Develop Reference

Table of Squares and Cubes

I am learning Python on my own and I am stuck on a problem from a book I purchased. I can only seem to get the numbers correctly, but I cannot get the title. I was wondering if this has anything to do with the format method to make it look better? This is what I have so far:
number = 0
square = 0
cube = 0
for number in range(0, 6):
square = number * number
cube = number * number * number
print(number, square, cube)
What I am returning:
0 0 0
1 1 1
2 4 8
3 9 27
4 16 64
5 25 125
I would like my desired output to be this:
number square cube
0 0 0
1 1 1
2 4 8
3 9 27
4 16 64
5 25 125

Here we can specify the width of the digits using format in paranthesis
print('number square cube')
for x in range(0, 6):
print('{0:6d}\t {1:7d}\t {2:3d}'.format(x, x*x, x*x*x))
This would result in
number square cube
0 0 0
1 1 1
2 4 8
3 9 27
4 16 64
5 25 125

You need to print the header row. I used tabs \t to space the numbers our properly and f-stings because they are awesome (look them up).
number = 0
square = 0
cube = 0
# print the header
print('number\tsquare\tcube')
for number in range(0, 6):
square = number * number
cube = number * number * number
# print the rows using f-strings
print(f'{number}\t{square}\t{cube}')
Output:
number square cube
0 0 0
1 1 1
2 4 8
3 9 27
4 16 64
5 25 125
The only thing this doesn't do is right-align the columns, you'd have to write some custom printing function for that which determines the proper width in spaces of the columns based on each item in that column. To be honest, the output here doesn't make ANY difference and I'd focus on the utility of your code rather than what it looks like when printing to a terminal.

There is a somewhat more terse method you might consider that also uses format, as you guessed. I think this is worth learning because the Format Specification Mini-Language can be useful. Combined with f-stings, you go from eight lines of code to 3.
print('number\tsquare\tcube')
for number in range(0, 6):
print(f'{number:>6}{number**2:>8}{number**3:>6}')
The numbers here (e.g.:>6) aren't special but are just there to get you the output you desired. The > however is, forcing the number to be right-aligned within the space available.

How can I recursively divide the grid in my maze?

I am trying to create a maze generator using recursive division. I use this link:
Maze generation - recursive division (how it works?) as my guide as to how to approach the problem.
Here is my code so far:
import random
# Maze: 0 - N : 4 x 4 Grid
# Grid: 0 - (2n + 1) : 9 x 9 Array
# TODO: Now, Find a way to save the previous walls and not just only one at a time
rows = 9
cols = 9
start = 2
end = 7
# -------------------------------------------------------------------------------------
# Lists for all even / odd numbers in given range
evens = [n for n in range(start, end+1) if n % 2 == 0]
odds = [m for m in range(start, end+1) if m % 2 != 0]
# Generate random even/odd integer value for walls/ passages respectively
# Walls: Not sure if 2 variables are necessary-----------------------------------------
wallX = random.choice(evens)
wallY = random.choice(evens)
# Passages
passageX = random.choice(odds)
passageY = random.choice(odds)
#--------------------------------------------------------------------------------------
# Random direction: True = Horizontal Slice, False = Vertical Slice
randomDirection = random.choice([True, False])
arr = [['0' for i in range(cols)] for j in range(rows)]
def displayBoard(arr):
print()
for i in range(len(arr)):
for j in range(len(arr[i])):
# Print just the edges
if i == 0 or i == 8 or j == 0 or j == 8:
print('*', end = ' ')
# Print wall
elif arr[i][j] == 1:
print('.', end = ' ')
else:
print (' ', end = ' ')
print()
# Function choose direction to slice
def chooseDir(arr):
for i in range(len(arr)):
for j in range(len(arr[i])):
# Horizontal Direction Slice
if randomDirection:
arr[wallX][j] = 1
arr[wallX][passageY] = 2
print(arr[i][j], end = ' ')
# Vertical Slice
else:
arr[i][wallY] = 1
arr[passageX][wallY] = 2
print(arr[i][j], end = ' ')
print()
displayBoard(arr)
print()
mazeX = 0
mazeY = 0
# Write the recursive division function:
def divide():
chooseDir(arr)
print()
divide()
What this produces is a grid that is randomly sliced at an even index (creating walls) and creates passages at odd indices.
Output: 1 = wall, 2 = passage made
0 0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 2 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0 0
* * * * * * * * *
* . *
* . *
* *
* . *
* . *
* . *
* . *
* * * * * * * * *
My issue is that I don't know how to write my recursive function. Do I call division on the two new cells created when a wall is made and continuously divide the "sub cells".
Or I know that a 4 x 4 cell grid will provide an array of 9 x 9 and I will have 16 cells total.
Then I can call division until a certain condition is met, increment to the next cell until all 16 were visited.
In both cases, I am not sure how to represent the new walls/cells created so that I can write the division function. Up until now, I've been using the grid coordinates.

