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I have a CSV file with records in below format.

area confidence boundary
159.1512 0.8084 POLYGON((87.7185491791352 22.5667828312016, 87.7184590804737 22.5667864311825, 87.7184519272495 22.5666320082646, 87.7185420258111 22.5666284082901, 87.7185491791352 22.5667828312016))

I am using the following python code which uses the boundary column to get the polygon shapefile from it.The code works fine however, when I add the exported the shapefile into ArcGIS the polygons are not falling on correct location.

import geopandas as gpd
import pandas as pd
from shapely.geometry import Polygon

df = pd.read_csv(r"/content/Test_Polygon.csv")

df.boundary = df.boundary.str.findall('\d+\.\d+').apply(lambda x: list(map(float, x))) #Extract numbers from boundary string

def givepoly(col):
    """Function to return a shapely polygon from a string of lat and longs"""
    return Polygon([[col[i+1], col[i]] for i in range(0, len(col)-1,2)])

df['geometry'] = df.apply(lambda row: givepoly(row['boundary']), axis=1)

df2 = gpd.GeoDataFrame(df, geometry='geometry', crs="EPSG:4326")
df2.drop('boundary', inplace=True, axis=1) #A shapefile cant have a list as data type
df2.to_file(r"/content/test3.shp")

enter image description here

Am not sure where am going wrong in this. I used the following link for reference : Converting CSV file with polygons in dictionary format into GeoDataFrame

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  • You have a tag for ArcPy but the code that you've presented does use that module.
    – PolyGeo
    Commented Feb 2 at 7:59
  • 1
    Traditionally the first thing to check is what coordinate order the library or application assumes - lon/lat or lat/lon.
    – user30184
    Commented Feb 2 at 8:09
  • In this piece of code, it is assumed that the coordinates are referring to EPSG:4326 (WGS84). Are you sure this is the case with your coordinates?
    – kowalski
    Commented Feb 2 at 8:10
  • The comment in the code "Function to return a shapely polygon from a string of lat and longs" but I believe that the geometry is lon-lats from around here openstreetmap.org/#map=5/22.567/87.719.
    – user30184
    Commented Feb 2 at 8:37
  • @user30184 Yes, after your earlier I comment I paid attention to the order and indeed the file I have contains the order of Long/Lat whereas code is for Lat/Long order however, am not sure how to fix the code so that it takes Long/Lat order.
    – gis
    Commented Feb 2 at 8:41

3 Answers 3

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A better solution is to use directly geopandas.GeoSeries.from_wkt

 df = pd.read_csv("Test_Polygon.csv")
 print(df.boundary[0])
 POLYGON ((87.7185491791352 22.5667828312016, 87.7184590804737 22.5667864311825, 87.7184519272495 22.5666320082646, 87.7185420258111 22.5666284082901, 87.7185491791352 22.5667828312016))
 import geopandas as gpd
 s = gpd.GeoSeries.from_wkt(df.boundary)
 gdf = gpd.GeoDataFrame(data=df, geometry=s)
 print(gdf.geometry[0])
 POLYGON ((87.7185491791352 22.5667828312016, 87.7184590804737 22.5667864311825, 87.7184519272495 22.5666320082646, 87.7185420258111 22.5666284082901, 87.7185491791352 22.5667828312016))

You can also use shapely

from shapely import wkt
df["geom"] = gpd.GeoSeries.from_wkt(df["boundary"])
gdf2 = gpd.GeoDataFrame(df, geometry="geom")
print(gdf2.geometry[0])
POLYGON ((87.7185491791352 22.5667828312016, 87.7184590804737 22.5667864311825, 87.7184519272495 22.5666320082646, 87.7185420258111 22.5666284082901, 87.7185491791352 22.5667828312016))

or

df['geom2']= df.boundary.apply(wkt.loads)
gdf = gpd.GeoDataFrame(df, geometry='geom2')
print(gdf.geometry[0])
POLYGON ((87.7185491791352 22.5667828312016, 87.7184590804737 22.5667864311825, 87.7184519272495 22.5666320082646, 87.7185420258111 22.5666284082901, 87.7185491791352 22.5667828312016))
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1

Swapping the values placed the geometry on correct location. Here's the modified the code:

import geopandas as gpd
import pandas as pd
from shapely.geometry import Polygon

df = pd.read_csv(r"/content/Test_Polygon.csv")

df.boundary = df.boundary.str.findall('\d+\.\d+').apply(lambda x: list(map(float, x))) #Extract numbers from boundary string

def givepoly(col):
    """Function to return a shapely polygon from a string of lat and longs"""
    return Polygon([[col[i], col[i + 1]] for i in range(0, len(col)-1,2)])

df['geometry'] = df.apply(lambda row: givepoly(row['boundary']), axis=1)

df2 = gpd.GeoDataFrame(df, geometry='geometry', crs="EPSG:4326")
df2.drop('boundary', inplace=True, axis=1) #A shapefile cant have a list as data type
df2.to_file(r"/content/test3.shp")

The only other concern that I have is when I add the shapefile to the map the geometries are displaying at a very small size.

enter image description here

Am not sure if this is an expected behaviour or code can be fixed so that geometry is displayed at an appropriate scale.

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I tried adding the above mentioned code however got the error : TypeError: Expected bytes or string, got list


def givepoly(col):
    """Function to return a shapely polygon from a string of lat and longs"""
    return Polygon([[col[i], col[i+1]] for i in range(0, len(col)-1,2)])

df['geometry'] = df.apply(lambda row: givepoly(row['boundary']), axis=1)

s = gpd.GeoSeries.from_wkt(df.boundary)
df2 = gpd.GeoDataFrame(df, geometry=s)
df2.drop('boundary', inplace=True, axis=1) #A shapefile cant have a list as data type
df2.to_file(r"/content/test5.shp")

Not sure where am I going wrong.

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