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Python 我想使用比较操作返回一个大于5小于20的值,但是当运行时,“5和sloperater_Python_Comparison - Fatal编程技术网

Python 我想使用比较操作返回一个大于5小于20的值,但是当运行时,“5和sloperater

Python 我想使用比较操作返回一个大于5小于20的值,但是当运行时,“5和sloperater,python,comparison,Python,Comparison,我正在寻找介于5和20之间的坡度,但我仍然在小于符号上得到无效语法 import arcpy arcpy.env.workspace = "E:/GEOG488_lecture10data" from arcpy.sa import* arcpy.CheckOutExtension('Spatial') sloperaster = arcpy.Raster('slope') Modslope = sloperaster > 5 and < 20

我正在寻找介于5和20之间的坡度,但我仍然在小于符号上得到无效语法

import arcpy    
arcpy.env.workspace = "E:/GEOG488_lecture10data"    
from arcpy.sa import*    
arcpy.CheckOutExtension('Spatial')    
sloperaster = arcpy.Raster('slope')    
Modslope = sloperaster > 5 and < 20 

这是一个简单的解决方案,它检查变量是否大于5和小于20,因为这不是有效的语法。使用sloperater>5和sloperater<20或5 20 > variable > 5