I want to remove item from a list called mom. I have another list called cut
mom= [[0,8,1], [0, 6, 2, 7], [0, 11, 12, 3, 9], [0, 5, 4, 10]]
cut =[0, 9, 8, 2]
How do I remove what in cut from mom, except for zero?
My desire result is
mom=[[0,1],[0,6,7],[0,11,12,3],[0,5,4,10]]
>>> [[e for e in l if e not in cut or e == 0] for l in mom]
[[0, 1], [0, 6, 7], [0, 11, 12, 3], [0, 5, 4, 10]]
This is how I'd do it with List comprehension.
mom= [[0,8,1], [0, 6, 2, 7], [0, 11, 12, 3, 9], [0, 5, 4, 10]]
cut =[0, 9, 8, 2]
mom = [[x for x in subList if x not in cut or x == 0 ] for subList in mom ]
The answers provided by Ingnacio and Dom are perfect. The same can be done in a more clear and easy to understand way. Try the following:
mom= [[0,8,1], [0, 6, 2, 7], [0, 11, 12, 3, 9], [0, 5, 4, 10]]
cut =[0, 9, 8, 2]
for e in mom:
for f in e:
if f in cut and f != 0:
e.remove(f) #used the remove() function of list
print(mom)
Much easier for a novice in Python. Isn't it?
Given the cut=[0,9,8,2] and
mom = [[0,8,1], [0, 6, 2, 7], [0, 11, 12, 3, 9], [0, 5, 4, 10]]
Assuming 0 element is removed from cut list
cut=[9,8,2]
result =[]
for e in mom:
result.append(list(set(e)-set(cut)))
o/p
result
[[0, 1], [0, 6, 7], [0, 11, 3, 12], [0, 10, 4, 5]]
Related
I'm trying to split a list into groups based on index pairs from another list, given:
>>> l = list(range(10))
>>> l
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> idx = [0, 5]
I need to break up the list resulting in:
>>> l[0:5]
[0, 1, 2, 3, 4]
>>> l[5:]
[5, 6, 7, 8, 9]
The list idx will at a minimum always be [0], but may be of size n; values inside idx will always be sorted ascending.
Currently I have:
>>> l = list(range(10))
>>> idx = [0, 5]
>>> idx.append(None)
>>> [l[idx[i]:idx[i + 1]] for i in range(len(idx) - 1)]
[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9]]
Is there a way to accomplish this without explicitly appending Non and iterating over a range?
Edit: for another example...
Given:
>>> l = list(range(14))
>>> l
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
>>> idx = [0, 5, 10]
Desired result:
[[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13]]
You could try about itertools.zip_longest:
from itertools import zip_longest
l = list(range(14))
idx = [0, 5, 10]
print([l[pre: next] for pre, next in zip_longest(idx,idx[1:])])
Result:
[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13]]
With numpy you can use numpy.split()
import numpy as np
res =[list(x) for x in np.split(l, idx) if x.size != 0]
print(res)
Output:
[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13]]
result = [l[curr_idx:curr_idx+idx[1]] for curr_idx in idx]
result
[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13]]
I am trying to solve this puzzle question:
There is only one five-digit number n, such that every one of the following ten numbers share exactly one digit in common in the same position as n. Find n.
I'm trying to get like this:
https://i.stack.imgur.com/hufIc.jpg
For each column, selects every element and compares it along the column if it matches the same number. And then, appends it to get which number were matched on every column.
should i add if's in here:
L = [[0, 1, 2, 6, 5],
[1, 2, 1, 7, 1],
[2, 3, 2, 5, 7],
[3, 4, 5, 4, 8],
[4, 5, 9, 7, 0],
[5, 6, 2, 3, 6],
[6, 7, 3, 2, 4],
[7, 8, 0, 8, 4],
[8, 9, 8, 7, 2],
[9, 9, 4, 1, 4]]
for c in range(10):
for r in range(5):
print(L[c][r], end=' ')
print()
The easiest way I can think of is to get all the values of a column in a variable and then check if any value occurs more than once. The solution would look something like:
L = [
[0, 1, 2, 6, 5],
[1, 2, 1, 7, 1],
[2, 3, 2, 5, 7],
[3, 4, 5, 4, 8],
[4, 5, 9, 7, 0],
[5, 6, 2, 3, 6],
[6, 7, 3, 2, 4],
[7, 8, 0, 8, 4],
[8, 9, 8, 7, 2],
[9, 9, 4, 1, 4]
]
for r in range(5):
val = [L[c][r] for c in range(10)]
for i in range(10):
if val[i] in val[:i]:
print(val[i], end=" ")
break
else:
print(0, end=" ")
print()
L = [
[0, 1, 2, 6, 5],
[1, 2, 1, 7, 1],
[2, 3, 2, 5, 7],
[3, 4, 5, 4, 8],
[4, 5, 9, 7, 0],
[5, 6, 2, 3, 6],
[6, 7, 3, 2, 4],
[7, 8, 0, 8, 4],
[8, 9, 8, 7, 2],
[9, 9, 4, 1, 4]
]
for col_index in range(5):
col= [row[col_index] for row in L]
duplicate = [x for x in col if col.count(x) > 1]
if len(duplicate) == 0:
print(0, end=" ")
else:
print(duplicate[0], end=" ")
print()
Is there an elegant way how to pad the last sublist with zeroes while creating sublists from a list of integers?
