[GIS] Split vector layer in PyQGIS

pyqgisqgisqgis-processingsplitting

I have a polygon shapefile that has several attributes.

I would like to know how can I create automatically a vector file for each one.

I already find split vector layer in QGIS (like picture):

but I want to use python.

I found a way through the web site at the following site, but this caused an error in QGIS.

https://docs.qgis.org/2.8/en/docs/user_manual/processing_algs/qgis/vector_general_tools/splitvectorlayer.html

>>> import processing
>>> processing.runalg('qgis:splitvectorlayer')
ERROR: Algorithm not found

So I need another way to split..

Best Answer

I'm not familiar with PyQGIS but this can be easily accomplished using ogr.

The following script shows an example:

from osgeo import ogr

fn = r""  # Vector Layer File path
driver = ogr.GetDriverByName('ESRI Shapefile')  # See OGR Vector Format for other options
dataSource = driver.Open(fn)
layer = dataSource.GetLayer()
sr = layer.GetSpatialRef()  # Spatial Reference

dst = r""  # Output directory
new_feat = ogr.Feature(layer.GetLayerDefn())  # Dummy feature

for id, feat in enumerate(layer):
    new_ds = driver.CreateDataSource(r"{}\feat_{}.shp".format(dst, id))
    new_lyr = new_ds.CreateLayer('feat_{}'.format(id), sr, ogr.wkbPolygon) 
    geom = feat.geometry().Clone()
    new_feat.SetGeometry(geom)
    new_lyr.CreateFeature(new_feat)

    del new_ds, new_lyr

You can, of course, change the name of the driver (in case your layer is not a shapefile) and the geometry of your new files (in case the input vector layer file is not a polygon).

More information about ogr vector formats here: http://www.gdal.org/ogr_formats.html

Related Solutions

[GIS] QGIS Modeler: Where is output from algorithm ‘Set style for vector layer’

I don't think the docs are very clear on this. From my understanding, the Set style for vector layer algorithm does not create a new output layer. The end result is the input layer with the style file applied. I believe this is why there is no "output" option for the algorithm.

From the QGIS 2.8 docs and checking the parameters from the Python console, both mention an Output:

import processing

processing.alghelp("qgis:setstyleforvectorlayer")
ALGORITHM: Set style for vector layer
    INPUT <ParameterVector>
    STYLE <ParameterFile>
    OUTPUT <OutputVector>

However, when you check the Console usage, you will see that there is no parameter for Output:

processing.runalg("qgis:setstyleforvectorlayer", input, style)

So if you want to add more than one style to a shapefile, you may need to create another set of the same algorithms within your model.

Hope this helps!

[GIS] Using PyQGIS processing runalg clipvectorsbypolygon

You may continue using the memory layer by loading it as an object. Before doing this, you need to know how the output from the algorithm is called, so typing these lines in the Python Console:

import processing
processing.alghelp('gdalogr:clipvectorsbypolygon')

you will see that the output's name is OUTPUT_LAYER:

ALGORITHM: Clip vectors by polygon
INPUT_LAYER <ParameterVector>
CLIP_LAYER <ParameterVector>
OPTIONS <ParameterString>
OUTPUT_LAYER <OutputVector>

This means that you will be able to use it if you add a new variable and call it in this way:

# change 'my_algorithm' name as you want
my_algorithm = general.runalg('gdalogr:clipvectorsbypolygon', 'lokale_orte.shp', 'kleine_kreise.shp', None, None)

# change 'my_output' name as you want
my_output = processing.getObject(my_algorithm['OUTPUT_LAYER'])

By now, my_output will be the object to use for calling the layer returned by the clip operation and you are ready to do something with it.

EDIT For getting your script working, you may try using the following code (slightly different from yours):

import sys
from qgis.core import *
import qgis.utils

qgs = QApplication([], True)
QgsApplication.setPrefixPath("/usr", True)
QgsApplication.initQgis()

# Prepare processing framework 
sys.path.append('/usr/share/qgis/python/plugins') # Dir where Processing is located
from processing.core.Processing import Processing
from processing.tools import *
Processing.initialize()

# load base layer
blayer = QgsVectorLayer("path/to/the/shapefile/lokale_orte.shp", "lokale_orte", "ogr")
if not blayer.isValid():
    print "Layer failed to load!"
else:
    print "success:  load layer"

# load clip layer
clayer = QgsVectorLayer("path/to/the/shapefile/kleine_kreise.shp", "kleine_kreise", "ogr")
if not clayer.isValid():
    print "Layer failed to load!"
else:
    print "success:  load layer"

# run clip
# change 'my_algorithm' name as you want
my_algorithm = general.runalg('gdalogr:clipvectorsbypolygon', 'lokale_orte.shp', 'kleine_kreise.shp', None, None)

# change 'my_output' name as you want
my_output = processing.getObject(my_algorithm['OUTPUT_LAYER'])

QgsApplication.exitQgis()
qgs.exit()

Best Answer

Related Solutions

[GIS] QGIS Modeler: Where is output from algorithm ‘Set style for vector layer’

[GIS] Using PyQGIS processing runalg clipvectorsbypolygon

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