Develop Reference

printing a number table of square

n: 8
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
How to print a number table like this in python with n that can be any number?
I am using a very stupid way to print it but the result is not the one expected:
n = int(input('n: '))
if n == 4:
print(' 0 1 2 3\n4 5 6 7\n8 9 10 11\n12 13 14 15')
if n == 5:
print(' 0 1 2 3 4\n5 6 7 8 9\n10 11 12 13 14\n15 16 17 18 19\n20 21 22 23 24')
if n == 6:
print(' 0 1 2 3 4 5\n6 7 8 9 10 11\n12 13 14 15 16 17\n18 19 20 21 22 23\n24 25 26 27 28 29\n30 31 32 33 34 35')
if n == 7:
print(' 0 1 2 3 4 5 6\n7 8 9 10 11 12 13\n14 15 16 17 18 19 20\n21 22 23 24 25 26 27\n28 29 30 31 32 33 34\n35 36 37 38 39 40 41\n42 43 44 45 46 47 48')
if n == 8:
print(' 0 1 2 3 4 5 6 7\n8 9 10 11 12 13 14 15\n16 17 18 19 20 21 22 23\n24 25 26 27 28 29 30 31\n32 33 34 35 36 37 38 39\n40 41 42 43 44 45 46 47\n48 49 50 51 52 53 54 55\n56 57 58 59 60 61 62 63')
if n == 9:
print(' 0 1 2 3 4 5 6 7 8\n9 10 11 12 13 14 15 16 17\n18 19 20 21 22 23 24 25 26\n27 28 29 30 31 32 33 34 35\n36 37 38 39 40 41 42 43 44\n45 46 47 48 49 50 51 52 53\n54 55 56 57 58 59 60 61 62\n63 64 65 66 67 68 69 70 71\n72 73 74 75 76 77 78 79 80')
if n == 10:
print(' 0 1 2 3 4 5 6 7 8 9\n10 11 12 13 14 15 16 17 18 19\n20 21 22 23 24 25 26 27 28 29\n30 31 32 33 34 35 36 37 38 39\n40 41 42 43 44 45 46 47 48 49\n50 51 52 53 54 55 56 57 58 59\n60 61 62 63 64 65 66 67 68 69\n70 71 72 73 74 75 76 77 78 79\n80 81 82 83 84 85 86 87 88 89\n90 91 92 93 94 95 96 97 98 99')
here is the result:
n: 8
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

I won't show you the code directly, here is some tips for you. Do you know % operator in python? And how to use it to break lines. As for the format, zfill function will help you. You may need to learn for or while statement to solve your problem

You can do this with a range loop and a list comprehension.
In order for the output to look right you need to figure out what the width of the largest value in the square will be. You then need to format each value to fit in that width (right-justified). Something like this:
def number_square(n):
w = len(str(n*n-1))
for r in range(n):
print(*[f'{c:>{w}}' for c in range(r*n, r*n+n)])
number_square(8)
Output:
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

Web Scraping School Project

I need help with a school project. The code that I have "#" I can't seem to get to work with the table I scraped. I need to change it into a data frame. Can anyone see what I'm missing and if I am missing a step.
Tertiary=pd.read_html("https://en.wikipedia.org/wiki/List_of_countries_by_tertiary_education_attainment")
Tertiary=pd.DataFrame(Tertiary[1])
#Tertiary=Tertiary.drop(["Non-OECD"], axis=1, inplace=True)
print(Tertiary.dtypes)
#Tertiary["Age25-64(%)"] = pd.to_numeric(Tertiary["Age25-64(%)"])
#Tertiary["Age"] = pd.to_numeric(Tertiary["Age"])
print(Tertiary.dtypes)
print()
#print(Tertiary.describe)
print()
#print(Tertiary.isnull().sum())
#print(Tertiary)

