list by swapping pairs of elements:
# Complete the minimumSwaps function below.
def minimumSwaps(arr):
counter = 0
val_2_indx = {val: arr.index(val) for val in arr}
for indx, x in enumerate(arr):
if x != indx+1:
arr[indx] = indx+1
s_indx = val_2_indx[indx+1]
arr[s_indx] = x
val_2_indx[indx+1] = indx
val_2_indx[x] = s_indx
counter += 1
return counter
def minimumSwaps(arr):
temp = [0] * (len(arr) + 1)
for pos, val in enumerate(arr):
temp[val] = pos
swaps = 0
for i in range(len(arr)):
if arr[i] != i+1:
swaps += 1
t = arr[i]
arr[i] = i+1
arr[temp[i+1]] = t
temp[t] = temp[i+1]
temp[i+1] = i
return swaps
The second function works much faster than the first one. However, I was told that dictionary is faster than list.
What's the reason here?
A list is a data structure, and a dictionary is a data structure. It doesn't make sense to say one is "faster" than the other, any more than you can say that an apple is faster than an orange. One might grow faster, you might be able to eat the other one faster, and they might both fall to the ground at the same speed when you drop them. It's not the fruit that's faster, it's what you do with it.
If your problem is that you have a sequence of strings and you want to know the position of a given string in the sequence, then consider these options:
You can store the sequence as a list. Finding the position of a given string using the .index method requires a linear search, iterating through the list in O(n) time.
You can store a dictionary mapping strings to their positions. Finding the position of a given string requires looking it up in the dictionary, in O(1) time.
So it is faster to solve that problem using a dictionary.
But note also that in your first function, you are building the dictionary using the list's .index method - which means doing n linear searches each in O(n) time, building the dictionary in O(n^2) time because you are using a list for something lists are slow at. If you build the dictionary without doing linear searches, then it will take O(n) time instead:
val_2_indx = { val: i for i, val in enumerate(arr) }
But now consider a different problem. You have a sequence of numbers, and they happen to be the numbers from 1 to n in some order. You want to be able to look up the position of a number in the sequence:
You can store the sequence as a list. Finding the position of a given number requires linear search again, in O(n) time.
You can store them in a dictionary like before, and do lookups in O(1) time.
You can store the inverse sequence in a list, so that lst[i] holds the position of the value i in the original sequence. This works because every permutation is invertible. Now getting the position of i is a simple list access, in O(1) time.
This is a different problem, so it can take a different amount of time to solve. In this case, both the list and the dictionary allow a solution in O(1) time, but it turns out it's more efficient to use a list. Getting by key in a dictionary has a higher constant time than getting by index in a list, because getting by key in a dictionary requires computing a hash, and then probing an array to find the right index. (Getting from a list just requires accessing an array at an already-known index.)
This second problem is the one in your second function. See this part:
temp = [0] * (len(arr) + 1)
for pos, val in enumerate(arr):
temp[val] = pos
This creates a list temp, where temp[val] = pos whenever arr[pos] == val. This means the list temp is the inverse permutation of arr. Later in the code, temp is used only to get these positions by index, which is an O(1) operation and happens to be faster than looking up a key in a dictionary.
I am doing this problem a friend gave me where you are given 2 arrays say (a[1,2,3,4] and b[8,7,9,2,1]) and you have to find not common elements.
Expected output is [3,4,8,7,9]. Code below.
def disjoint(e,f):
c = e[:]
d = f[:]
for i in range(len(e)):
for j in range(len(f)):
if e[i] == f[j]:
c.remove(e[i])
d.remove(d[j])
final = c + d
print(final)
print(disjoint(a,b))
I tried with nested loops and creating copies of given arrays to modify them then add them but...
def disjoint(e,f):
c = e[:] # list copies
d = f[:]
for i in range(len(e)):
for j in range(len(f)):
if e[i] == f[j]:
c.remove(c[i]) # edited this line
d.remove(d[j])
final = c + d
print(final)
print(disjoint(a,b))
when I try removing common element from list copies, I get different output [2,4,8,7,9]. why ??
This is my first question in this website. I'll be thankful if anyone can clear my doubts.
Using sets you can do:
a = [1,2,3,4]
b = [8,7,9,2,1]
diff = (set(a) | set(b)) - (set(a) & set(b))
(set(a) | set(b)) is the union, set(a) & set(b) is the intersection and finally you do the difference between the two sets using -.
Your bug comes when you remove the elements in the lines c.remove(c[i]) and d.remove(d[j]). Indeed, the common elements are e[i]and f[j] while c and d are the lists you are updating.
To fix your bug you only need to change these lines to c.remove(e[i]) and d.remove(f[j]).
Note also that your method to delete items in both lists will not work if a list may contain duplicates.
Consider for instance the case a = [1,1,2,3,4] and b = [8,7,9,2,1].
