[GIS] How to create a linestring from a geopandas data frame

geopandaslinestringpython

I am trying to create a linestring from a geopandas dataframe with points as geometries

date_time   latitude    longitude   year    data01  hour    minute  data02
0   2019-11-08 08:21:38 36.830444   -2.466583   2019    312 8   21  0
1   2019-11-08 08:32:15 36.830250   -2.466611   2019    312 8   32  0
2   2019-11-08 08:41:51 36.830417   -2.466778   2019    312 8   41  0
3   2019-11-08 08:51:53 36.830361   -2.466833   2019    312 8   51  0
4   2019-11-08 09:01:50 36.830528   -2.466806   2019    312 9   1   0

using the this code:

    date_time   latitude    longitude   year    data01  hour    minute  data02  geometry
0   2019-11-08 08:21:38 36.830444   -2.466583   2019    312 8   21  0   POINT (-2.466583 36.830444)
1   2019-11-08 08:32:15 36.830250   -2.466611   2019    312 8   32  0   POINT (-2.466611 36.83025)
2   2019-11-08 08:41:51 36.830417   -2.466778   2019    312 8   41  0   POINT (-2.466778 36.830417)
3   2019-11-08 08:51:53 36.830361   -2.466833   2019    312 8   51  0   POINT (-2.466833 36.830361)
4   2019-11-08 09:01:50 36.830528   -2.466806   2019    312 9   1   0   POINT (-2.466806 36.830528)

geo_df = geo_df.groupby(['date_time'])['geometry'].apply(lambda x: LineString(x.tolist()))
geo_df = GeoDataFrame(geo_df, geometry='geometry')

Error: ValueError: LineStrings must have at least 2 coordinate tuples

How can I do that ? or can I create the linestring using just the latitude and longitude columns?

Even if use geopandas to convert to linestring geometry the columns look no good for that

geometry = [LineString(xy) for xy in zip(df.longitude, df.latitude)]
geo_df = GeoDataFrame(df, geometry=geometry)
geo_df

 ValueError: Input (-2.466583, 36.830444) is the wrong shape for a LineString

Best Answer

With all your solutions, you try to create a LineString with only one point

Since all the dates (hours) are different:

geo_df = geo_df.groupby(['date_time'])
print("number of groups:", geo_df.ngroups)
number of groups: 5 # and 5 original geometries
for group in gdf.groupby(['date_time']).geometry:
    print(group)
(Timestamp('2019-11-08 08:21:38'), 0    POINT (-2.466583 36.830444)
 Name: geometry, dtype: object)
(Timestamp('2019-11-08 08:32:15'), 1    POINT (-2.466611 36.83025)
Name: geometry, dtype: object)
(Timestamp('2019-11-08 08:41:51'), 2    POINT (-2.466778 36.830417)
Name: geometry, dtype: object)
(Timestamp('2019-11-08 08:51:53'), 3    POINT (-2.466833 36.830361)
Name: geometry, dtype: object)
(Timestamp('2019-11-08 09:01:50'), 4    POINT (-2.466806 36.830528)
Name: geometry, dtype: object)

Only one point per group -> Error: ValueError: LineStrings must have at least 2 coordinate tuples)

And

geometry = [xy for xy in zip(df.longitude, df.latitude)]
print(geometry)
[(-2.466583, 36.830444), (-2.466611, 36.83025), (-2.4667779999999997, 36.830417), (-2.4668330000000003, 36.830360999999996), (-2.466806, 36.830528)
# geometry is a list of point coordinates
Point(geometry[0]).wkt
'POINT (-2.466583 36.830444)'
#but
LineString(geometry[0])
...

ValueError: Input (-2.466583, 36.830444) is the wrong shape for a LineString

And I suppose you want to create a line with all the points, therefore

 print(LineString(geometry).wkt)
 LINESTRING (-2.466583 36.830444, -2.466611 36.83025, -2.466778 36.830417, -2.466833 36.830361, -2.466806 36.830528)

or directly

print(LineString(df.geometry).wkt)
LINESTRING (-2.466583 36.830444, -2.466611 36.83025, -2.466778 36.830417, -2.466833 36.830361, -2.466806 36.830528)

