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Write a program to solve a Sudoku puzzle by filling the empty cells where 0 represent empty cell.
Rules:
All the number in sudoku must appear exactly once in diagonal running from top-left to bottom-right.
All the number in sudoku must appear exactly
once in diagonal running from top-right to bottom-left.
All the number in sudoku
must appear exactly once in a 3*3 sub-grid.
However number can repeat in row or column.
Constraints:
N = 9; where N represent rows and column of grid.
What I have tried:
N = 9
def printing(arr):
for i in range(N):
for j in range(N):
print(arr[i][j], end=" ")
print()
def isSafe(grid, row, col, num):
for x in range(9):
if grid[x][x] == num:
return False
cl = 0
for y in range(8, -1, -1):
if grid[cl][y] == num:
cl += 1
return False
startRow = row - row % 3
startCol = col - col % 3
for i in range(3):
for j in range(3):
if grid[i + startRow][j + startCol] == num:
return False
return True
def solveSudoku(grid, row, col):
if (row == N - 1 and col == N):
return True
if col == N:
row += 1
col = 0
if grid[row][col] > 0:
return solveSudoku(grid, row, col + 1)
for num in range(1, N + 1, 1):
if isSafe(grid, row, col, num):
grid[row][col] = num
print(grid)
if solveSudoku(grid, row, col + 1):
return True
grid[row][col] = 0
return False
if (solveSudoku(grid, 0, 0)):
printing(grid)
else:
print("Solution does not exist")
Input:
grid =
[
[0, 3, 7, 0, 4, 2, 0, 2, 0],
[5, 0, 6, 1, 0, 0, 0, 0, 7],
[0, 0, 2, 0, 0, 0, 5, 0, 0],
[2, 8, 3, 0, 0, 0, 0, 0, 0],
[0, 5, 0, 0, 7, 1, 2, 0, 7],
[0, 0, 0, 3, 0, 0, 0, 0, 3],
[7, 0, 0, 0, 0, 6, 0, 5, 0],
[0, 2, 3, 0, 3, 0, 7, 4, 2],
[0, 5, 0, 0, 8, 0, 0, 0, 0]
]
Output:
8 3 7 0 4 2 1 2 4
5 0 6 1 8 6 3 6 7
4 1 2 7 3 5 5 0 8
2 8 3 5 4 8 6 0 5
6 5 0 0 7 1 2 1 7
1 4 7 3 2 6 8 4 3
7 8 1 4 1 6 3 5 0
6 2 3 7 3 5 7 4 2
0 5 4 2 8 0 6 8 1
Basically I am stuck on the implementation of checking distinct number along diagonal. So question is how I can make sure that no elements gets repeated in those diagonal and sub-grid.
I have a pandas dataframe like this:
pd.DataFrame({
'customer_id': ['100', '200', '300', '400', '500', '600'],
'Month1': [1, 1, 1, 1, 1, 1],
'Month2': [1, 0, 1, 1, 1, 1],
'Month3': [0, 0, 0, 0, 1, 1],
'Month4': [0, 0, 0, 0, 0, 1]})
This is showing a boolean value for when a customer defaults on a loan. The first month with 0 means the customer defaulted that month. I want an output that displays the month number the customer defaulted on the loan.
Output:
pd.DataFrame({
'customer_id': ['100', '200', '300', '400', '500', '600'],
'Month1': [1, 1, 1, 1, 1, 1],
'Month2': [1, 0, 1, 1, 1, 1],
'Month3': [0, 0, 0, 0, 1, 1],
'Month4': [0, 0, 0, 0, 0, 1],
'default_month': [3, 2, 3, 3, 4, np.nan]})
You can check whether all the 'Month' columns in a row are not 0, using all(axis=1) and ne(0) and return np.nan which means that the person has not yet defaulted (i.e. your row 5).
Then using eq(0)and idxmax you can check which is the first value of a row that equals to 0 and grab that column name.
import numpy as np
m = df.filter(like='Month')
df['default_month'] = np.where((m.ne(0)).all(1),np.nan,
m.eq(0).idxmax(1))
df
customer_id Month1 Month2 Month3 Month4 default_month
0 100 1 1 0 0 Month3
1 200 1 0 0 0 Month2
2 300 1 1 0 0 Month3
3 400 1 1 0 0 Month3
4 500 1 1 1 0 Month4
5 600 1 1 1 1 NaN
Here's some code to get your result:
first we need a list of months in reverse order. From your data, I just pulled them directly from the index.
months = list(df.columns[1:5])
months.reverse()
months is now ['Month4', 'Month3', 'Month2', 'Month1'].
We iterate backwards so when we find an earlier month of default, it overwrites
for (i,m) in enumerate(months):
mask = df[m] == 0 # Check for a default
df.loc[mask,'default_month'] = len(months) - i
This returns the output you are looking for.
