Python in Hydrology and Hydraulics: geometry

Wednesday, May 30, 2018

Pandas - Operations between rows - distance between 2 points

If we have a table with a column with xy coordinates, for example:

We can get the difference between consecutive rows by using Pandas SHIFT function on columns.
".shift(-1)" will roll the rows 1 position backwards, and ".shift(1)" or simply ".shift()" will roll down your column by 1 position of the rows.

In our example, df1['x'].shift() will return:

0 NaN
1 455395.996360
2 527627.076641
3 536278.269190
4 553932.441097
5 568699.553239
6 569709.130272
7 573016.302437
8 575141.096777
9 580107.934566

if we want to calculate the euclidean distance between consecutive points, we can use the shift associated with numpy functions numpy.sqrt and numpy.power as following:

df1['diff']= np.sqrt(np.power(df1['x'].shift()-df1['x'],2)+
np.power(df1['y'].shift()-df1['y'],2))

Resulting in:

0 NaN
1 89911.101224
2 21323.016099
3 204394.524574
4 37767.197793
5 46692.771398
6 13246.254235
7 2641.201366
8 15153.187527
9 15853.974422

Thursday, August 31, 2017

Calculate the centroid of a polygon with python

In this post I will show a way to calculate the centroid of a non-self-intersecting closed polygon.

I used the following formulas,

$C_{\mathrm {x} }={\frac {1}{6A}}\sum _{i=0}^{n-1}(x_{i}+x_{i+1})(x_{i}\ y_{i+1}-x_{i+1}\ y_{i})$

$C_{\mathrm {y} }={\frac {1}{6A}}\sum _{i=0}^{n-1}(y_{i}+y_{i+1})(x_{i}\ y_{i+1}-x_{i+1}\ y_{i})$

$A={\frac {1}{2}}\sum _{i=0}^{n-1}(x_{i}\ y_{i+1}-x_{i+1}\ y_{i})\;$

as shown in https://en.wikipedia.org/wiki/Centroid#Centroid_of_a_polygon .

In the following code, the function receives the coordinates as a list of lists or as a list of tuples.

from math import sqrt

def centroid(lstP):
    sumCx = 0
    sumCy = 0
    sumAc= 0
    for i in range(len(lstP)-1):
        cX = (lstP[i][0]+lstP[i+1][0])*(lstP[i][0]*lstP[i+1][1]-lstP[i+1][0]*lstP[i][1])
        cY = (lstP[i][1]+lstP[i+1][1])*(lstP[i][0]*lstP[i+1][1]-lstP[i+1][0]*lstP[i][1])
        pA = (lstP[i][0]*lstP[i+1][1])-(lstP[i+1][0]*lstP[i][1])
        sumCx+=cX
        sumCy+=cY
        sumAc+=pA
        print cX,cY,pA
    ar = sumAc/2.0
    print ar
    centr = ((1.0/(6.0*ar))*sumCx,(1.0/(6.0*ar))*sumCy)
    return centr