Everything works fine for me.
import pandas as pd
df = pd.read_html("https://en.wikipedia.org/wiki/List_of_countries_by_tertiary_education_attainment")
table = pd.DataFrame(df[1])
print(table)
print(table.columns)
Output:
Country Age 25–64 (%) Age Year Non-OECD
Country Age 25–64 (%) 25–34 (%) 35–44 (%) 45–54 (%) 55–64 (%) Year Non-OECD
0 Australia 42 48 46 38 33 2014 NaN
1 Austria 30 38 33 27 21 2014 NaN
2 Belgium 37 44 42 34 26 2014 NaN
3 Brazil 14 15 14 14 11 2013 NaN
4 Canada 54 58 61 51 45 2014 NaN
5 Chile 21 27 24 17 14 2013 NaN
6 China 17 27 15 7 2 2018 NaN
7 Colombia 22 28 23 18 16 2014 NaN
8 Costa Rica 18 21 19 17 17 2014 NaN
9 Czech Republic 22 30 21 20 15 2014 NaN
10 Denmark 36 42 41 33 29 2014 NaN
11 Estonia 38 40 39 35 36 2014 NaN
12 Finland 42 40 50 44 34 2014 NaN
13 France 32 44 39 26 20 2013 NaN
14 Germany 27 28 29 26 25 2014 NaN
15 Greece 28 39 27 26 21 2014 NaN
16 Hungary 23 32 25 20 17 2014 NaN
17 Iceland 37 41 42 36 29 2014 NaN
18 Indonesia 8 10 9 8 4 2011 NaN
19 Ireland 41 51 49 34 24 2014 NaN
20 Israel 49 46 53 48 47 2014 NaN
21 Italy 17 24 19 13 12 2014 NaN
22 Japan 48 59 53 47 35 2014 NaN
23 Latvia 30 39 31 27 23 2014 NaN
24 Lithuania 37 53 38 30 28 2014 NaN
25 Luxembourg 46 53 56 40 32 2014 NaN
26 Mexico 19 25 17 16 13 2014 NaN
27 Netherlands 34 44 38 30 27 2014 NaN
28 New Zealand 36 40 41 32 29 2014 NaN
29 Norway 42 49 49 36 32 2014 NaN
30 Poland 27 43 32 18 14 2014 NaN
31 Portugal 22 31 26 17 13 2014 NaN
32 Russia 54 58 55 53 50 2013 NaN
33 Saudi Arabia 22 26 22 18 14 2013 NaN
34 Slovakia 20 30 21 15 14 2014 NaN
35 Slovenia 29 38 35 24 18 2014 NaN
36 South Africa 7 5 7 8 7 2012 NaN
37 South Korea 45 68 56 33 17 2014 NaN
38 Spain 35 41 43 30 21 2014 NaN
39 Sweden 39 46 46 32 30 2014 NaN
40 Switzerland 40 46 45 38 31 2014 NaN
41 Turkey 17 25 16 10 10 2014 NaN
42 Taiwan[3] 45 X X X X 2015 NaN
43 United Kingdom 42 49 46 38 35 2014 NaN
44 United States 44 46 47 43 41 2014 NaN
__
MultiIndex([( 'Country', 'Country'),
('Age 25–64 (%)', 'Age 25–64 (%)'),
( 'Age', '25–34 (%)'),
( 'Age', '35–44 (%)'),
( 'Age', '45–54 (%)'),
( 'Age', '55–64 (%)'),
( 'Year', 'Year'),
( 'Non-OECD', 'Non-OECD')],
)

Calculating the Derivatives of Curvatures on surface

I want to calculate the derivatives of curvatures on surface using vtk and python. I first calculate the curvatures using:
curvatures = vtk.vtkCurvatures()
curvatures.SetInputConnection(reader.GetOutputPort())
curvatures.SetCurvatureTypeToGaussian()
and calculate the derivative of curvatures using:
Derivativers = vtk.vtkCellDerivatives()
Derivativers.SetInputConnection(curvatures.GetOutputPort())
It seems that the results are the same with vtkCurvatures and vtkCellDerivatives.
What should I do to get the derivative of curvature on a surface. Many thanks!

