I am struggling in one of the Pattern matching problems in Python
When input = 3, below is the expected output (input value is the number of columns it should print)
Expected output:
1
2 6
3 7 9
4 8
5
I am somehow moving in a wrong direction, hence would need some help in it.
This is the code I have tried so far:
def display():
n = 5
i = 1
# Outer loop for how many lines we want to print
while(i<=n):
k = i
j = 1
# Inner loop for printing natural number
while(j <= i):
print (k,end=" ")
# Logic to print natural value column-wise
k = k + n - j
j = j + 1
print("\r")
i = i + 1
#Driver code
display()
But it is giving me output as this:
1
2 6
3 7 10
4 8 11 13
5 9 12 14 15
Anybody who can help me with this?
n=10
for i in range(1,2*n):
k=i
for j in range(2*n-i if i>n else i):
print(k,end=' ')
k = k + 2*n - 2*j - 2
print()
Result
1
2 20
3 21 37
4 22 38 52
5 23 39 53 65
6 24 40 54 66 76
7 25 41 55 67 77 85
8 26 42 56 68 78 86 92
9 27 43 57 69 79 87 93 97
10 28 44 58 70 80 88 94 98 100
11 29 45 59 71 81 89 95 99
12 30 46 60 72 82 90 96
13 31 47 61 73 83 91
14 32 48 62 74 84
15 33 49 63 75
16 34 50 64
17 35 51
18 36
19
>
Here's a way, I started from scratch and not for code, much more easy for me
def build(nb_cols):
values = list(range(1, nb_cols ** 2 + 1))
res = []
for idx in range(nb_cols):
row_values, values = values[-(idx * 2 + 1):], values[:-(idx * 2 + 1)]
res.append([' '] * (nb_cols - idx - 1) + row_values + [' '] * (nb_cols - idx - 1))
for r in zip(*reversed(res)):
print(" ".join(map(str, r)))
Here's a recursive solution:
def col_counter(start, end):
yield start
if start < end:
yield from col_counter(start+1, end)
yield start
def row_generator(start, col, N, i=1):
if i < col:
start = start + 2*(N - i)
yield start
yield from row_generator(start, col, N, i+1)
def display(N):
for i, col_num in enumerate(col_counter(1, N), 1):
print(i, *row_generator(i, col_num, N))
Output:
>>> display(3)
1
2 6
3 7 9
4 8
5
>>> display(4)
1
2 8
3 9 13
4 10 14 16
5 11 15
6 12
7
>>> display(10)
1
2 20
3 21 37
4 22 38 52
5 23 39 53 65
6 24 40 54 66 76
7 25 41 55 67 77 85
8 26 42 56 68 78 86 92
9 27 43 57 69 79 87 93 97
10 28 44 58 70 80 88 94 98 100
11 29 45 59 71 81 89 95 99
12 30 46 60 72 82 90 96
13 31 47 61 73 83 91
14 32 48 62 74 84
15 33 49 63 75
16 34 50 64
17 35 51
18 36
19
Here is the solution using simple loops
def display(n):
nrow = 2*n -1 #Number of rows
i = 1
noofcols = 1 #Number of columns in each row
t = 1
while (i <= nrow):
print(i,end=' ')
if i <= n:
noofcols = i
else:
noofcols = 2*n - i
m =i
if t < noofcols:
for x in range(1,noofcols):
m = nrow + m -(2*x-1)
print(m, end=' ')
i = i+1
print()
Related
I have a DataFrame (df) and I need to split it into n number of Dataframes based on the column numbers. But, it has to follow the Triangular Series pattern:
df1 = df[[0]]
df2 = df[[1,2]]
df3 = df[[3,4,5]]
df4 = df[[6,7,8,9]]
etc.
