If I have the following:
import numpy as np
a = np.array([[0, 1],
[1, 3],
[4, 4]])
And want to update the column value if a column condition is met. For example if the 2nd column value is greater than 2, then replace only that column value with 9.
a = [[0, 1],
[1, 9],
[4, 9]]
I would have thought this would work, but it updates all the values in that row.
a[a[:,1] > 2] = 9
But it replaces all the values in the row.
a =[[0, 1],
[9, 9],
[9, 9]]
I'm guessing I'm missing some understanding of how the boolean indexing is being created here.
You need:
import numpy as np
a = np.array([[0, 1],
[1, 3],
[4, 4]])
a[:,1]= np.where(a[:,1]>2, 9, a[:,1])
print(a)
Output:
array([[0, 1],
[1, 9],
[4, 9]])
why your code not working
try printing out print(a[a[:,1] > 2])
it will give output as:
[[1 3]
[4 4]]
It will check for 2nd index if it is greater than 2 it will return an entire row.
Related
Hi i want to classify indexes of same rows in 2D numpy array. Is there any function to do it ?
Something like this :
a= [[1,2,3] , [2,3,4] , [5,6,7] ,[1,2,3] ,[1,2,3] , [2,3,4]]
then f(a) returns same row indexes [[0,3,4],[1,5],[2]]
I would appreciate for your solutions
Here's one to output list of arrays of row indices -
def classify_rows(a):
sidx = np.lexsort(a.T)
b = a[sidx]
m = ~(b[1:] == b[:-1]).all(1)
return np.split(sidx, np.flatnonzero(m)+1)
If you need a list of lists as output -
def classify_rows_list(a):
sidx = np.lexsort(a.T)
b = a[sidx]
m = np.concatenate(( [True], ~(b[1:] == b[:-1]).all(1), [True]))
l = sidx.tolist()
idx = np.flatnonzero(m)
return [l[i:j] for i,j in zip(idx[:-1],idx[1:])]
Sample run -
In [78]: a
Out[78]:
array([[1, 2, 3],
[2, 3, 4],
[5, 6, 7],
[1, 2, 3],
[1, 2, 3],
[2, 3, 4]])
In [79]: classify_rows(a)
Out[79]: [array([0, 3, 4]), array([1, 5]), array([2])]
In [80]: classify_rows_list(a)
Out[80]: [[0, 3, 4], [1, 5], [2]]
I have two lists of lists as follows. To merge them, I usually do the following:
>>>from itertools import imap, ilist
>>>a = [1,2,3]
>>>b = [4,5,6]
>>> c = list(imap(list,izip(a,b)))
>>> c
[[1, 4]], [2, 5], [3, 6]]
However, now I have a list of list as follows:
[[1,2,3],
[4,5,6],
[7,8,9],
]
How do I iterate through each list and pass it to the izip function to obtain the following output:
[[1,4,7],[2,5,8],[3,6,9]]
Answer for edited Question:
>>> input_list=[[1,2,3],
[4,5,6],
[7,8,9],
]
Using map and zip:
>>> map(list,zip(*input_list))
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
Using imap and izip:
>>> list(imap(list,list(izip(*input_list))))
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
Answer for previous question:
By using list comprehension and two for loops:
input_list =[[[1],[2],[3]],
[[4],[5],[6]],
[[7],[8],[9]],
]
out_list = [[] for i in range(len(input_list))]
for each_row in input_list:
for i in range(len(each_row)):
out_list[i].extend(each_row[i])
print out_list
Output:
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
I think u need something like this:
input_list =[[1,2,3],
[4,5,6],
[7,8,9],
]
result = []
for i in range(len(input_list)):
temp = []
for list in input_list:
temp.append(list[i])
result.append(temp)
print result
result will be:
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
I want my program to take numbers from 1 to X and randomly distribute those numbers between X/2 numbers of lists Y times. I don't want a number to be repeated during one shuffle, also I don't want the lists to repeat at all. So if there is list [1,2] there shouldn't be another list containing 1 or 2 in same shuffle and there shouldn't be another [1,2] or [2,1] in the whole result.