You asked "Do I call division on the two new cells created when a wall is made and continuously divide the "sub cells"". Yes, that is the essence of recursion. Take a big problem and make smaller problem(s). Repeat with the smaller problems. Eventually, the problems are small enough to easily solve. Then put all the small problems back together to solve the original problem.
For the maze, the first wall split the maze into two smaller mazes. Use the same algorithm to split each of them and there are not 4 smaller mazes. Repeat until the sub-mazes are too small to split any more.
Your code that splits the maze should go in a function. If the sub-maze is big enough to split, the function splits the sub-maze and then calls itself on the two smaller sub-mazes.

Split every row containing long text into multiple rows in pandas

I have a DataFrame which has a string column such as below:
id text label
1 this is long string with many words 1
2 this is a middle string 0
3 short string 1
and i want to convert this DataFrame to another DataFrame based on the string length i.e. (df['text'].str.len > 3) :
id text label
1 this is long 1
1 string with many 1
1 words 1
2 this is a 0
2 middle string 0
3 short string 1
this is my code:
pd.concat(df['text'].str.len() > 200)
but it is wrong.

You could
In [1257]: n = 3
In [1279]: df.set_index(['label', 'id'])['text'].str.split().apply(
lambda x: pd.Series([' '.join(x[i:i+n]) for i in range(0, len(x), n)])
).stack().reset_index().drop('level_2', 1)
Out[1279]:
label id 0
0 1 1 this is long
1 1 1 string with many
2 1 1 words
3 0 2 this is a
4 0 2 middle string
5 1 3 short string
Details
label text id
0 1 this is long string with many words 1
1 0 this is a middle string 2
2 1 short string 3

IIUC
v=df.text.str.split(' ')
s=pd.DataFrame({'text':v.sum(),'label':df.label.repeat(v.str.len())})
s['New']=s.groupby(s.index).cumcount()
s.groupby([s.New//3,s.index.get_level_values(level=0)]).agg({'text':lambda x : ' '.join(x),'label':'first'}).sort_index(level=1)
Out[1785]:
text label
New
0 0 this is long 1
1 0 string with many 1
2 0 words 1
0 1 this is a 0
1 1 middle string 0
0 2 short string 1

This is one solution, using a couple of for loops to split your text into sets of 3:
array = []
for ii,row in df.iterrows():
if row['text'].split() > 3:
jj = 0
while jj < len(row['text'].split()):
array.append(
pd.Series(
{'id':row['id'],'label':row['label'],
'text':row['text'].split()[jj:jj+3]}
)
)
jj += 3
else:
array.append(row)