I have following table and data.
create table test ( id bigserial not null,
geo geometry not null );
insert
into
test(geo)
values ('MULTIPOLYGON (((0 0, 0 0, 0 7, 0 7, 0 0)), ((0 0, 0 7, 7 7, 7 0, 0 0)), ((0 0, 7 0, 7 0, 0 0, 0 0)), ((7 7, 7 7, 7 0, 7 0, 7 7)), ((0 7, 0 7, 7 7, 7 7, 0 7)), ((0 0, 7 0, 7 7, 0 7, 0 0)))'),
('POLYGON ((0 0, 5 0, 5 5, 0 5, 0 0))'),
('POLYGON ((2 2, 5 2, 5 5, 2 5, 2 2))');
select * from test;
id|geo |
--|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
5|MULTIPOLYGON (((0 0, 0 0, 0 7, 0 7, 0 0)), ((0 0, 0 7, 7 7, 7 0, 0 0)), ((0 0, 7 0, 7 0, 0 0, 0 0)), ((7 7, 7 7, 7 0, 7 0, 7 7)), ((0 7, 0 7, 7 7, 7 7, 0 7)), ((0 0, 7 0, 7 7, 0 7, 0 0)))|
6|POLYGON ((0 0, 5 0, 5 5, 0 5, 0 0)) |
7|POLYGON ((2 2, 5 2, 5 5, 2 5, 2 2)) |
following query (Q) should return all rows
select
*
from
test t
where
st_contains('MULTIPOLYGON (((0 0, 0 0, 0 7, 0 7, 0 0)), ((0 0, 0 7, 7 7, 7 0, 0 0)), ((0 0, 7 0, 7 0, 0 0, 0 0)), ((7 7, 7 7, 7 0, 7 0, 7 7)), ((0 7, 0 7, 7 7, 7 7, 0 7)), ((0 0, 7 0, 7 7, 0 7, 0 0)))',
geo);
id|geo |
--|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
5|MULTIPOLYGON (((0 0, 0 0, 0 7, 0 7, 0 0)), ((0 0, 0 7, 7 7, 7 0, 0 0)), ((0 0, 7 0, 7 0, 0 0, 0 0)), ((7 7, 7 7, 7 0, 7 0, 7 7)), ((0 7, 0 7, 7 7, 7 7, 0 7)), ((0 0, 7 0, 7 7, 0 7, 0 0)))|
6|POLYGON ((0 0, 5 0, 5 5, 0 5, 0 0)) |
Because following constraint returns true.
select
st_contains('MULTIPOLYGON (((0 0, 0 0, 0 7, 0 7, 0 0)), ((0 0, 0 7, 7 7, 7 0, 0 0)), ((0 0, 7 0, 7 0, 0 0, 0 0)), ((7 7, 7 7, 7 0, 7 0, 7 7)), ((0 7, 0 7, 7 7, 7 7, 0 7)), ((0 0, 7 0, 7 7, 0 7, 0 0)))',
'POLYGON ((2 2, 5 2, 5 5, 2 5, 2 2))');
What is wrong here with the query Q above?
The input geometry is invalid, and so is the result as per the doc:
So ST_Contains(A,B) implies ST_Within(B,A) except in the case of
invalid geometries where the result is always false regardless or not
defined.
WITH test(geo) as (
values ('MULTIPOLYGON (((0 0, 0 0, 0 7, 0 7, 0 0)), ((0 0, 0 7, 7 7, 7 0, 0 0)), ((0 0, 7 0, 7 0, 0 0, 0 0)), ((7 7, 7 7, 7 0, 7 0, 7 7)), ((0 7, 0 7, 7 7, 7 7, 0 7)), ((0 0, 7 0, 7 7, 0 7, 0 0)))'),
('POLYGON ((0 0, 5 0, 5 5, 0 5, 0 0))'),
('POLYGON ((2 2, 5 2, 5 5, 2 5, 2 2))'))
select st_isvalid(geo), st_isvalidreason(geo) from test;
st_isvalid | st_isvalidreason
------------+-------------------------------------------
f | Too few points in geometry component[0 7]
t | Valid Geometry
t | Valid Geometry
That being said, you may want to read carefully the doc on st_contains and st_covers as there are subtleties when the geometries share an edge.
DataFrame
df=pd.DataFrame({'occurance':[1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0],'value':[45, 3, 2, 12, 14, 32, 1, 1, 6, 4, 9, 32, 78, 96, 12, 6, 3]})
df
Expected output
df=pd.DataFrame({'occurance':[1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0],'value':[45, 3, 2, 12, 14, 32, 1, 1, 6, 4, 9, 32, 78, 96, 12, 6, 3],'group':[1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 4, 100, 5, 5, 5, 5]})
df
I need to transform the dataframe into the output. I am after a wild card that will determine 1 is the start of a new group and a group consists of only 1 followed by n zeroes. If a group criteria is not met, then group it as 100.
