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Python Welsh-Powell graph coloring

I have implemented algorithm for Welsh-Powell graph coloring.
Task is that you have txt document with lines of 2 numbers, in which the first one represent vertex and the second one represent its neighbor.
Looks like this:
1 2
1 4
1 5
1 8
1 9
1 11
2 3
2 4
2 5
2 6
2 7
2 8
2 10
2 11
2 12
3 4
3 5
3 6
3 7
3 8
3 9
3 11
4 6
4 9
4 10
4 11
4 12
5 6
5 7
5 10
5 11
5 12
6 7
6 10
6 11
6 12
7 8
7 10
7 11
7 12
8 10
8 11
8 12
9 10
9 12
10 12
Output should be also txt document with two numbers in line, where the first one is vertex and the second one is number of color like this:
1 2
2 1
3 2
4 3
5 3
6 4
7 5
8 3
9 1
10 2
11 6
12 6
The number of used colors is not important, as it shouldnt be more than all vertexes.
This is graphColoring.py
def file_to_graph(filename):
file1 = open(filename, 'r')
lines = file1.readlines()
graph = [[], []]
for line in lines:
if line == '':
continue
row = line.split(' ')
from_vertex = int(row[0])
to_vertex = int(row[1])
found_from_vertex = -1
found_to_vertex = -1
for i in range(len(graph[0])):
if graph[0][i][0] == from_vertex:
graph[0][i][1] += 1
found_from_vertex = i
if graph[0][i][0] == to_vertex:
graph[0][i][1] += 1
found_to_vertex = i
if found_from_vertex == -1:
graph[0].append([from_vertex, 1])
if found_to_vertex == -1:
graph[0].append([to_vertex, 1])
graph[1].append((from_vertex, to_vertex))
return graph
def get_color(colored_vertexes, vertex):
for colored_vertex in colored_vertexes:
if colored_vertex[0] == vertex:
return colored_vertex[1]
return -1
def color_graph(graph, max_colors=-1):
sorted_vertexes = sorted(graph[0], key=lambda x: x[1])
colored_vertexes = []
for vertex_with_edges in reversed(sorted_vertexes):
colored_vertexes.append([vertex_with_edges[0], 0])
actual_color = 1
while True:
for vertex_with_edge_count in reversed(sorted_vertexes):
color = 0
colored_vertex_index = -1
vertex_name = vertex_with_edge_count[0]
for i in range(len(colored_vertexes)):
if colored_vertexes[i][0] == vertex_name:
color = colored_vertexes[i][1]
colored_vertex_index = i
break
if color != 0:
continue
color_of_neighbours = []
for (from_vertex, to_vertex) in graph[1]:
if from_vertex == vertex_name:
color_of_neighbours.append(get_color(colored_vertexes, to_vertex))
if to_vertex == vertex_name:
color_of_neighbours.append(get_color(colored_vertexes, from_vertex))
if actual_color not in color_of_neighbours:
colored_vertexes[colored_vertex_index][1] = actual_color
actual_color += 1
is_colored = True
for colored_vertex in colored_vertexes:
if colored_vertex[1] == 0:
is_colored = False
break
if is_colored:
break
return colored_vertexes
And this is main.py
from graphColoring import file_to_graph, color_graph
import sys
if __name__ == '__main__':
graph = file_to_graph(sys.argv[1])
colored_vertexes = color_graph(graph)
for colored_vertex in sorted(colored_vertexes, key=lambda x: x[0]):
print(str(colored_vertex[0]) + " " + str(colored_vertex[1]))
EDIT//: I received a solution for problem with input, but I dont have expected output. Instead of expected output above, i received this:
1 2
2 1
3 2
4 4
5 6
6 5
7 4
8 5
9 1
10 3
11 3
12 2
//
I should use "python main.py ./input.txt" to input txt document, but after running the program it says:
line 7, in <module>
graph = file_to_graph(sys.argv[1])
~~~~~~~~^^^
IndexError: list index out of range
I˙m not so good at programming and all of this is really hard for me to understand so I really appreciate some help to make this functional.

For me it is working. And it also gives your desired output
1 1
2 2
3 1
4 2
I think it is about how you run the script. Because when I run
python .\main.py I also get
Traceback (most recent call last):
File "C:\Users\fhdwnig\Desktop\main.py", line 5, in <module>
graph = file_to_graph(sys.argv[1])
~~~~~~~~^^^
you have to run it like this:
python .\main.py .\input.txt
(on windows)
The script is missing the input file.