If you want to use groupby to create separate lines choose another column to regroup (look at Turning GeoDataFrame of x,y coordinates into Linestrings using GROUPBY? or Convert Points to Lines Geopandas)

with data02 modified (2 groups, 0 and 1)

date_time,latitude,longitude,year,data01,hour,minute,data02,wtkcolumn
2019-11-08 08:21:38,36.830444,-2.466583,   2019,    312, 8 ,  21 , 0 ,  POINT (-2.466583 36.830444)
2019-11-08 08:32:15,36.830250,-2.466611,   2019,    312, 8  , 32,  0,   POINT (-2.466611 36.83025)
2019-11-08 08:41:51 ,36.830417,-2.466778,   2019,    312, 8 ,  41 , 1 ,  POINT (-2.466778 36.830417)
2019-11-08 08:51:53 ,36.830361,-2.466833,   2019,    312, 8,   51,  1 ,  POINT (-2.466833 36.830361)
2019-11-08 09:01:50, 36.830528, -2.466806,   2019,    312, 9 ,  1  , 1,   POINT (-2.466806 36.830528)

list(geo_df.groupby(['data02'])['geometry'].apply(lambda x: LineString(x.tolist())))
[<shapely.geometry.linestring.LineString object at 0x11bc81668>, <shapely.geometry.linestring.LineString object at 0x11bc81400>]

Related Solutions

Python Geopandas – GDAL/Geopandas Data Object Compatibility in Python

The GDAL memory driver is written for GDAL and obviously exactly for improving the performance https://www.gdal.org/drv_memory.html as you wrote. The documentation seems to suggest that other processes can't open a memory datastore.

Memory This driver implements read and write access layers of features contained entirely in memory. This is primarily useful as a high performance, and highly malleable working data store. All update options, geometry types, and field types are supported.

There is no way to open an existing Memory datastore. It must be created with CreateDataSource() and populated and used from that handle. When the datastore is closed all contents are freed and destroyed.

GDAL memory rasters https://www.gdal.org/frmt_mem.html can be opened. The difference is probably that for memory raster it is possible to define the offset where the data begins in the memory:

It is possible to open an existing array in memory. To do so, construct a dataset name with the following format:

MEM:::option=value[,option=value...]

For example:

MEM:::DATAPOINTER=342343408,PIXELS=100,LINES=100,BANDS=3,DATATYPE=Byte, PIXELOFFSET=3,LINEOFFSET=300,BANDOFFSET=1

or

MEM:::DATAPOINTER=0x1467BEF0,PIXELS=100,LINES=100,BANDS=3,DATATYPE=Byte, PIXELOFFSET=3,LINEOFFSET=300,BANDOFFSET=1

DATAPOINTER: address of the first pixel of the first band. The address can be represented as a hexadecimal or decimal value. Hexadecimal values must be prefixed with '0x'. Some implementations (notably Windows) doesn't print hexadecimal pointer values with a leading '0x', so the prefix must be added. You can use CPLPrintPointer to create a string with format suitable for use as a DATAPOINTER.

Python – Converting CSV Containing GeoJSON Format Geometries to DataFrame with GeoPandas and Shapely

The shape function from shapely needs a dictionary, so you need to parse the strings to dictionaries before calling shape. If you have these imports

import geopandas as gpd
import json
import pandas as pd
from shapely.geometry import shape

and you define this function

def parse_geom(geom_str):
    try:
        return shape(json.loads(geom_str))
    except (TypeError, AttributeError):  # Handle NaN and empty strings
        return None

Then this should give you a GeoDataframe

df["geom 1"] = df["geom 1"].apply(parse_geom)
gdf = gpd.GeoDataFrame(df, geometry="geom 1")

Worked example:

import geopandas as gpd
import pandas as pd
import json
from shapely.geometry import shape

def parse_geom(geom_str):
    try:
        return shape(json.loads(geom_str))
    except (TypeError, AttributeError):  # Handle NaN and empty strings
        return None

df = pd.read_csv('path/to/your.csv')
df["geom 1"] = df["geom 1"].apply(parse_geom)
gdf = gpd.GeoDataFrame(df, geometry="geom 1")
print(gdf.head())

Best Answer

Related Solutions

Python Geopandas – GDAL/Geopandas Data Object Compatibility in Python

Python – Converting CSV Containing GeoJSON Format Geometries to DataFrame with GeoPandas and Shapely

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