I tried in the line of;
bs=df[df.occurance.eq(1).any(1)&df.occurance.shift(-1).eq(0).any(1)].squeeze()
bs
This even when broken down could only bool select start and nothing more.
Any help?
Create mask by compare 1 and next 1 in mask, then filter occurance for all values without them, create cumulative sum by Series.cumsum and last add 100 values by Series.reindex:
m = df.occurance.eq(1) & df.occurance.shift(-1).eq(1)
df['group'] = df.loc[~m, 'occurance'].cumsum().reindex(df.index, fill_value=100)
print (df)
occurance value group
0 1 45 1
1 0 3 1
2 0 2 1
3 0 12 1
4 1 14 2
5 0 32 2
6 0 1 2
7 0 1 2
8 0 6 2
9 0 4 2
10 1 9 3
11 0 32 3
12 1 78 100
13 1 96 4
14 0 12 4
15 0 6 4
16 0 3 4
aMat=df1000.iloc[:,1:].values
print(aMat)
By using the above code I got the below mentioned data matrix from a dataset:
[[1 2 5 2 4]
[1 2 1 2 2]
[1 2 4 2 4]
[1 5 1 1 4]
[1 4 4 2 5]]
The data set only can hold 1,2,3,4 and 5 value. So I want to count number of 1 present in first column, number of 2 present in first column, number of 3 present in first column, number of 4 present in first column, number of 5 present in first column, number of 1 present in second column,.............so on. Means at the end the list will look like this:
[[5,0,0,0,0],[0,3,0,1,1],[2,0,0,2,5],[1,4,0,0,0],[0,1,0,3,1]]
Please help
Let's try:
df = pd.DataFrame([[1, 2, 5, 2, 4],
[1, 2, 1, 2, 2],
[1, 2, 4, 2, 4],
[1, 5, 1, 1, 4],
[1, 4, 4, 2, 5]])
df.apply(pd.Series.value_counts).reindex([1,2,3,4,5]).fillna(0).to_numpy('int')
Output:
array([[5, 0, 2, 1, 0],
[0, 3, 0, 4, 1],
[0, 0, 0, 0, 0],
[0, 1, 2, 0, 3],
[0, 1, 1, 0, 1]])
Or, transposed:
df.apply(pd.Series.value_counts).reindex([1,2,3,4,5]).fillna(0).T.to_numpy('int')
Output:
array([[5, 0, 0, 0, 0],
[0, 3, 0, 1, 1],
[2, 0, 0, 2, 1],
[1, 4, 0, 0, 0],
[0, 1, 0, 3, 1]])
You can use np.bincount with apply_along_axis.
a = df.to_numpy()
np.apply_along_axis(np.bincount, 0, a, minlength=a.max()+1).T[:, 1:]
array([[5, 0, 0, 0, 0],
[0, 3, 0, 1, 1],
[2, 0, 0, 2, 1],
[1, 4, 0, 0, 0],
[0, 1, 0, 3, 1]], dtype=int64)
May using stack
df.stack().groupby(level=1).value_counts().unstack(fill_value=0).reindex(columns=[1,2,3,4,5],fill_value=0)
Out[495]:
1 2 3 4 5
0 5 0 0 0 0
1 0 3 0 1 1
2 2 0 0 2 1
3 1 4 0 0 0
4 0 1 0 3 1
Method from collections
pd.DataFrame(list(map(collections.Counter,a.T))).fillna(0)#.values
Out[527]:
1 2 4 5
0 5.0 0.0 0.0 0.0
1 0.0 3.0 1.0 1.0
2 2.0 0.0 2.0 1.0
3 1.0 4.0 0.0 0.0
4 0.0 1.0 3.0 1.0
My attempt with get_dummies and sum:
pd.get_dummies(df.stack()).sum(level=1)
1 2 4 5
0 5 0 0 0
1 0 3 1 1
2 2 0 2 1
3 1 4 0 0
4 0 1 3 1
If you need the column 3 with all zeros, use reindex:
pd.get_dummies(df.stack()).sum(level=1).reindex(columns=range(1, 6), fill_value=0)
1 2 3 4 5
0 5 0 0 0 0
1 0 3 0 1 1
2 2 0 0 2 1
3 1 4 0 0 0
4 0 1 0 3 1
Or, if you fancy a main course of numpy with a side dish of broadcasting:
# edit courtesy #user3483203
np.equal.outer(df.values, np.arange(1, 6)).sum(0)
array([[5, 0, 0, 0, 0],
[0, 3, 0, 1, 1],
[2, 0, 0, 2, 1],
[1, 4, 0, 0, 0],
[0, 1, 0, 3, 1]])