change this line
sorted_vertexes = sorted(graph[0], key=lambda x: x[1])
to this one
sorted_vertexes = sorted(sorted(graph[0][::-1],key=lambda x:x[0], reverse=True), key=lambda x: x[1])

Drop rows if value in column changes

Assume I have the following pandas data frame:
my_class value
0 1 1
1 1 2
2 1 3
3 2 4
4 2 5
5 2 6
6 2 7
7 2 8
8 2 9
9 3 10
10 3 11
11 3 12
I want to identify the indices of "my_class" where the class changes and remove n rows after and before this index. The output of this example (with n=2) should look like:
my_class value
0 1 1
5 2 6
6 2 7
11 3 12

My approach:
# where class changes happen
s = df['my_class'].ne(df['my_class'].shift(-1).fillna(df['my_class']))
# mask with `bfill` and `ffill`
df[~(s.where(s).bfill(limit=1).ffill(limit=2).eq(1))]
Output:
my_class value
0 1 1
5 2 6
6 2 7
11 3 12

One of possible solutions is to:
Make use of the fact that the index contains consecutive integers.
Find index values where class changes.
For each such index generate a sequence of indices from n-2
to n+1 and concatenate them.
Retrieve rows with indices not in this list.
The code to do it is:
ind = df[df['my_class'].diff().fillna(0, downcast='infer') == 1].index
df[~df.index.isin([item for sublist in
[ range(i-2, i+2) for i in ind ] for item in sublist])]

my_class = np.array([1] * 3 + [2] * 6 + [3] * 3)
cols = np.c_[my_class, np.arange(len(my_class)) + 1]
df = pd.DataFrame(cols, columns=['my_class', 'value'])
df['diff'] = df['my_class'].diff().fillna(0)
idx2drop = []
for i in df[df['diff'] == 1].index:
idx2drop += range(i - 2, i + 2)
print(df.drop(idx_drop)[['my_class', 'value']])
Output:
my_class value
0 1 1
5 2 6
6 2 7
11 3 12

Print a number table in a simple format

I am stuck trying to print out a table in Python which would look like this (first number stands for amount of numbers, second for amount of columns):
>>> print_table(13,4)
0 1 2 3
4 5 6 7
8 9 10 11
12 13
Does anyone know a way to achieve this?

This is slightly more difficult than it sounds initially.
def numbers(n, r):
print('\n'.join(' '.join(map(str, range(r*i, min(r*(i + 1), n + 1)))) for i in range(n//r + 1)))
numbers(13, 4)
#>>> 0 1 2 3
4 5 6 7
8 9 10 11
12 13

def numbers(a,b):
i=0;
c=0;
while i<=a:
print(i,end="") #prevents printing a new line
c+=1
if c>=b:
print("\n") #prints a new line when the number of columns is reached and then reset the current column number
c=0;
I think it should work

def num2(n=10, r=3):
print('\n'.join(' '.join(tuple(map(str, range(n+1)))[i:i+r]) for i in range(0, n+1, r)))
<<<
0 1 2
3 4 5
6 7 8
9 10

write a program that prints a nested loop in Python

I'm trying to print a nested loops that looks like this:
1 2 3 4
5 6 7 8
9 10 11 12
This is what I have so far:
def main11():
for n in range(1,13)
print(n, end=' ')
however, this prints the numbers in one line: 1 2 3 4 5 6 7 8 9 10 11 12