You can simplify your code to make it works:
def disjoint(e,f):
c = e.copy() # [:] works also, but I think this is clearer
d = f.copy()
for i in e: # no need for index. just walk each items in the array
for j in f:
if i == j: # if there is a match, remove the match.
c.remove(i)
d.remove(j)
return c + d
print(disjoint([1,2,3,4],[8,7,9,2,1]))
Try it online!
There are a lot of more effecient way to achieve this. Check this stack overflow question to discover them: Get difference between two lists. My favorite way is to use set (like in #newbie's answer). What is a set? Lets check the documentation:
A set object is an unordered collection of distinct hashable objects. Common uses include membership testing, removing duplicates from a sequence, and computing mathematical operations such as intersection, union, difference, and symmetric difference. (For other containers see the built-in dict, list, and tuple classes, and the collections module.)
emphasis mine
Symmetric difference is perfect for our need!
Returns a new set with elements in either the set or the specified iterable but not both.
Ok here how to use it in your case:
def disjoint(e,f):
return list(set(e).symmetric_difference(set(f)))
print(disjoint([1,2,3,4],[8,7,9,2,1]))
Try it online!
How to find the minimum number of ways in which elements taken from a list can sum towards a given number(N)
For example if list = [1,3,7,4] and N=14 function should return 2 as 7+7=14
Again if N= 11, function should return 2 as 7+4 =11. I think I have figured out the algorithm but unable to implement it in code.
Pls use Python, as that is the only language I understand(at present)
Sorry!!!
Since you mention dynamic programming in your question, and you say that you have figured out the algorithm, i will just include an implementation of the basic tabular method written in Python without too much theory.
The idea is to have a tabular structure we will use to compute all possible values we need without having to doing the same computations many times.
The basic formula will try to sum values in the list till we reach the target value, for every target value.
It should work, but you can of course make some optimization like trying to order the list and/or find dividends in order to construct a smaller table and have faster termination.
Here is the code:
import sys
# num_list : list of numbers
# value: value for which we want to get the minimum number of addends
def min_sum(num_list, value):
list_len = len(num_list)
# We will use the tipycal dynamic programming table construct
# the key of the list will be the sum value we want,
# and the value will be the
# minimum number of items to sum
# Base case value = 0, first element of the list is zero
value_table = [0]
# Initialize all table values to MAX
# for range i use value+1 because python range doesn't include the end
# number
for i in range(1, value+1):
value_table.append(sys.maxsize);
# try every combination that is smaller than <value>
for i in range(1, value+1):
for j in range(0, list_len):
if (num_list[j] <= i):
tmp = value_table[i-num_list[j]]
if ((tmp != sys.maxsize) and (tmp + 1 < value_table[i])):
value_table[i] = tmp + 1
return value_table[value]
## TEST ##
num_list = [1,3,16,5,3]
value = 22
print("Min Sum: ",min_sum(num_list,value)) # Outputs 3
it would be helpful if you include your Algorithm in Pseudocode - it will very much look like Python :-)
Another aspect: your first operation is a multiplication with one item from the list (7) and one outside of the list (2), whereas for the second opration it is 7+4 - both values in the list.
Is there a limitation for which operation or which items to use (from within or without the list)?
I have a list of about 20-30 items [strings].
I'm able to print them out in my program just fine - but I'd like to save some space, and merge items that are shorter...
So basically, if I have 2 consecutive items that the combined lengths are less than 30, I want to join those to items as a single entry in the list - with a / between them
I'm not coming up with a simple way of doing this.
I don't care if I do it in the same list, or make a new list of items... it's all happening inside 1 function...
You need to loop through the list and keep joining items till they satisfy your requirement (size 30). Then add them to a new list when an element grows that big.
l=[] # your new list
buff=yourList[0] if len(yourList)>0 else "" # hold strings till they reach desired length
for i in range(1,len(yourList)):
# check if concatenating will exceed the size or not
t=yourList[i]
if (len(buff) + len(t) + 1) <= 30:
buff+="/"+t
else:
l.append(buff)
buff=t
l.append(buff) # since last element is yet to be inserted
You can extend method of list as follows:
a = [1,2,3]
b = [4,5,6]
a.append('/')
a.extend(b)
You just need to check the size of two list a and b as per your requirements.
I hope I understood your problem !
This code worked for me, you can check to see if that's what you wanted, it's a bit lenghty but it works.
list1 = yourListOfElements
for elem in list1:
try: # Needs try/except otherwise last iteration would throw an indexerror
listaAUX = [] # Auxiliar list to check length and join smaller elements. You can probably eliminate this using list slicing
listaAUX.append(elem)
listaAUX.append(list1[list1.index(elem)+1])
if len(listaAUX[0]) + len(listaAUX[1]) < 30:
concatenated = '/'.join(listaAUX)
print(concatenated)
else:
print(elem)
except IndexError:
print(elem)