Consider the dataframe df
df = pd.DataFrame(
np.arange(100).reshape(10, 10),
columns=list('ABCDEFGHIJ')
)
df
A B C D E F G H I J
0 0 1 2 3 4 5 6 7 8 9
1 10 11 12 13 14 15 16 17 18 19
2 20 21 22 23 24 25 26 27 28 29
3 30 31 32 33 34 35 36 37 38 39
4 40 41 42 43 44 45 46 47 48 49
5 50 51 52 53 54 55 56 57 58 59
6 60 61 62 63 64 65 66 67 68 69
7 70 71 72 73 74 75 76 77 78 79
8 80 81 82 83 84 85 86 87 88 89
9 90 91 92 93 94 95 96 97 98 99
i_s, j_s = np.arange(4).cumsum(), np.arange(1, 5).cumsum()
df1, df2, df3, df4 = [
df.iloc[:, i:j] for i, j in zip(i_s, j_s)
]
Verify
pd.concat(dict(enumerate([df.iloc[:, i:j] for i, j in zip(i_s, j_s)])), axis=1)
0 1 2 3
A B C D E F G H I J
0 0 1 2 3 4 5 6 7 8 9
1 10 11 12 13 14 15 16 17 18 19
2 20 21 22 23 24 25 26 27 28 29
3 30 31 32 33 34 35 36 37 38 39
4 40 41 42 43 44 45 46 47 48 49
5 50 51 52 53 54 55 56 57 58 59
6 60 61 62 63 64 65 66 67 68 69
7 70 71 72 73 74 75 76 77 78 79
8 80 81 82 83 84 85 86 87 88 89
9 90 91 92 93 94 95 96 97 98 99
first get Triangular Number Series, then apply it to dataframe
n = len(df.columns.tolist())
end = 0
i = 0
res = []
while end < n:
begin = end
end = i*(i+1)/2
res.append(begin,end)
idx = map( lambda x:range(x),res)
for i in idx:
df[i]
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I'm trying to make a triangle that looks like this
10
11 12
13 14 15
16 17 18 19
20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36 37
38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54
I am trying to use two for loops with one nested. Here is as close as I have gotten so far.
for j in range(11):
print(end='\n')
for i in range(j+1):
print(i+j,'',end='')
print(end='\n')
I'm pretty sure I need to create a variable, but not really sure how to incorporate it into the loop.
Here you go:
>>> a=range(10, 55)
>>> for i in range(10):
... print(' '.join(repr(e) for e in a[:i+1]))
... a = a[i+1:]
...
10
11 12
13 14 15
16 17 18 19
20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36 37
38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54
How about short and simple like this:
k=10
for i in range(9):
for j in range(i+1):
print(k, end='')
k+=1
print('')
Here is another single for loop based solution:
number = 10
for line_length in range(9):
print(*range(number, number + line_length + 1))
number += line_length + 1
Giving:
10
11 12
13 14 15
16 17 18 19
20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36 37
38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54
I like Maelstrom's short and sweet answer, but if you want to look at it mathematically, you might do something like this instead:
>>> for i in range(1, 10):
... j = 10 + i * (i - 1) // 2
... print(*range(j, j + i)) # This line edited per lvc's comment
...
10
11 12
13 14 15
16 17 18 19
20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36 37
38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54
Use one variable to keep track of the current number and one to keep track of the tier you are on
num = 10;
tier = 1;
tiers = 10;
for i in range(tiers):
for j in range(tier):
print(num + " ");
num = num + 1;
print("\n");
tier = tier + 1
You can change the triangle height by adjusting the triangle_height variable and the starting element by changing print_number.
print_number = 10
triangle_height = 9
for level_element_count in range(triangle_height):
print('\n')
while level_element_count > -1:
print(print_number, '', end='')
print_number += 1
level_element_count -= 1
print('\n')
Just for fun.
This is related to a common pattern where you divide a given sequence (here, the numbers from 10 to 54, inclusive) into non-overlapping 'windows', to do some analysis on, say, 10 values at a time. The twist here is that each window is one element larger than the last.
This looks like a job for itertools!
import itertools as it
def increasing_windows(i, start=1, step=1):
'''yield non-overlapping windows from iterable `i`,
increasing in size from `start` by `step`.