I think your code is correct as it is. But we need to be sure that the curvature point data array is the currently active scalar array. I have attached a input data file that you can save with name 'Test.vtk'. It has two point data arrays -- PointIds (a scalar array) and PointNormals( a vector array). Then we will calculate Gaussian curvatures which will become the third array of scalars of the point data. We will print names of all the point data arrays irrespective of whether they are scalars or vectors. Then we will explicitly set the 'Gauss_Curvature' scalar array as the Active Scalar. We will compute Cell Derivatives which will create a Cell Data Vector array called 'ScalarGradient' which will be the gradient of the curvatures. This will be saved in a file 'Output.vtk'
import vtk
rd = vtk.vtkPolyDataReader()
rd.SetFileName('Test.vtk')
curv = vtk.vtkCurvatures()
curv.SetInputConnection(rd.GetOutputPort())
curv.SetCurvatureTypeToGaussian()
curv.Update()
pd = curv.GetOutput()
for i in range(pd.GetPointData().GetNumberOfArrays()):
print(pd.GetPointData().GetArrayName(i))
# This will print the following:
# PointIds
# PointNormals
# Gauss_Curvature
# To set the active scalar to Gauss_Curvature
pd.GetPointData().SetActiveScalars('Gauss_Curvature')
curvdiff = vtk.vtkCellDerivatives()
curvdiff.SetInputData(pd)
curvdiff.SetVectorModeToComputeGradient()
curvdiff.Update()
writer = vtk.vtkPolyDataWriter()
writer.SetFileName('Output.vtk')
writer.SetInputConnection(curvdiff.GetOutputPort())
writer.Write()
gives me the following outputs -- first for the curvature and then the gradient. Notice that the color scale in the two figures are different. So the curvature and derivative values are different although the color scheme makes them look similar.
In case you want to reproduce the results, the input vtk file is as below
# vtk DataFile Version 4.2
vtk output
ASCII
DATASET POLYDATA
POINTS 72 double
2.0927648978 0.33091989273 -0.39812666792 1.6450815105 0.64303293033 -1.236079764 1.7000810807 1.2495041516 -0.44287861593
1.0622264471 1.4540269048 -1.1853937884 0.8533187462 0.72833963362 -1.8409362444 0.161573121 1.415272931 -1.6182009866
-0.4682233113 2.0970647997 -0.17539653223 0.30090053169 1.9778473 -0.80327873468 -0.62604403311 1.746197318 -1.0984268611
0.62604948422 1.746195345 1.0984268742 0.4682298575 2.0970633231 0.17539654742 -0.30089435724 1.9778482191 0.80327874624
1.3794219731 1.1031586743 1.2360880686 1.9321437012 0.84755424016 0.44288858377 1.3329709879 1.6469225081 0.39813606858
-1.3329658439 1.6469266769 -0.39813605266 -1.3794185207 1.1031629885 -1.2360880529 -1.9321410548 0.84756028031 -0.44288857482
-0.16156870247 1.4152734137 1.6182009959 -1.0622219128 1.4540302146 1.1853938087 -0.85331647216 0.72834227646 1.8409362479
-1.7000771766 1.2495094572 0.44287862867 -2.0927638628 0.33092642637 0.39812667143 -1.6450795106 0.64303805991 1.2360797754
0.10502897512 0.5677157381 2.0771002606 -0.54417928828 -0.19289519204 2.0770984773 0.43913323132 -0.37482057542 2.077101172
1.0574135878 0.37481822068 1.8409414841 1.3064404335 -0.56771795917 1.6182050108 1.7903331906 0.19289323113 1.1854016225
-0.72812102639 -1.6469234624 1.18539471 -0.20411225533 -1.1031605232 1.8409380189 -1.1448850389 -0.84755547744 1.6181982897
