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Fix spelling errors. (#561)

* neighour     -> neighbour
 * excentricity -> eccentricity
Bas Couwenberg 5 年之前
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ca90310887

+ 1 - 1
imagery/i.ortho.photo/i.ortho.camera/i.ortho.camera.html

@@ -54,7 +54,7 @@ fiducial marks. In the ideal case of no deviations in the camera (see camera
 calibration certificate) the center is the origin and the values are 0 for
 calibration certificate) the center is the origin and the values are 0 for
 both X and Y of Point of Symmetry. But usually the principal point does not
 both X and Y of Point of Symmetry. But usually the principal point does not
 fall on the intersection of the radii at the center of the picture. This
 fall on the intersection of the radii at the center of the picture. This
-excentricity is usually of the order of a few micrometers. <p>
+eccentricity is usually of the order of a few micrometers. <p>
 
 
 You are then asked to enter the X and Y photo coordinates of each fiducial
 You are then asked to enter the X and Y photo coordinates of each fiducial
 as follows.
 as follows.

+ 1 - 1
imagery/i.segment/region_growing.c

@@ -474,7 +474,7 @@ int region_growing(struct globals *globals)
 				pathflag = FALSE;
 				pathflag = FALSE;
 
 
 			    if (Rk_bestn.id < 0) {
 			    if (Rk_bestn.id < 0) {
-				G_debug(4, "Rk's best neighour is negative");
+				G_debug(4, "Rk's best neighbour is negative");
 				pathflag = FALSE;
 				pathflag = FALSE;
 			    }
 			    }
 
 

+ 1 - 1
lib/vector/Vlib/remove_areas.c

@@ -119,7 +119,7 @@ Vect_remove_small_areas_ext(struct Map_info *Map, double thresh,
 	}
 	}
 	G_debug(3, "num neighbours = %d", AList->n_values);
 	G_debug(3, "num neighbours = %d", AList->n_values);
 
 
-	/* Go through the list of neighours and find that with the longest boundary */
+	/* Go through the list of neighbours and find that with the longest boundary */
 	dissolve_neighbour = 0;
 	dissolve_neighbour = 0;
 	length = -1.0;
 	length = -1.0;
 	for (i = 0; i < AList->n_values; i++) {
 	for (i = 0; i < AList->n_values; i++) {