So far I have this oneliner and need to fill the last sublist with 2 zeroes
[lst[x:x+3] for x in range(0, len(lst), 3)]
for example
lst =[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
result should be:
[1,2,3][4,5,6][7,8,9][10,0,0]
With itertools.zip_longest, consuming the same iterator created off of the list, and fill in the missing values as 0 :
[[*i] for i in itertools.zip_longest(*[iter(lst)] * 3, fillvalue=0)]
Example:
In [1219]: lst =[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
In [1220]: [[*i] for i in itertools.zip_longest(*[iter(lst)] * 3, fillvalue=0)]
Out[1220]: [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 0, 0]]
Without itertools:
lst = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
print([lst[x:x+3]+[0]*(x-len(lst)+3) for x in range(0, len(lst), 3)])
Prints:
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 0, 0]]
I am working on the following problem:
This function returns a list of all possible sublists in L of length n without skipping elements in L. The sublists in the returned list should be ordered in the way they appear in L, with those sublists starting from a smaller index being at the front of the list.
Example 1, if L = [10, 4, 6, 8, 3, 4, 5, 7, 7, 2] and n = 4 then your function should return the list [[10, 4, 6, 8], [4, 6, 8, 3], [6, 8, 3, 4], [8, 3, 4, 5], [3, 4, 5, 7], [4, 5, 7, 7], [5, 7, 7, 2]]
My solution works but how can I make it shorter? What is a better way to do this?
def getSublists(L, n):
newN = n
myList = []
for i in range(len(L)):
orginalLen = L[i:n]
if(len(orginalLen) == n):
myList.append(L[i:n])
n = n + 1
else:
myList.append(L[i:n])
n = n + 1
if(newN == 1):
print(myList)
else:
print(myList[:len(myList)-(n-1)])
getSublists([10, 4, 6, 8, 3, 4, 5, 7, 7, 2],4)
getSublists([1], 1)
getSublists([0, 0, 0, 0, 0], 2)
OUTPUT
[[10, 4, 6, 8], [4, 6, 8, 3], [6, 8, 3, 4], [8, 3, 4, 5], [3, 4, 5, 7], [4, 5, 7, 7], [5, 7, 7, 2]]
[[1]]
[[0, 0], [0, 0], [0, 0], [0, 0]]
l = [1,2,3,4,5,6,87,9]
n = ..
print [l[i:i+n] for i in range(len(l)-n+1)]
maybe you need.
In one line:
get_sublists = lambda ls, n: [ls[x:x+n] for x in range(len(ls)-n+1)]
get_sublists([10, 4, 6, 8, 3, 4, 5, 7, 7, 2], 4)
[[10, 4, 6, 8], [4, 6, 8, 3], [6, 8, 3, 4], [8, 3, 4, 5], [3, 4, 5, 7], [4, 5, 7, 7], [5, 7, 7, 2]]
def get_sublists(L, n):
return [ L[i:i+n] for i in range(len(L)-n) ]
I completed the program a little better understanding of the reader.
def getSublists(L, n):
new_list = []
for i in range(len(L)-n+1):
a = L[i:i+n]
new_list.append(a)
return new_list
answer:
[[10, 4, 6, 8],
[4, 6, 8, 3],
[6, 8, 3, 4],
[8, 3, 4, 5],
[3, 4, 5, 7],
[4, 5, 7, 7],
[5, 7, 7, 2]]
This is pretty readable I think, to understand the concept. The idea here is to iterate through the numbers from 0 to the length of L, minus 4. And just take the sublist of L from your current index i, to i+4. Iterating to length-4 ensures you don't try to access an index out of bounds!
>>> for i in range(len(L)-4+1):
print L[i:i+4]
[10, 4, 6, 8]
[4, 6, 8, 3]
[6, 8, 3, 4]
[8, 3, 4, 5]
[3, 4, 5, 7]
[4, 5, 7, 7]
[5, 7, 7, 2]
I have a lists of lists in variable lists something like this:
[7, 6, 1, 8, 3]
[1, 7, 2, 4, 2]
[5, 6, 4, 2, 3]
[0, 3, 3, 1, 6]
[3, 5, 2, 14, 3]
[3, 11, 9, 1, 1]
[1, 10, 2, 3, 1]
When I write lists[1] I get vertically:
6
7
6
3
5
11
10
but when I loop it:
for i in list:
print(i)
I get this horizontally.
7
6
1
8
3
etc...
So, how it works? How can I modify loop to go and give me all vertically?
Short answer:
for l in lists:
print l[1]
Lists of lists
list_of_lists = [ [1, 2, 3], [4, 5, 6], [7, 8, 9]]
for list in list_of_lists:
for x in list:
print x
Here is how you would print out the list of lists columns.
lists = [[7, 6, 1, 8, 3],
[1, 7, 2, 4, 2],
[5, 6, 4, 2, 3],
[0, 3, 3, 1, 6],
[3, 5, 2, 14, 3],
[3, 11, 9, 1, 1],
[1, 10, 2, 3, 1]]
for i in range(0, len(lists[1])):
for j in range(0, len(lists)):
print lists[j][i],
print "\n"