0.26564737208 -1.7461967516 1.236085002 -0.23207016686 -2.0970637037 0.44288263714 0.75978960067 -1.9778489401 0.39813448025
1.1992202745 -1.4152750453 1.0984284306 1.5819944619 -1.4540310306 0.17539958384 1.8633106814 -0.72834386503 0.80328466622
-1.825278792 -0.33092031521 1.0984201446 -2.0502257619 -0.64303229501 0.17538963068 -1.5624229303 -1.2495043655 0.80327527281
-0.26565282447 -1.7461959014 -1.2360850131 0.23206361633 -2.0970644256 -0.44288265596 -0.7597957797 -1.977846564 -0.39813449851
-1.1992246997 -1.4152712955 -1.0984284473 -1.5819990123 -1.4540260972 -0.17539960215 -1.8633129661 -0.72833804688 -0.80328468018
0.20410881451 -1.1031611451 -1.8409380327 1.1448823984 -0.84755903977 -1.6181983017 0.72811588321 -1.6469257176 -1.1853947189
2.0502237661 -0.64303869999 -0.17538964133 1.5624190405 -1.2495092418 -0.80327529169 1.8252777661 -0.33092600698 -1.0984201511
-0.43913440065 -0.37481918558 -2.0771011678 -0.10502720377 0.56771608521 -2.0771002475 0.54417868626 -0.19289687027 -2.0770984714
-1.3064422115 -0.56771386838 -1.6182050202 -1.7903325818 0.19289882961 -1.185401614 -1.057412421 0.3748215375 -1.8409414839
-0.76083174443 1.3178134523 -1.9919051229 -0.7608358562 -1.3178110596 -1.9919051353 -2.4621262785 3.8465962003e-06 -0.47023127203
1.5216839818 -2.3645462409e-06 -1.991898872 2.4621262803 -3.846902628e-06 0.47023127288 1.2310617434 -2.1322669408 -0.47022115796
-1.2310684033 -2.1322631023 0.47022113869 -1.5216839821 2.3661982943e-06 1.9918988726 0.76083174316 -1.3178134534 1.9919051234
0.76083585779 1.317811059 1.9919051359 -1.2310617441 2.1322669425 0.47022115881 1.2310684021 2.1322631008 -0.47022113785
POLYGONS 140 560
3 12 14 9
3 27 69 24
3 70 21 19
3 1 53 63
3 2 14 13
3 38 36 37
3 28 68 36
3 39 67 23
3 64 38 51
3 13 14 12
3 20 24 18
3 34 35 33
3 40 41 39
3 16 58 17
3 20 18 19
3 26 27 24
3 11 6 70
3 10 14 71
3 22 39 23
3 6 10 7
3 3 5 7
3 29 64 13
3 41 30 32
3 57 45 47
3 54 61 57
3 66 30 41
3 50 43 42
3 30 33 31
3 33 35 36
3 65 37 35
3 37 36 35
3 26 68 28
3 68 33 36
3 27 28 29
3 28 36 38
3 29 28 38
3 38 37 51
3 61 48 42
3 37 65 52
3 66 34 30
3 43 65 35
3 32 30 31
3 30 34 33
3 40 39 22
3 41 32 39
3 66 41 46
3 32 67 39
3 67 32 25
3 33 68 31
3 32 31 25
3 31 26 25
3 27 26 28
3 26 31 68
3 64 29 38
3 12 69 27
3 18 9 11
3 69 12 9
3 18 24 69
3 20 67 25
3 26 24 25
3 24 20 25
3 13 12 29
3 12 27 29
3 18 11 19
3 11 9 10
3 69 9 18
3 9 14 10
3 70 6 15
3 11 10 6
3 10 71 7
3 71 14 2
3 70 15 21
3 6 8 15
3 21 17 22
3 15 8 16
3 20 23 67
3 19 11 70
3 21 23 19
3 23 20 19
3 22 17 62
3 22 23 21
3 15 17 21
3 62 40 22
3 58 57 47
3 62 17 58
3 62 47 40
3 58 16 59
3 15 16 17
3 6 7 8
3 16 60 59
3 48 54 56
3 8 5 60
3 2 1 3
3 7 5 8
3 3 1 4
3 2 3 71
3 71 3 7
3 3 4 5
3 5 55 60
3 49 50 48
3 8 60 16
3 60 55 59
3 4 55 5
3 54 57 59
3 1 63 4
3 56 55 4
3 49 48 56
3 44 45 42
3 63 56 4
3 48 61 54
3 56 54 55
3 54 59 55
3 59 57 58
3 47 62 58
3 40 46 41
3 57 61 45
3 47 45 46
3 43 34 44
3 47 46 40
3 46 44 66
3 37 52 51
3 42 48 50
3 42 43 44
3 43 35 34
3 45 44 46
3 44 34 66
3 61 42 45
3 50 65 43
3 65 50 52
3 56 63 49
3 51 52 53
3 49 63 53
3 50 49 52
3 49 53 52
3 2 0 1
3 1 0 53
3 0 51 53
3 0 64 51
3 13 64 0
3 2 13 0
POINT_DATA 72
SCALARS PointIds vtkIdType
LOOKUP_TABLE default