You can do that using string formatting:
for i in range(1,13):
print '{:2}'.format(i),
if i%4==0: print
[OUTPUT]
1 2 3 4
5 6 7 8
9 10 11 12

Modulus Operator (%)
for n in range(1,13):
print(n, end=' ')
if n%4 == 0:
print

for offset in range(3):
for i in range(1,5):
n = offset*4 + i
print(n, end=' ')
print()
Output:
1 2 3 4
5 6 7 8
9 10 11 12
Or if you want it nicely formatted the way you have in your post:
for offset in range(3):
for i in range(1,5):
n = offset*4 + i
print("% 2s"%n, end=' ')
print()
Output:
1 2 3 4
5 6 7 8
9 10 11 12

Most of the time when you write a for loop, you should check if this is the right implementation.
From the requirements you have, I would write something like this:
NB_NB_INLINE = 4
MAX_NB = 12
start = 1
while start < MAX_NB:
print( ("{: 3d}" * NB_NB_INLINE).format(*tuple( j+start for j in range(NB_NB_INLINE))) )
start += NB_NB_INLINE

Aligning items from 2 lists(simple python)

I was wondering how I could align every item in one list, to the corresponding index in the second list. Here is my code so far:
letters=['a','ab','abc','abcd','abcde','abcdef','abcdefg','abcdefgh','abcdefghi','abcdefghij']
numbers=[1,2,3,4,5,6,7,8,9,10]
for x in range(len(letters)):
print letters[x]+"----------",numbers[x]
This is the output I get:
a---------- 1
ab---------- 2
abc---------- 3
abcd---------- 4
abcde---------- 5
abcdef---------- 6
abcdefg---------- 7
abcdefgh---------- 8
abcdefghi---------- 9
abcdefghij---------- 10
This is the output I want:
a---------- 1
ab--------- 2
abc-------- 3
abcd------- 4
abcde------ 5
abcdef----- 6
abcdefg---- 7
abcdefgh--- 8
abcdefghi-- 9
abcdefghij- 10

You could use string formatting:
for left, right in zip(letters, numbers):
print '{0:-<12} {1}'.format(left, right)
And the output:
a----------- 1
ab---------- 2
abc--------- 3
abcd-------- 4
abcde------- 5
abcdef------ 6
abcdefg----- 7
abcdefgh---- 8
abcdefghi--- 9
abcdefghij-- 10

Something like this using string.formatting:
def solve(letters,numbers):
it=iter(range( max(numbers) ,0,-1))
for x,y in zip(letters,numbers):
print "{0}{1} {2}".format(x,"-"*next(it),y)
....:
In [38]: solve(letters,numbers)
a---------- 1
ab--------- 2
abc-------- 3
abcd------- 4
abcde------ 5
abcdef----- 6
abcdefg---- 7
abcdefgh--- 8
abcdefghi-- 9
abcdefghij- 10

letters=['a','ab','abc','abcd','abcde','abcdef','abcdefg','abcdefgh','abcdefghi','abcdefghij']
for c,x in enumerate(letters, start=1):
print x+("-"*(10-c))+" %s" % c
a--------- 1
ab-------- 2
abc------- 3
abcd------ 4
abcde----- 5
abcdef---- 6
abcdefg--- 7
abcdefgh-- 8
abcdefghi- 9
abcdefghij 10

letters=['a','ab','abc','abcd','abcde','abcdef','abcdefg','abcdefgh','abcdefghi','abcdefghij']
numbers=[1,2,3,4,5,6,7,8,9,10]
for x in range(len(letters)):
print '{0:11}{1}'.format(letters[x],numbers[x]).replace(' ','-');
a----------1
ab---------2
abc--------3
abcd-------4
abcde------5
abcdef-----6
abcdefg----7
abcdefgh---8
abcdefghi--9
abcdefghij-10