This is what I came up with, however, it keeps on repeating numbers. Any adice?
import random
def Shuffler():
amount = int(raw_input("Numbers up to: "))
times = int(raw_input("Number of shuffles: "))
numberslist = range(1,amount+1)
twos = []
thisshuffle = []
final = []
while len(final) < (amount/2)*times:
twos = []
thisshuffle = []
while len(twos) < 2:
if len(numberslist)!=0:
randomnumber = random.choice(numberslist)
if (randomnumber in twos) or (randomnumber in thisshuffle):
numberslist.remove(randomnumber)
else:
twos.append(randomnumber)
thisshuffle.append(randomnumber)
numberslist.remove(randomnumber)
else:
numberslist = range(1,amount+1)
if (twos or list(reversed(twos))) not in final:
final.append(twos)
k=0
for i in range(times): #this shit prints shit
print "%s:" % (i+1), final[k:k+amount/2]
print
k = k + amount/2
Shuffler()
Shuffler()
As ccf pointed out, your requirements aren't trivial. A few more steps and you'd have a sudoku generator :)
I tried a few solutions but they either didn't produce random output or were rather inefficient. Ccf's solution is clearly well written but it seems to have the same problem; it produces ordered output (e.g. [1, 2], [1, 3], [1, 4], [1, 5], [1, 6]).
#cff - wouldn't it be better to use itertools.combinations rather than itertools.permutations to avoid generating repetitions?
Here's a "solution" that's quite similar to ccf's (doesn't produce random output either):
import itertools
def Shuffler():
amount = int(raw_input("Numbers up to: "))
times = int(raw_input("Number of shuffles: "))
rng = range(1, amount+1)
perms = list(itertools.combinations(rng, 2))
lst_single = []
lst_all = []
for p in perms:
if len(lst_all) >= times:
for i, lst in enumerate(lst_all):
print str(i+1) + ": ", lst
break
if len(lst_single) == amount/2:
lst_all.append(lst_single)
lst_single = []
elif p[0] < p[1]:
p = list(p)
lst_single.append(p)
Shuffler()
Output
Numbers up to: 6
Number of shuffles: 3
1: [[1, 2], [1, 3], [1, 4]]
2: [[1, 6], [2, 3], [2, 4]]
3: [[2, 6], [3, 4], [3, 5]]
And here's a slightly hackish solution that seems to produce the output you want but in an inefficient way. It relies on a set for filtering out unwanted combinations but still wastes resources producing them in the first place.
import random
def Shuffler():
amount = int(raw_input("Numbers up to: "))
times = int(raw_input("Number of shuffles: "))
rng = range(1, amount+1)
final = []
lst_len = amount/2
combos_unique = set()
while len(combos_unique) < lst_len*times:
combo_rand = random.sample(rng, 2)
if combo_rand[0] < combo_rand[1]:
combos_unique.add(tuple(combo_rand))
tmp = []
for combo in combos_unique:
tmp.append(list(combo))
if len(tmp) >= lst_len:
final.append(tmp)
tmp = []
for i, lst in enumerate(final):
print str(i+1) + ": ", lst
Shuffler()
Output
Numbers up to: 6
Number of shuffles: 3
1: [[2, 6], [4, 6], [5, 6]]
2: [[4, 5], [1, 3], [1, 6]]
3: [[3, 4], [2, 4], [3, 5]]
You don't want any repeated numbers in one shuffle, and any repeated list ... and so on. This is not an easy task. Plus another fact is that, the unique sets of numbers are fixed, which cannot be set too high. For example, if you set "Numbers up to:" 5, and "Number of shuffles: " 20, for sure you will get repeated numbers.