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
NORMALS PointNormals double
0.94738623196 0.18727650058 -0.25958975291 0.78313719053 0.35367076216 -0.51148131227 0.83545291047 0.50824408436 -0.20906072109
0.47898857295 0.62402000487 -0.61738884061 0.34465195337 0.40584589543 -0.84646567573 0.15649087604 0.66776200195 -0.72773931766
-0.15609353126 0.97764567412 -0.14086782943 0.059136449433 0.91410106494 -0.40115099829 -0.27742338135 0.85504231805 -0.43810832201
0.27739675558 0.85505949665 0.43809165386 0.1561128187 0.97764027026 0.14088395868 -0.05910174957 0.91410169764 0.40115467037
0.6978536347 0.50139725414 0.51146954756 0.85786633279 0.46941626794 0.20907826874 0.63588503517 0.72681646701 0.25959207486
-0.63587825439 0.72682167945 -0.25959409059 -0.69785435483 0.50138010962 -0.51148537136 -0.85787788306 0.46940090711 -0.20906536337
-0.15651072102 0.66775823558 0.72773850593 -0.47897825964 0.62400201859 0.61741502054 -0.34463970914 0.40587327082 0.84645753521
-0.8354604399 0.50822639825 0.20907362693 -0.94738511041 0.18728497536 0.25958773195 -0.78315168201 0.35366155935 0.51146548701
-0.0042059530133 0.19834561529 0.98012311821 -0.16967339936 -0.10281294988 0.98012266318 0.17387129188 -0.095532679284 0.98012360499
0.52381294065 0.095528092331 0.84645991446 0.65654796659 -0.19833859028 0.72774073074 0.77988819356 0.10280486141 0.61741846913
-0.30091501143 -0.72680688867 0.6174155023 -0.17918019275 -0.50140137515 0.8464579845 -0.50004323045 -0.46941965564 0.72773755886
0.085293841035 -0.8550588207 0.51146786196 -0.022407612419 -0.97764266904 0.20907584885 0.31149794024 -0.91410144601 0.25959117785
0.60180306843 -0.66776342878 0.4380925359 0.76860432156 -0.62401813766 0.14088563004 0.82118451863 -0.40586818759 0.40115707731
-0.87920342686 -0.18729607405 0.43808847833 -0.92471829912 -0.35362213463 0.14088098936 -0.76208606823 -0.50823351579 0.40115273655
-0.08527862936 -0.85505086473 -0.51148369876 0.022426691996 -0.97764498848 -0.20906295701 -0.31150584284 -0.91409817614 -0.25959320919
-0.60177487516 -0.6677778914 -0.43810921854 -0.76861864432 -0.62400413386 -0.14086951598 -0.82120131764 -0.40583781416 -0.40115341765
0.17915035572 -0.50139827931 -0.84646613373 0.5000564084 -0.4694043424 -0.72773838139 0.30092542605 -0.72682480482 -0.61738933507
0.92471333385 -0.3536415386 -0.14086487277 0.7620681671 -0.50826324483 -0.40114907784 0.8792018588 -0.18726442077 -0.43810515655
-0.17386795802 -0.09554490851 -0.98012300434 0.0041936861197 0.19834885542 -0.98012251507 0.16968233044 -0.10280393451 -0.9801220627
-0.65654129971 -0.19835764899 -0.72774155087 -0.77990892195 0.10280480888 -0.61739229403 -0.52379534825 0.095552395563 -0.84646805779
-0.3035934934 0.52568869256 -0.79465866212 -0.30345974591 -0.52576590314 -0.79465866742 -0.98224561547 8.0181630296e-06 -0.18759944248
0.60705977261 7.7220189155e-05 -0.79465616874 0.98224726985 9.1023794904e-07 0.18759078032 0.49111663606 -0.85065410882 -0.18759540755
-0.49112361722 -0.85065199031 0.18758673723 -0.60706560357 -2.8418937296e-07 0.79465171803 0.30352977695 -0.52573221141 0.79465421184
0.30352929122 0.52573248433 0.79465421681 -0.49112519531 0.85065107744 0.18758674521 0.49113052171 0.85064609373 -0.18759539936

