I want to form an array within an array. So that if I call Array_1[0] will give me the first value and Array_1[1] will give me the second value. In this instance, the first value will be 1 and the second value will be 2. Similar to tuples within a list. I have written the following code but I end up just with just one array containing all elements.
import numpy as np
list_A = [1,3,5]
list_B = [2,4,6]
Array_1 = np.array([])
for i,j in zip(list_A,list_B):
Array_1 = np.append(arr = Array_1, values = [i,j])
print(Array_2.astype(int))
#output: [1,2,3,4,5,6]
#Desired output: [[1,2],[3,4],[5,6]]
Any ideas if the desired output can be achieved through a numpy array. (btw not using a list and tuples)
Related
I am looking around a way to get the subset from an integer array based on certain range
For example
Input
array1=[3,5,4,12,34,54]
#Now getting subset for every 3 element
Output
subset= [(3,5,4), (12,34,54)]
I know it could be simple, but didn't find the right way to get this output
Appreciated for the help
Thanks
Consider using a list comprehension:
>>> array1 = [3, 5, 4, 12, 34, 54]
>>> subset = [tuple(array1[i:i+3]) for i in range(0, len(array1), 3)]
>>> subset
[(3, 5, 4), (12, 34, 54)]
Links to other relevant documentation:
tuples
ranges
arr = [1,2,3,4,5,6]
sets = [tuple(arr[i:i+3]) for i in range(0, len(arr), 3)]
print(sets)
We are taking a range of values from the array that we make into a tuple. The range is determined by the for loop which iterates at a step of three so that a tuple only is create after every 3 items.
you can use code:
from itertools import zip_longest
input_list = [3,5,4,12,34,54]
iterables = [iter(input_list)] * 3
slices = zip_longest(*iterables, fillvalue=None)
output_list =[]
for slice in slices:
my_list = [slice]
# print(my_list)
output_list = output_list + my_list
print(output_list)
You could use the zip_longest function from itertools
https://docs.python.org/3.0/library/itertools.html#itertools.zip_longest
I have a list like this:
a = [[4.0, 4, 4.0], [3.0, 3, 3.6], [3.5, 6, 4.8]]
I want an outcome like this (EVERY first element in the list):
4.0, 3.0, 3.5
I tried a[::1][0], but it doesn't work
You can get the index [0] from each element in a list comprehension
>>> [i[0] for i in a]
[4.0, 3.0, 3.5]
Use zip:
columns = zip(*rows) #transpose rows to columns
print columns[0] #print the first column
#you can also do more with the columns
print columns[1] # or print the second column
columns.append([7,7,7]) #add a new column to the end
backToRows = zip(*columns) # now we are back to rows with a new column
print backToRows
You can also use numpy:
a = numpy.array(a)
print a[:,0]
Edit:
zip object is not subscriptable. It need to be converted to list to access as list:
column = list(zip(*row))
You could use this:
a = ((4.0, 4, 4.0), (3.0, 3, 3.6), (3.5, 6, 4.8))
a = np.array(a)
a[:,0]
returns >>> array([4. , 3. , 3.5])
You can get it like
[ x[0] for x in a]
which will return a list of the first element of each list in a
Compared the 3 methods
2D list: 5.323603868484497 seconds
Numpy library : 0.3201274871826172 seconds
Zip (Thanks to Joran Beasley) : 0.12395167350769043 seconds
D2_list=[list(range(100))]*100
t1=time.time()
for i in range(10**5):
for j in range(10):
b=[k[j] for k in D2_list]
D2_list_time=time.time()-t1
array=np.array(D2_list)
t1=time.time()
for i in range(10**5):
for j in range(10):
b=array[:,j]
Numpy_time=time.time()-t1
D2_trans = list(zip(*D2_list))
t1=time.time()
for i in range(10**5):
for j in range(10):
b=D2_trans[j]
Zip_time=time.time()-t1
print ('2D List:',D2_list_time)
print ('Numpy:',Numpy_time)
print ('Zip:',Zip_time)
The Zip method works best.
It was quite useful when I had to do some column wise processes for mapreduce jobs in the cluster servers where numpy was not installed.
If you have access to numpy,
import numpy as np
a_transposed = a.T
# Get first row
print(a_transposed[0])
The benefit of this method is that if you want the "second" element in a 2d list, all you have to do now is a_transposed[1]. The a_transposed object is already computed, so you do not need to recalculate.