The issue with your code, I see, is in this if statement:
if (twos or list(reversed(twos))) not in final:
final.append(twos)
(twos or list(reversed(twos))) is logical OR, the result is twos, because twos is not empty. I suggest you change the if statement to:
if (twos not in final) and (list(reversed(twos)) not in final):
final.append(twos)
The following code (python 2.7x) uses permutations and shuffle to generate numbers list. Next, make the list unique (e.g., no [1,2] and [2,1] in same list). then, divide them into groups based on number of shuffles specified by the user. Press any letter, script will exit. Hope it helps:
from itertools import permutations
from random import shuffle
def Shuffler():
try:
amount = input("Numbers up to: ")
p = list(permutations(range(1, amount + 1), 2))
p_uniq = [list(x) for x in p if x[::-1] in p and x[0]<=x[1]]
shuf_max = len(p_uniq) /(amount / 2)
times = shuf_max + 1 # set a higher value to trigger prompt
while times > shuf_max:
shuffle(p_uniq) # shuffle the unique list in place
times = input("Number of shuffles (MAX %s): " % (shuf_max))
else:
for i, group in enumerate(list(zip(*[iter(p_uniq[: (amount /2) * times + 1])]* (amount/2)))):
print "%i: " % (i + 1), list(group)
Shuffler()
except:
print 'quitting...'
Shuffler()
Output：
Numbers up to: 10
Number of shuffles (MAX 9): 8
1: [[6, 7], [1, 9], [2, 5], [5, 9], [9, 10]]
2: [[1, 10], [3, 8], [4, 10], [8, 10], [1, 5]]
3: [[1, 4], [6, 8], [3, 6], [2, 4], [4, 7]]
4: [[2, 10], [5, 8], [3, 9], [1, 7], [4, 9]]
5: [[1, 2], [7, 9], [1, 3], [6, 9], [1, 6]]
6: [[2, 9], [4, 8], [3, 5], [8, 9], [7, 10]]
7: [[2, 7], [2, 3], [7, 8], [3, 7], [3, 10]]
8: [[3, 4], [2, 6], [5, 6], [5, 7], [4, 6]]
Numbers up to:
Define a function createSquareMat_sumpos(...) which receives one argument, dim, an integer number greater than 0, and it returns a list of lists which represents a matrix of dimension dim x dim, so that each element in the matrix has as content the sum of the column + the row position.
As an example, the following code fragment:
print (createSquareMat_sumpos(2))
should produce the output:
[[0, 1], [1, 2]]
def createSquareMat_sumpos(dim):
list=[[]]*dim
for i in range(len(list)):
list[i]=[i,(i+1)] #### not sure what to write here ####
return list
this works for
print (createSquareMat_sumpos(2)) and gives [[0, 1], [1, 2]]
but print (createSquareMat_sumpos(3)) I need it to give me [[0, 1, 2], [1, 2, 3], [2, 3, 4]]
instead of
[[0, 1], [1, 2], [2, 3]]
Im trying ti implement a function evenrow() that takes a two dimensional list of integers and returns True if each row of the table sums up to an even number and False otherwise (i.e.., if some row sums up to an odd number)
usage
>>> evenrow([[1, 3], [2, 4], [0, 6]])
True
>>> evenrow([[1, 3], [3, 4], [0, 5]])
False
This is what I got so far:
def evenrow(lst):
for i in range(len(lst)-1):
if sum(lst[i])%2==0: # here is the problem, it only iterates over the first item in the lst [1, 3] - i cant figure this out - range problem?
return True
else:
False
How do I get the loop to iterate over every item [1, 3], [2, 4], [0, 6] in the list and not just the first?
well I have gotten this far now:
def evenrow(lst):
for i in range(len(lst)-1):
if sum(lst[i]) %2 >0:
return False
else:
return True
and i get the following answer when executing different lists:
>>> evenrow([[1, 3], [2, 4], [0, 6]])
True
>>> evenrow([[1, 3], [3, 4], [0, 5]])
False
>>> evenrow([[1, 3, 2], [3, 4, 7], [0, 6, 2]])
True
>>> evenrow([[1, 3, 2], [3, 4, 7], [0, 5, 2]])
True
(the last one is not correct though - should be False) I just dont get why this is not working...
You are returning too early. You should check for all the pairs, only returning True afterwards, or return False if a odd sum is encountered.
Spoiler alert:
def evenrow(lst):
for i in range(len(lst)-1):
if sum(lst[i]) % 2 != 0: # here is the problem, it only iterates over the first item in the lst [1, 3] - i cant figure this out - range problem?
return False
return True
This will achieve the goal.