Fix bug in metaflow version when public=True (#2539)

`metaflow_version.get_version(public=True)` doesn't return the public
version if executed within a metaflow step. This is because, we return
the version in the `INFO` file directly without any parsing. This PR
addresses that.

Flow to reproduce the issue:
```python
from metaflow import FlowSpec, step, conda

class HelloFlow(FlowSpec):
    @conda()
    @step
    def start(self):
        from metaflow import metaflow_version
        version = metaflow_version.get_version(public=True)
        print(f"Metaflow version: {version}")
        self.next(self.end)

    @step
    def end(self):
        print("HelloFlow is all done.")


if __name__ == "__main__":
    HelloFlow()
```

Author: talsperre

metaflow/metaflow_version.py

@@ -133,6 +133,16 @@ def read_info_version():
     return None
 
 
+def make_public_version(version_string):
+    """
+    Takes a complex version string and returns a public, PEP 440-compliant version.
+    It removes local version identifiers (+...) and development markers (-...).
+    """
+    base_version = version_string.split("+", 1)[0]
+    public_version = base_version.split("-", 1)[0]
+    return public_version
+
+
 def get_version(public=False):
     """Tracks the version number.
 
@@ -170,6 +180,11 @@ def get_version(public=False):
     )  # Version info is cached in INFO file; includes extension info
 
     if version:
+        # If we have a version from the INFO file, use it directly.
+        # However, if we are asked for a public version, we parse it to make sure
+        # that no local information is included.
+        if public:
+            version = make_public_version(version)
         _version_cache[public] = version
         return version