below are 2 lst1 and lst2 and expected output is in output as below.
lst1 = ['q','r','s','t','u','v','w','x','y','z']
lst2 =['1','2','3']
Output expected
[['q','1'], ['r','2'], ['s','3'], ['t','1'],['u','2'],['v','3'],['w','1'],['x','2'],['y','3'],
['z','1']]"
This is a very simple approach to this problem.
lst1 = ['q','r','s','t','u','v','w','x','y','z']
lst2 = ['1','2','3']
new_list = []
for x in range(len(lst1)):
new_list.append([lst1[x], lst2[x % 3]])
print(new_list) # [['q', '1'], ['r', '2'], ['s', '3'], ['t', '1'], ['u', '2'], ['v', '3'], ['w', '1'], ['x', '2'], ['y', '3'], ['z', '1']]
You could also use list comprehension in this case, like so:-
new_list = [[lst1[x], lst2[x % 3]] for x in range(len(lst1))]
You can use zip() and itertools.cycle().
from itertools import cycle
lst1 = ['q','r','s','t','u','v','w','x','y','z']
lst2 =['1','2','3']
result = [[letter, number] for letter, number in zip(lst1, cycle(lst2))]
print(result)
Expected output:
[['q', '1'], ['r', '2'], ['s', '3'], ['t', '1'], ['u', '2'], ['v', '3'], ['w', '1'], ['x', '2'], ['y', '3'], ['z', '1']]
Another solution would be to additonally use map().
result = list(map(list, zip(lst1, cycle(lst2))))
In case you wanna use tuples you could just do
from itertools import cycle
lst1 = ['q','r','s','t','u','v','w','x','y','z']
lst2 =['1','2','3']
result = list(zip(lst1, cycle(lst2)))
print(result)
which would give you
[('q', '1'), ('r', '2'), ('s', '3'), ('t', '1'), ('u', '2'), ('v', '3'), ('w', '1'), ('x', '2'), ('y', '3'), ('z', '1')]
Related
I have these lists in Python
text_0 = ['Weight','Weight', 'Weight']
text_1 = [['x','y','z'],['x','y','z'],['x','y','z']]
text_2 = [['1','2,','3'],['4','5','6'],['7','8','9']]
I would like would to create a new list like this
new_list = ['Weight','x','1','y','2','z','3','Weight','x','4','y','5','z','6','Weight','x','7','y','8','z','9']
How do I do this in Python?
In your case we can do with python
sum([[x] + [ items for item in zip(y,z) for items in item]
for x, y, z in zip(text_0,text_1,text_2)],[])
['Weight', 'x', '1', 'y', '2', 'z', '3', 'Weight', 'x', '4', 'y', '5', 'z', '6', 'Weight', 'x', '7', 'y', '8', 'z', '9']
I have 3 lists:
color = ['red','orange','purple','black']
number = ['2','4','6','77']
letter = ['K','R','U','Z']
what I want to end up with is:
mylist=[['K','R','U','Z','red','2'],['K','R','U','Z','orange','4'],['K','R','U','Z','purple','6'], ['K','R','U','Z','black','77']]
I tried:
for i in range(4):
letter.append(color[i])
letter.append(number[i])
This does not give me what I need.
You can use a list comprehension for that :
color = ['red','orange','purple','black']
number = ['2','4','6','77']
letter = ['K','R','U','Z']
mylist = [ letter + [c, n] for c, n in zip(color, number) ]
print(mylist)
Not so beautiful but pretty simple in my opinion:
color = ['red', 'orange', 'purple', 'black']
number = ['2', '4', '6', '77']
letter = ['K', 'R', 'U', 'Z']
my_list = []
for i in range(len(color)):
new_list = letter[:]
my_list.append(new_list)
new_list.append(color[i])
new_list.append(number[i])
print(my_list)
Output:
[['K', 'R', 'U', 'Z', 'red', '2'], ['K', 'R', 'U', 'Z', 'orange', '4'], ['K', 'R', 'U', 'Z', 'purple', '6'], ['K', 'R', 'U', 'Z', 'black', '77']]
I have 2 matrices:
list_alpha = [['a'],
['b'],
['c'],
['d'],
['e']]
list_beta = [['1', 'a', 'e', 'b'],
['2', 'd', 'X', 'X'],
['3', 'a', 'X', 'X'],
['4', 'd', 'a', 'c'],
And my goal is if a letter from list_alpha is in a sublist of list_beta, then the first element of that line in list_beta (the #) is added to the correct line in list_alpha.
So my output would be:
final_list = [['a', '1', '3', '4'],
['b', '1'],
['c', '4'],
['d', '2', '4'],
['e', '1']]
But I'm pretty new to python and coding in general and I'm not sure how to do this. Is there a way to code this? Or do I have to change the way the data is stored in either list?
Edit:
Changing list_alpha to a dictionary helped!
Final code:
dict_alpha = {'a': [], 'b': [], 'c': [], 'd': [], 'e':[]}
list_beta = [['1', 'a', 'e', 'b'],
['2', 'd', 'X', 'X'],
['3', 'a', 'X', 'X'],
['4', 'd', 'a', 'c'],
['5', 'X', 'X', 'e'],
['6', 'c', 'X', 'X']]
for letter in dict_alpha:
for item in list_beta:
if letter in item:
dict_alpha.get(letter).append(item[0])
print(dict_alpha)
You can use dict_alpha as same as list_alpha , then fix your for loop.