BFS to find all fixed-length cycles

I have to find all cycles of length 3 in a given graph. I've implemented it using BFS, but so far it works only for relatively small inputs. It still works for bigger ones and gives correct answer, but the time it takes to find an answer is extremely high. Is there any way to improve the following code to make it more efficient?
num_res = 0
adj_list = []
cycles_list = []
def bfs_cycles(start):
queue = [(start, [start])]
depth = 0
while queue and depth <= 3:
(vertex, path) = queue.pop(0)
current_set = set(adj_list[vertex]) - set(path)
if start in set(adj_list[vertex]):
current_set = current_set.union([start])
depth = len(path)
for node in current_set:
if node == start:
if depth == 3 and sorted(path) not in cycles_list:
cycles_list.append(sorted(path))
yield path + [node]
else:
queue.append((node, path + [node]))
if __name__ == "__main__":
num_towns, num_pairs = [int(x) for x in input().split()]
adj_list = [[] for x in range(num_towns)]
adj_matrix = [[0 for x in range(num_towns)] for x in range(num_towns)]
# EDGE LIST TO ADJACENCY LIST
for i in range(num_pairs):
cur_start, cur_end = [int(x) for x in input().split()]
adj_list[cur_start].append(cur_end)
adj_list[cur_end].append(cur_start)
num_cycles = 0
for i in range(num_towns):
my_list = list(bfs_cycles(i))
num_cycles += len(my_list)
print(num_cycles)
Examples of inputs:
6 15
5 4
2 0
3 1
5 1
4 1
5 3
1 0
4 0
4 3
5 2
2 1
3 0
3 2
5 0
4 2
(output: 20; works ok)
52 1051
48 5
41 28
12 4
33 27
12 5
1 0
15 12
50 8
33 8
38 28
26 10
13 7
39 18
31 11
48 19
41 19
40 25
47 45
27 16
46 25
42 6
5 4
51 2
30 21
41 27
26 25
33 11
45 26
16 7
23 15
17 6
45 22
32 6
29 8
36 20
30 1
36 25
41 6
46 4
46 40
18 8
38 1
28 5
43 22
21 11
39 14
31 29
18 9
50 35
32 17
48 27
49 40
16 1
49 47
41 12
30 28
33 14
48 12
37 20
49 20
48 8
48 6
27 17
46 44
31 12
17 9
32 27
14 11
40 23
36 19
38 10
42 2
35 22
26 23
29 23
30 11
11 7
47 12
30 13
38 34
48 11
46 8
42 31
30 4
35 17
50 2
51 1
12 10
44 25
47 17
45 24
25 2
45 11
39 21
39 31
9 6
16 3
10 6
15 11
37 2
23 6
41 40
34 26
45 33
35 23
45 36
11 4
38 7
36 6
10 3
33 12
39 12
41 24
47 8
33 5
44 18
45 8
48 41
44 37
11 3
16 6
21 10
20 0
44 36
29 4
43 33
48 4
46 35
33 6
42 12
45 19
12 8
37 15
43 41
36 11
12 11
50 37
9 7
51 30
36 0
33 17
36 35
50 36
49 37
50 16
46 21
36 22
49 15
46 28
50 27
20 10
23 0
36 29
35 33
42 17
31 16
48 47
48 23
17 2
40 14
10 5
45 7
48 42
39 32
51 4
42 8
38 19
34 10
50 5
51 36
46 26
42 38
20 12
44 32
34 4
49 6
50 45
37 10
45 41
38 11
42 30
21 20
43 23
42 26
33 1
17 7
26 6
16 12
44 16
21 9
36 30
39 24
26 4
47 10
18 7
36 12
26 17
28 13
18 11
23 7
44 4
43 26
26 16
22 21
37 0
36 28
34 5
22 17
41 20
31 8
27 25
12 2
42 11
29 28
39 33
34 12
30 2
22 8
40 15
42 9
28 7
44 41
41 35
44 17
12 7
13 10
23 20
48 38
43 12
32 19
43 30
50 1
10 1
17 12
32 2
26 14
29 12
32 5
7 6
36 16
49 7
31 1
45 17
33 29
28 11
32 0
49 32
42 36
16 4
45 20
21 14
39 15
34 18
13 8
27 15
19 11
37 36
36 14
28 4
36 13
17 11
38 13
35 28
50 10
39 28
40 2
35 8
32 24
47 34
45 27
41 21
21 4
47 27
48 1
35 30
21 5
20 14
27 26
17 1
28 17
43 7
31 6
20 3
34 21
8 2
21 1
32 9
29 1
45 43
50 39
19 15
22 12
48 7
46 18
45 35
50 42
51 17
37 6
24 23
29 3
39 20
51 50
38 6
50 11
38 14
25 24
14 7
45 44
28 14
50 49
42 28
36 7
35 25
13 4
46 1
48 21
51 11
39 11
17 5
31 0
49 36
40 4
37 21
35 1
23 4
43 4
46 36
38 20
37 27
30 0
44 34
49 10
48 14
48 45
38 31
47 29
40 16
51 20
34 17
51 19
24 9
24 5
5 1
15 13
26 2
19 12
50 14
42 7
35 14
46 20
43 28
8 3
38 37
28 1
21 0
51 5
17 16
38 17
34 30
46 12
17 14
50 9
16 13
30 27
45 0
41 16
41 32
48 18
30 8
51 47
11 8
40 13
34 32
23 11
51 28
42 35
36 2
13 11
28 8
15 10
39 35
27 1
50 7
41 23
46 39
38 9
44 10
46 38
6 4
44 27
36 21
35 9
45 30
44 7
37 1
44 28
9 1
32 31