Number Pyramid Nested for Loop

I'm wondering if you could help me out. I'm trying to write a nested for loop in Python 3 that displays a number pyramid that looks like;
1
1 2 1
1 2 4 2 1
1 2 4 8 4 2 1
1 2 4 8 16 8 4 2 1
1 2 4 8 16 32 16 8 4 2 1
1 2 4 8 16 32 64 32 16 8 4 2 1
1 2 4 8 16 32 64 128 64 32 16 8 4 2 1
Can anybody help me out? It would be much appreciated!
This is what I have so far:
col = 1
for i in range(-1, 18, col*2):
for j in range(1, 0, 1):
print(" ", end = "")
for j in range(i, 0, -2):
print(j, end = " ")
print()
So, I can only get half of the pyramid to display.
I guess the main problems I'm having is:
How do i get the output to display an increasing and then decreasing value (ie. 1, 2, 4, 2, 1)?

An alternate way using list comprehensions.
Always break the problem down into digestable chunks. Each line is a mirror of itself, so lets just deal with first making out set of numbers we need.
This generates a list of strings that hold all powers of two which is what this is generating
lines = []
for i in range(1,9):
lines.append([str(2**j) for j in range(i)])
But if we just print this list, a) its going to only have half, and b) its going to mush the numbers together. We need to buffer the numbers with spaces. Fortunately, the last row will have the largest digits for any column, so:
Firstly, how long does each line need to end up being (we need this later) and also, what is the longest number in each column. We can use len as we cast the numbers to strings above.
b = len(lines[-1])
buffers = [len(x) for x in lines[-1]]
Now I have everything I need to print the strings (we stopped using numbers above):
So, for each line, find out how long it is, and expand the array it to the length of the longest line by filling the left of the array with empty strings (for this we're still pretending we're only printing the left half of the triangle):
for line in lines:
l = len(line)
line = [" "]*(b-len(line)) + line
With each line now buffered, we'll make a new array that we will print from. By zip()ing together the line and the buffer, we can easily right justify (String.rjust()) numberic strings, expanded out to the length required.
out = []
for x,y in zip(line,buffers):
out.append(x.rjust(y))
Remmeber until now, we've still just been working with the left half of the pyramid. So we take the output array, reverse it (array[::-1]) and then take every element but the first (array[1:]) and join it all together with a string and print it out.
print(" ".join(out+out[::-1][1:]))
Voila! The completed code:
lines = []
for i in range(1,9):
lines.append([str(2**j) for j in range(i)])
b = len(lines[-1])
buffers = [len(x) for x in lines[-1]]
for line in lines:
l = len(line)
line = [" "]*(b-len(line)) + line
out = []
for x,y in zip(line,buffers):
out.append(x.rjust(y))
print(" ".join(out+out[::-1][1:]))
Output:
1
1 2 1
1 2 4 2 1
1 2 4 8 4 2 1
1 2 4 8 16 8 4 2 1
1 2 4 8 16 32 16 8 4 2 1
1 2 4 8 16 32 64 32 16 8 4 2 1
1 2 4 8 16 32 64 128 64 32 16 8 4 2 1

height = 8
maxHeight = height - 1
for i in range(height):
k, Max = 1, i * 2 + 1
print(maxHeight * " ", end="")
maxHeight -= 1
for j in range(Max):
print("%5d" % k, end="")
if (j < (Max // 2)):
k *= 2
else:
k //= 2
print()
Output:
1
1 2 1
1 2 4 2 1
1 2 4 8 4 2 1
1 2 4 8 16 8 4 2 1
1 2 4 8 16 32 16 8 4 2 1
1 2 4 8 16 32 64 32 16 8 4 2 1
1 2 4 8 16 32 64 128 64 32 16 8 4 2 1