'''
pos = 0
for size in it.count(start, step):
yield it.islice(i, pos, pos+size)
pos += size
for line in it.islice(increasing_windows(range(10, 55)), 9):
print(*line)
Try this
counter = 10
for i in range(10):
output = ""
for j in range(i):
output = output + " " + str(counter)
counter += 1
print(output)
Output:
10
11 12
13 14 15
16 17 18 19
20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36 37
38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54
Explanation:
First loop controls the width of the triangle and second loop controls the content and hence the height. We need to convert an integer to string and concatenate. We create proper output in a string variable in each iteration of second loop and then display it once it gets finished.The key thing is to iterate second loop according to the first one, i.e. loop it as much as first does
You could write a generator:
def number_triangle(start, nrows):
current = start
for length in range(1, nrows+1):
yield range(current, current+length)
current += length
>>> for row in number_triangle(10, 9):
... print(*row)
10
11 12
13 14 15
16 17 18 19
20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36 37
38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54
>>> for row in number_triangle(1, 12):
... print(*row)
1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
16 17 18 19 20 21
22 23 24 25 26 27 28
29 30 31 32 33 34 35 36
37 38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54 55
56 57 58 59 60 61 62 63 64 65 66
67 68 69 70 71 72 73 74 75 76 77 78
Or you could have an infinite generator and leave it up to the caller to control how many rows to generate:
def number_triangle(start=0):
length = 1
while True:
yield range(start, start+length)
start += length
length += 1
>>> nt = number_triangle()
>>> for i in range(15):
... print(*next(nt))
0
1 2
3 4 5
6 7 8 9
10 11 12 13 14
15 16 17 18 19 20
21 22 23 24 25 26 27
28 29 30 31 32 33 34 35
36 37 38 39 40 41 42 43 44
45 46 47 48 49 50 51 52 53 54
55 56 57 58 59 60 61 62 63 64 65
66 67 68 69 70 71 72 73 74 75 76 77
78 79 80 81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100 101 102 103 104
105 106 107 108 109 110 111 112 113 114 115 116 117 118 119
No need for nested loop:
a = range(10, 55)
flag = 0
current = 0
for i, e in enumerate(a):
print e,
if flag == i:
current += 1
flag = i + 1 + current
print '\n',
Suppose I had this large data frame:
In [31]: df
Out[31]:
A B C D E F G H I J ... Q R S T U V W X Y Z
0 0 1 2 3 4 5 6 7 8 9 ... 16 17 18 19 20 21 22 23 24 25
1 26 27 28 29 30 31 32 33 34 35 ... 42 43 44 45 46 47 48 49 50 51
2 52 53 54 55 56 57 58 59 60 61 ... 68 69 70 71 72 73 74 75 76 77
[3 rows x 26 columns]
which you can create using
alphabet = [chr(letter_i) for letter_i in range(ord('A'), ord('Z')+1)]
df = pd.DataFrame(np.arange(3*26).reshape(3, 26), columns=alphabet)
What's the best way to drop all columns between column 'D' and 'R' using the labels of the columns?
I found one ugly way to do it:
df.drop(df.columns[df.columns.get_loc('D'):df.columns.get_loc('R')+1], axis=1)
Here's my entry:
>>> df.drop(df.columns.to_series()["D":"R"], axis=1)
A B C S T U V W X Y Z
0 0 1 2 18 19 20 21 22 23 24 25
1 26 27 28 44 45 46 47 48 49 50 51
2 52 53 54 70 71 72 73 74 75 76 77
By converting df.columns from an Index to a Series, we can take advantage of the ["D":"R"]-style selection:
>>> df.columns.to_series()["D":"R"]
D D
E E
F F
G G
H H
I I
J J
... ...
Q Q
R R
dtype: object
Here you are:
print df.ix[:,'A':'C'].join(df.ix[:,'S':'Z'])
Out[1]:
A B C S T U V W X Y Z
0 0 1 2 18 19 20 21 22 23 24 25
1 26 27 28 44 45 46 47 48 49 50 51
2 52 53 54 70 71 72 73 74 75 76 77
Here's another way ...
low, high = df.columns.get_slice_bound(('D', 'R'), 'left')
drops = df.columns[low:high+1]
print df.drop(drops, axis=1)
A B C S T U V W X Y Z
0 0 1 2 18 19 20 21 22 23 24 25
1 26 27 28 44 45 46 47 48 49 50 51
2 52 53 54 70 71 72 73 74 75 76 77
Use numpy for more flexibility ... numpy allows comparison of letters (probably by comparing on ASCII bit level, or something):
import numpy as np
array = (['A','B','C','D'])
array > 'B'
print(array)
print(array>'B')
gives:
['A' 'B' 'C' 'D']
array([False, False, True, True], dtype=bool)
More difficult selections are also easily possible:
b[np.logical_and(b>'B', b<'D')]
gives:
array(['C'],
dtype='|S1')
I came across this problem: http://codegolf.com/numeric-diamonds which requires you to print numbers in this format:
1
4 2
7 5 3
8 6
9
1
7 2
13 8 3
19 14 9 4
25 20 15 10 5
31 26 21 16 11 6
32 27 22 17 12
33 28 23 18
34 29 24
35 30
36
1
11 2
21 12 3
31 22 13 4
41 32 23 14 5
51 42 33 24 15 6
61 52 43 34 25 16 7
71 62 53 44 35 26 17 8
81 72 63 54 45 36 27 18 9
91 82 73 64 55 46 37 28 19 10
92 83 74 65 56 47 38 29 20
93 84 75 66 57 48 39 30
94 85 76 67 58 49 40
95 86 77 68 59 50
96 87 78 69 60
97 88 79 70
98 89 80
99 90
100
I was able to find the numbers that go into a particular row just fine, but my logic for arranging them is quite obscure.
for i in range(2*n - 1):
m = <list of numbers for this row>
row_width = (n+i)*(n/3) if i < n else row_width - (n/3)
print ''.join(str(i).rjust(n/2 + 1) for i in m).strip().rjust(row_width))
That's ok from a code golf perspective, but is there a clean, pythonic way to arrange numbers from a 2D array in this format?