39 16
4 0
44 13
24 0
17 15
15 1
32 8
39 22
42 34
24 6
49 18
36 1
51 42
38 5
14 12
33 3
51 45
24 18
37 32
46 6
44 12
23 10
32 12
50 26
29 20
41 30
6 0
48 31
39 8
21 19
47 6
47 16
18 3
46 27
11 10
36 3
47 2
17 10
43 6
36 8
4 1
14 9
42 1
44 1
46 22
44 23
40 26
30 17
21 17
42 29
45 16
49 45
11 6
35 7
46 42
14 10
26 13
49 44
19 18
26 12
46 2
50 41
43 20
38 24
48 30
34 29
25 19
32 11
46 16
30 25
38 15
50 38
51 23
47 28
14 5
40 12
21 8
47 36
38 32
32 15
28 21
45 10
44 8
34 0
32 14
43 25
32 21
38 2
27 2
24 17
33 31
49 26
22 13
13 1
32 20
43 0
46 0
45 29
40 32
48 44
45 34
29 2
39 27
14 8
26 3
40 19
45 38
40 11
34 6
43 39
40 8
35 0
18 0
47 25
21 18
24 8
18 4
25 14
20 11
18 17
24 14
27 23
47 15
38 21
19 2
6 1
46 11
51 38
6 3
31 17
3 0
13 2
41 1
51 14
19 5
39 2
41 22
16 9
22 3
13 0
42 21
24 16
44 31
51 25
40 33
46 29
47 31
51 35
35 18
43 1
47 22
20 18
48 29
39 23
31 25
32 25
22 10
46 24
32 3
46 13
24 15
34 13
50 18
41 4
41 2
43 27
29 10
30 20
32 7
50 20
42 10
42 24
15 7
48 25
41 39
32 1
40 36
20 7
32 13
27 3
34 7
48 34
47 39
39 36
40 5
19 0
25 20
38 12
27 14
44 3
36 4
37 4
33 28
37 23
34 9
46 45
25 9
30 16
34 14
46 37
28 26
26 22
18 5
16 0
36 27
45 42
38 33
37 22
27 0
44 15
49 42
34 23
29 11
30 12
17 8
48 28
10 4
36 15
44 14
23 19
43 18
27 5
40 1
18 12
34 20
50 23
9 3
35 4
46 15
37 11
27 4
19 3
45 1
47 1
48 17
9 2
39 26
33 10
38 30
45 25
48 24
29 17
37 28
34 31
51 21
43 8
31 4
20 16
39 25
31 13
24 3
50 43
13 9
32 23
40 18
45 40
37 35
47 38
42 13
51 26
43 31
49 23
18 15
15 0
43 9
7 2
48 46
35 11
42 23
47 40
3 1
25 6
46 3
42 19
28 9
15 3
43 3
35 10
42 41
51 46
9 4
46 34
28 0
6 5
45 14
26 11
48 13
33 23
40 9
23 21
18 16
28 12
43 29
35 31
30 14
36 34
49 38
49 22
24 11
23 14
45 13
49 21
48 16
51 10
39 4
50 46
50 48
43 17
31 18
38 23
2 0
41 0
30 19
20 1
29 19
48 32
30 15
40 22
51 12
50 40
24 4
39 10
31 20
7 0
40 17
41 31
37 29
33 32
30 3
40 6
51 15
46 19
31 28
34 22
31 5
33 7
29 14
34 24
44 6
24 2
44 40
35 6
37 18
47 0
43 42
49 30
49 25
19 1
25 3
49 5
40 10
25 21
48 15
35 19
50 6
36 17
44 33
21 13
15 4
36 32
28 6
49 35
47 9
49 46
47 14
25 4
44 29
38 25
23 12
51 41
20 5
39 34
15 6
47 23
21 6
47 11
22 7
41 29
34 2
43 38
6 2
3 2
40 20
40 24
37 16
32 26
49 31
49 16
50 13
31 2
26 1
5 0
19 16
45 32
42 40
16 5
15 8
38 27
12 6
47 4
39 6
31 19
26 9
47 18
42 32
4 2
42 20
46 10
27 6
41 7
49 2
49 28
20 9
46 33
16 11
14 4
34 1
33 2
30 6
47 44
41 8
23 17
33 25
23 5
24 13
33 20
44 35
47 46
47 7
41 25
45 5
28 23
31 15
31 10
39 9
40 7
45 6
43 11
35 26
51 34
44 38
45 3
24 19
51 22
47 42
34 15
37 33
29 9
49 3
14 3
23 2
39 7
46 23
40 31
33 16
44 43
41 36
37 17
43 40
32 18
46 32
26 18
4 3
39 5
44 11
28 20
44 21
41 26
39 38
36 5
7 3
39 0
27 18
26 20
18 2
50 28
37 26
40 27
17 4
50 3
39 30
32 29
50 34
18 1
20 4
36 23
25 15
49 0
45 39
39 1
37 5
23 16
47 20
27 20
38 4
46 43
34 27
15 5
31 23
39 29
46 7
38 35
41 14
45 9
25 22
10 9
35 21
19 14
37 8
47 35
9 0
35 13
21 16
50 32
37 7
19 8
22 5
51 24
51 9
29 0
51 39
44 19
42 5
31 9
40 30
51 37
25 12
26 0
32 16
25 1
41 13
47 43
25 18
35 29
50 44
45 23
44 20
50 47
22 2
45 4
34 19
48 33
34 16
18 10
29 18
37 13
45 2
43 14
48 10
15 2
28 22
29 16
45 15
19 17
35 16
46 9
9 5
35 27
30 5
49 39
32 28
42 3
48 37
43 32
44 30
37 30
14 2
47 32
20 8
18 13
25 5
44 5
29 15
49 11
42 14
30 29
42 27
19 6
51 49
51 13
12 1
40 34
23 13
27 11
51 43
27 24
19 13
26 19
16 10
23 1
46 5
35 15
30 10
48 3
19 9
25 23
16 14
23 3
34 11
27 9
32 30
39 19
50 33
45 21
50 12
13 3
50 15
25 16
49 14
41 17
47 19
43 36
13 12
30 7
49 48
14 0
24 7
49 27
30 26
47 21
14 6
30 22
22 9
29 5
23 22
51 40
42 37
29 6
8 5
51 29
22 4
28 19
21 3
45 12
47 26
43 35
48 43
20 2
24 21
33 22
24 20
41 5
35 3
43 15
43 34
19 10
47 41
49 8
29 21
51 31
43 19
50 17
47 24
(output: 11061; takes around 10 seconds)