This could be the other 9 line solution.
Generate power of two's numbers as series
Find the offset need to add in each rows
Print the empty space for the each row before printing the palindromic list.
Ie. (offset * (n - i)) times " "(empty space)
Build palindromic series by slice operation ie. temp + temp[::-1][1:]
Print the palindromic series and offset spaces relative to the length of the number you are printing.
Code:
n = 8
numbers = [2**x for x in range(n)] # Generate interseted series.
offset = len(str(numbers[-1:])) -1 # Find the max offset for the tree.
for i in range(1, n+1): # Iterate n times. 1 to n+1 helps eazy slicing.
temp = numbers[:i] # Slice series to get first row numbers.
print(' ' * (offset * (n - i)), end=" ") # Prefix spaces, multiples of offset.
for num in temp + temp[::-1][1:]: # Generate palindromic series for the row.
print(num, end=" " * (offset - len(str(num)))) # Adjust offset for the number.
print('')
output:
1
1 2 1
1 2 4 2 1
1 2 4 8 4 2 1
1 2 4 8 16 8 4 2 1
1 2 4 8 16 32 16 8 4 2 1
1 2 4 8 16 32 64 32 16 8 4 2 1
1 2 4 8 16 32 64 128 64 32 16 8 4 2 1

Grouping list of integers in a range into chunks

Given a set or a list (assume its ordered)
myset = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
I want to find out how many numbers appear in a range.
say my range is 10. Then given the list above, I have two sets of 10.
I want the function to return [10,10]
if my range was 15. Then I should get [15,5]
The range will change. Here is what I came up with
myRange = 10
start = 1
current = start
next = current + myRange
count = 0
setTotal = []
for i in myset:
if i >= current and i < next :
count = count + 1
print str(i)+" in "+str(len(setTotal)+1)
else:
current = current + myRange
next = myRange + current
if next >= myset[-1]:
next = myset[-1]
setTotal.append(count)
count = 0
print setTotal
Output
1 in 1
2 in 1
3 in 1
4 in 1
5 in 1
6 in 1
7 in 1
8 in 1
9 in 1
10 in 1
12 in 2
13 in 2
14 in 2
15 in 2
16 in 2
17 in 2
18 in 2
19 in 2
[10, 8]
notice 11 and 20 where skipped. I also played around with the condition and got wired results.
EDIT: Range defines a range that every value in the range should be counted into one chuck.
think of a range as from current value to currentvalue+range as one chunk.
EDIT:
Wanted output:
1 in 1
2 in 1
3 in 1
4 in 1
5 in 1
6 in 1
7 in 1
8 in 1
9 in 1
10 in 1
11 in 2
12 in 2
13 in 2
14 in 2
15 in 2
16 in 2
17 in 2
18 in 2
19 in 2
[10, 10]

With the right key function, thegroupbymethod in the itertoolsmodule makes doing this fairly simple:
from itertools import groupby
def ranger(values, range_size):
def keyfunc(n):
key = n/(range_size+1) + 1
print '{} in {}'.format(n, key)
return key
return [len(list(g)) for k, g in groupby(values, key=keyfunc)]
myset = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
print ranger(myset, 10)
print ranger(myset, 15)

You want to use simple division and the remainder; the divmod() function gives you both:
def chunks(lst, size):
count, remainder = divmod(len(lst), size)
return [size] * count + ([remainder] if remainder else [])
To create your desired output, then use the output of chunks():
lst = range(1, 21)
size = 10
start = 0
for count, chunk in enumerate(chunks(lst, size), 1):
for i in lst[start:start + chunk]:
print '{} in {}'.format(i, count)
start += chunk
count is the number of the current chunk (starting at 1; python uses 0-based indexing normally).
This prints:
1 in 1
2 in 1
3 in 1
4 in 1
5 in 1
6 in 1
7 in 1
8 in 1
9 in 1
10 in 1
11 in 2
12 in 2
13 in 2
14 in 2
15 in 2
16 in 2
17 in 2
18 in 2
19 in 2
20 in 2

If you don't care about what numbers are in a given chunk, you can calculate the size easily:
def chunk_sizes(lst, size):
complete = len(lst) // size # Number of `size`-sized chunks
partial = len(lst) % size # Last chunk
if partial: # Sometimes the last chunk is empty
return [size] * complete + [partial]
else:
return [size] * complete