Don't know about "pythonic", but this looks quite clean to me:
size = 10
maxlen = len(str(size * size))
m = size * 2 - 1
matrix = [[' ' * maxlen] * m for _ in range(m)]
for n in range(size * size):
r = n // size
c = n % size
matrix[c + r][size - r - 1 + c] = '{0:{1}}'.format(n + 1, maxlen)
print '\n'.join(''.join(row) for row in matrix)
So the function M returns the values of the number in each row (e.g. for n=10, then M(n) = [[1],[11,2],...]]. The cell function make every number take up as much space as the largest number does(learned for selected answer).
def M(n):
t = 2 * n - 1
N = [range(i * n + 1, (i + 1) * n + 1) for i in range(n)]
M = []
for k in xrange(t):
M.append([])
for i in reversed(xrange(n)):
for j in reversed(xrange(n)):
if i + j == k:
M[k].append(N[i][j])
return M
def magic_square(n):
t = 2 * n - 1
maxlen = len(str(n * n))
gap = ' ' * maxlen
cell = lambda num: '{0:{1}}'.format(num, maxlen)
for m in M(n):
padding = gap * ((t - (2 * len(m) - 1)) / 2)
print padding + gap.join(map(cell, m)) + padding
def print_diamond(n):
if n == 1:
print 1
return
maxlen = len(str(n*n))
gap = maxlen * ' '
first = 0
for i in xrange(2*n-1):
if i < n:
first = i*n + 1
print (abs(i-n)-1)*gap + gap.join(map(lambda x: '{0:{1}}'.format(x, maxlen), xrange(first, i, 1-n)))
else:
first += 1
print (abs(i-n)+1)*gap + gap.join(map(lambda x: '{0:{1}}'.format(x, maxlen), xrange(first, n*(i+2-n)-1, 1-n)))
I am trying to teach myself python using interactivepython.org. I have come across a problem that I can not figure out. I have the slope and the spacing correct. I need it to print one less number every time. Could anybody help a newbie out?...
My Code:
for j in range(11):
for i in range(j):
print(str(i), end=" ")
print()
print("")
Output:
0
0 1
0 1 2
0 1 2 3
0 1 2 3 4
0 1 2 3 4 5
0 1 2 3 4 5 6
0 1 2 3 4 5 6 7
0 1 2 3 4 5 6 7 8
0 1 2 3 4 5 6 7 8 9
Desired Output:
10
11 12
13 14 15
16 17 18 19
20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36 37
38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54
The exercise is about nesting for loops...I know there are other ways to do this.
This should do it:
start = 10
width = 9
for i in range(1, width+1):
for _ in range(i):
print (start, end=" ")
start += 1
print('\n')
Output:
10
11 12
13 14 15
16 17 18 19
20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36 37
38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54
Well this one again for you Brandon Shockley :)
code:
x = 9
lines = 10
for i in range(lines):
for j in range(i):
x+=1
print x,
print ''
output:
10
11 12
13 14 15
16 17 18 19
20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36 37
38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 5
Hope This helps :)
You can do it like this
current, levels = 10, 9
for i in range(levels):
for j in range(i + 1):
print(current, end = " ")
current += 1
print("\n")
Output
10
11 12
13 14 15
16 17 18 19
20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36 37
38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54
>>> lst = list(range(54,9, -1))
>>> for j in range(11):
... for i in range(j):
... if len(lst):
... print(lst.pop(), end=" ")
... print(" ")
inc = 10
for j in range(10):
for i in range(j):
print(inc, end=" ")
inc += 1
print()
print("")
Is anything wrong with this?
Single loop, in Python 2.x (can't remove the space after each print)
c = 1
j = 0
for i in range(10, 55):
print str(i) + ',',
j += 1
if j == c:
print
c += 1
j = 0
Using join
start = 10
for i in range(1, 10):
print(' '.join(map(str, range(start, start + i))))
start += i
A pointlessly compact version using join and some maths:
print('\n'.join(' '.join(map(str,
range(10 + i * (i+1) / 2, 10 + (i+1) * (i+2) / 2))) for i in range(9)))
And confusingly, this also works (in python 2):
j = 9
for i in range(10):
for j in range(j + 1, j + i + 1):
print j,
print