A few problems in your code:
the operation sorted(path) not in cycles_list has O(n) complexity, where n is the size of cycles_list
queue.pop(0) has O(n) complexity, where n is the size of the queue. You should use the collections.deque structure, not a list here.
As a general note, unless you really need to solve the question using specifically BFS (e.g. because some asked you to use this method), a simple combination of loops would do the job better. Pseudocode:
num_loops = 0
for a in nodes:
for b in neighbors(a)
if b > a:
for c in neighbors(b):
if c > b and a in neighbors(c):
num_loops += 1
The b > a and c > b checks are added to count each loop only once.

For a small number of steps like 3, you can just check for each node if you can walk away from and back to the node within 3 steps.
This works reasonably fast:
import fileinput
graph = {}
# Recursive function to find a goal in a number of steps
def count_unique_walks(start, goal, length, visited=[]):
if length == 0:
# Out of steps
return 1 if start == goal else 0
if start in visited:
# Already been here
return 0
result = 0
for neighbor in graph[start]:
if neighbor < start and neighbor != goal:
# Count only unique cycles
continue
result += count_unique_walks(neighbor, goal, length-1, visited+[start])
return result
# Read input
for line in fileinput.input():
a, b = map(int, line.split())
if a not in graph:
graph[a] = set()
graph[a].add(b)
if b not in graph:
graph[b] = set()
graph[b].add(a)
# Sum up the cycles of each node
result = 0
for node in graph:
result += count_unique_walks(node, node, 3)
print result

performing differences between rows in pandas based on columns values

I have this dataframe, I'm trying to create a new column where I want to store the difference of products sold based on code and date.
for example this is the starting dataframe:
date code sold
0 20150521 0 47
1 20150521 12 39
2 20150521 16 39
3 20150521 20 38
4 20150521 24 38
5 20150521 28 37
6 20150521 32 36
7 20150521 4 43
8 20150521 8 43
9 20150522 0 47
10 20150522 12 37
11 20150522 16 36
12 20150522 20 36
13 20150522 24 36
14 20150522 28 35
15 20150522 32 31
16 20150522 4 42
17 20150522 8 41
18 20150523 0 50
19 20150523 12 48
20 20150523 16 46
21 20150523 20 46
22 20150523 24 46
23 20150523 28 45
24 20150523 32 42
25 20150523 4 49
26 20150523 8 49
27 20150524 0 39
28 20150524 12 33
29 20150524 16 30
... ... ... ...
150 20150606 32 22
151 20150606 4 34
152 20150606 8 33
153 20150607 0 31
154 20150607 12 30
155 20150607 16 30
156 20150607 20 29
157 20150607 24 28
158 20150607 28 26
159 20150607 32 24
160 20150607 4 30
161 20150607 8 30
162 20150608 0 47
I think this could be a solution...
full_df1=full_df[full_df.date == '20150609'].reset_index(drop=True)
full_df1['code'] = full_df1['code'].astype(float)
full_df1= full_df1.sort(['code'], ascending=[False])
code date sold
8 32 20150609 33
7 28 20150609 36
6 24 20150609 37
5 20 20150609 39
4 16 20150609 42
3 12 20150609 46
2 8 20150609 49
1 4 20150609 49
0 0 20150609 50
full_df1.set_index('code')['sold'].diff().reset_index()
that gives me back this output for a single date 20150609 :
code difference
0 32 NaN
1 28 3
2 24 1
3 20 2
4 16 3
5 12 4
6 8 3
7 4 0
8 0 1
is there a better solution to have the same result in a more pythonic way?
I would like to create a new column [difference] and store the data there having as result 4 columns [date, code, sold, difference]

This exactly the kind of thing that panda's groupby functionality is built for, and I highly recommend reading and working through this documentation: panda's groupby documentation
This code replicates what you are asking for, but for every date.
df = pd.DataFrame({'date':['Mon','Mon','Mon','Tue','Tue','Tue'],'code':[10,21,30,10,21,30], 'sold':[12,13,34,10,15,20]})
df['difference'] = df.groupby('date')['sold'].diff()
df
code date sold difference
0 10 Mon 12 NaN
1 21 Mon 13 1
2 30 Mon 34 21
3 10 Tue 10 NaN
4 21 Tue 15 5
5 30 Tue 20 5