For example:
dict_alpha = [['a'],
['b'],
['c'],
['d'],
['e']]
list_beta = [['1', 'a', 'e', 'b'],
['2', 'd', 'X', 'X'],
['3', 'a', 'X', 'X'],
['4', 'd', 'a', 'c'],
['5', 'X', 'X', 'e'],
['6', 'c', 'X', 'X']]
for al in dict_alpha:
for bt in list_beta:
for i in range(1, len(bt)):
if (bt[i] == al[0]):
al.append(bt[0])
print(dict_alpha)
Output:
[['a', '1', '3', '4'],
['b', '1'],
['c', '4', '6'],
['d', '2', '4'],
['e', '1', '5']]
Hope to helpful!
array([
['192', '895'],
['14', '269'],
['1', '23'],
['1', '23'],
['50', '322'],
['19', '121'],
['17', '112'],
['12', '72'],
['2', '17'],
['5,250', '36,410'],
['2,546', '17,610'],
['882', '6,085'],
['571', '3,659'],
['500', '3,818'],
['458', '3,103'],
['151', '1,150'],
['45', '319'],
['44', '335'],
['30', '184']
])
How can I remove some of the rows and left the array like:
Table3=array([
['192', '895'],
['14', '269'],
['1', '23'],
['50', '322'],
['17', '112'],
['12', '72'],
['2', '17'],
['5,250', '36,410'],
['882', '6,085'],
['571', '3,659'],
['500', '3,818'],
['458', '3,103'],
['45', '319'],
['44', '335'],
['30', '184']
])
I removed the index 2,4,6. I am not sure how should I do it. I have tried few ways, but still can't work.
It seems like you actually deleted indices 2, 5, and 10 (not 2, 4 and 6). To do this you can use np.delete, pass it a list of the indices you want to delete, and apply it along axis=0:
Table3 = np.delete(arr, [[2,5,10]], axis=0)
>>> Table3
array([['192', '895'],
['14', '269'],
['1', '23'],
['50', '322'],
['17', '112'],
['12', '72'],
['2', '17'],
['5,250', '36,410'],
['882', '6,085'],
['571', '3,659'],
['500', '3,818'],
['458', '3,103'],
['151', '1,150'],
['45', '319'],
['44', '335'],
['30', '184']],
dtype='<U6')
How do I convert this list of lists:
[['0', '1'], ['0', '2'], ['0', '3'], ['1', '4'], ['1', '6'], ['1', '7'], ['1', '9'], ['2', '3'], ['2', '6'], ['2', '8'], ['2', '9']]
To this list of tuples:
[(0, [1, 2, 3]), (1, [0, 4, 6, 7, 9]), (2, [0, 3, 6, 8, 9])]
I am unsure how to implement this next step? (I can't use dictionaries,
sets, deque, bisect module. You can though, and in fact should, use .sort or sorted functions.)
Here is my attempt:
network= [['10'], ['0 1'], ['0 2'], ['0 3'], ['1 4'], ['1 6'], ['1 7'], ['1 9'], ['2 3'], ['2 6'], ['2 8'], ['2 9']]
network.remove(network[0])
friends=[]
for i in range(len(network)):
element= (network[i][0]).split(' ')
friends.append(element)
t=len(friends)
s= len(friends[0])
lst=[]
for i in range(t):
a= (friends[i][0])
if a not in lst:
lst.append(int(a))
for i in range(t):
if a == friends[i][0]:
b=(friends[i][1])
lst.append([b])
print(tuple(lst))
It outputs:
(0, ['1'], ['2'], ['3'], 0, ['1'], ['2'], ['3'], 0, ['1'], ['2'], ['3'], 1, ['4'], ['6'], ['7'], ['9'], 1, ['4'], ['6'], ['7'], ['9'], 1, ['4'], ['6'], ['7'], ['9'], 1, ['4'], ['6'], ['7'], ['9'], 2, ['3'], ['6'], ['8'], ['9'], 2, ['3'], ['6'], ['8'], ['9'], 2, ['3'], ['6'], ['8'], ['9'], 2, ['3'], ['6'], ['8'], ['9'])
I am very close it seems, not sure what to do??
A simpler method:
l = [['0', '1'], ['0', '2'], ['0', '3'], ['1', '4'], ['1', '6'], ['1', '7'], ['1', '9'], ['2', '3'], ['2', '6'], ['2', '8'], ['2', '9']]
a=set(i[0] for i in l)
b=list( (i,[]) for i in a)
[b[int(i[0])][1].append(i[1]) for i in l]
print(b)
Output:
[('0', ['1', '2', '3']), ('1', ['4', '6', '7', '9']), ('2', ['3', '6', '8', '9'])]
Alternate Answer (without using set)
l = [['0', '1'], ['0', '2'], ['0', '3'], ['1', '4'], ['1', '6'], ['1', '7'], ['1', '9'], ['2', '3'], ['2', '6'], ['2', '8'], ['2', '9']]
a=[]
for i in l:
if i[0] not in a:
a.append(i[0])
b=list( (i,[]) for i in a)
[b[int(i[0])][1].append(i[1]) for i in l]
print(b)
also outputs
[('0', ['1', '2', '3']), ('1', ['4', '6', '7', '9']), ('2', ['3', '6', '8', '9'])]
You can use Pandas:
import pandas as pd
import numpy as np
l = [['0', '1'], ['0', '2'], ['0', '3'], ['1', '4'], ['1', '6'], ['1', '7'], ['1', '9'], ['2', '3'], ['2', '6'], ['2', '8'], ['2', '9']]
df = pd.DataFrame(l, dtype=np.int)
s = df.groupby(0)[1].apply(list)
list(zip(s.index, s))
Output:
[(0, [1, 2, 3]), (1, [4, 6, 7, 9]), (2, [3, 6, 8, 9])]