Description
Finding the first element in a 2-D list can be rephrased as find the first column in the 2d list. Because your data structure is a list of rows, an easy way of sampling the value at the first index in every row is just by transposing the matrix and sampling the first list.
Try using
for i in a :
print(i[0])
i represents individual row in a.So,i[0] represnts the 1st element of each row.
I am doing a K-means project and I have to do it by hand, which is why I am trying to figure out what is the best ways to group things according to their last values into a list or a dictionary. Here is what I am talking about
list_of_tuples = [(honey,1),(bee,2),(tree,5),(flower,2),(computer,5),(key,1)]
Now my ultimate goal is to be able to sort out the list and have 3 different lists each with its respected element
"""This is the goal"""
list_1 = [honey,key]
list_2 = [bee,flower]
list_3 = [tree, computer]
I can use a lot of if statements and a for loop, but is there a more efficient way to do it?
If you're not opposed to using something like pandas, you could do something along these lines:
import pandas as pd
list_1, list_2, list_3 = pd.DataFrame(list_of_tuples).groupby(1)[0].apply(list).values
Result:
In [19]: list_1
Out[19]: ['honey', 'key']
In [20]: list_2
Out[20]: ['bee', 'flower']
In [21]: list_3
Out[21]: ['tree', 'computer']
Explanation:
pd.DataFrame(list_of_tuples).groupby(1) groups your list of tuples by the value at index 1, then you extract the values as lists of index 0 with [0].apply(list).values. This gives you an array of lists as below:
array([list(['honey', 'key']), list(['bee', 'flower']),
list(['tree', 'computer'])], dtype=object)
Something to the effect can be achieved with a dictionary and a for loop, using the second element of the tuple as a key value.
list_of_tuples = [("honey",1),("bee",2),("tree",5),("flower",2),("computer",5),("key",1)]
dict_list = {}
for t in list_of_tuples:
# create key and a single element list if key doesn't exist yet
# append to existing list otherwise
if t[1] not in dict_list.keys():
dict_list[t[1]] = [t[0]]
else:
dict_list[t[1]].append( t[0] )
list_1, list_2, list_3 = dict_list.values()
Given a matrix mat and an array arr, for each row of the matrix if elements of Column 1 are equal to the corresponding element of the array, then print the corresponding value of Column 2 of the matrix.
mat = np.array([['abc','A'],['def','B'],['ghi','C'],['jkl','D']])
arr = np.array(['abc','dfe','ghi','kjl'])
This can be solved via numpy.where.
Extract the first row of the matrix using mat[:,0], and compare it to arr using np.where(mat[:,0] == arr) to extract the indexes.
and use those indexes to get the elements you want from mat
In [1]: import numpy as np
...:
...: mat = np.array([['abc','A'],['def','B'],['ghi','C'],['jkl','D']])
...:
...: arr = np.array(['abc','dfe','ghi','kjl'])
In [2]: print(mat[np.where(mat[:,0] == arr)])
[['abc' 'A']
['ghi' 'C']]
Output should be `['A', 'C']``
So above code can be modified as
print(mat[np.where(mat[:,0]=arr)][:,1]
# output ['A' 'C']
I need to print only duplicate numbers in a list and need to multiply by count. the code is as follows , the output should be ,
{1:3, 2:2, 3:2} need to multiply each numbers by count and print as separate answers:
answer1 = 1*3, answer2 = 2*2 , answer3 = 3*2
Current attempt:
from collections import Counter
alist = [1,2,3,5,1,2,1,3,1,2]
a = dict(Counter(a_list))
print(a)
Counter already does the heavy lifting. So for the rest, what about generating a list of the values occuring more than once, formatting the output as you wish ? (sorting the keys seems necessary so indexes match the keys order):
from collections import Counter
a_list = [1,2,3,5,1,2,1,3,1,2]
a = ["{}*{}".format(k,v) for k,v in sorted(Counter(a_list).items()) if v > 1]
print(a)
result:
['1*4', '2*3', '3*2']
If you want the numerical result instead:
a = [k*v for k,v in sorted(Counter(a_list).items()) if v > 1]
result (probably more useful):
[4, 6, 6]
Assigning to separate variables (answer1,answer2,answer3 = a) is not a very good idea. Keep a indexed list