Package is showing in conda list but can't find in anaconda channels

Literalomicus 0
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I was installing Natasha, and so I used pip with conda to install Natasha and it's dependencies. They all show in conda list. What it says if I use conda pip install Natasha (I'm using conda so I can make virtualenv with packages I got on conda-forge). Is long, I'll summarize.

Requirement already satisfied: Natasha in c:\getfunds\lib\site-packages (1.4.0)
Requirement already satisfied: razdel>=0.5.0 in c:\getfunds\lib\site-packages (from Natasha) (0.5.0)
Requirement already satisfied: ipymarkup>=0.8.0 in c:\getfunds\lib\site-packages (from Natasha) (0.9.0)
Requirement already satisfied: yargy>=0.14.0 in c:\getfunds\lib\site-packages (from Natasha) (0.15.0)
Requirement already satisfied: navec>=0.9.0 in c:\getfunds\lib\site-packages (from Natasha) (0.10.0)
Requirement already satisfied: pymorphy2 in c:\getfunds\lib\site-packages (from Natasha) (0.9.1)
Requirement already satisfied: slovnet>=0.3.0 in c:\getfunds\lib\site-packages (from Natasha) (0.5.0)
Requirement already satisfied: pymorphy2-dicts-ru<3.0,>=2.4 in c:\getfunds\lib\site-packages (from pymorphy2->Natasha) (2.4.417127.4579844)

When I try to make the virtualenv, I get a PackagesNotFound error even though all of the packages show up in conda list with pypi_0. I think it's a channels problem, but I don't know what to add. My channel list :

conda virtualenv anaconda pip python
1 Answers

I found out what it was! It wasn't a channel issue ( I mean maybe it could be solved by adding the right channel, but there is some kind of incompatible difference between pypi and conda-forge the pypi in my channel list are null links) I had tried setting: conda config --set pip_interop_enabled True

And maybe that was a perquisite, I haven't tried disabling it and testing this following step.I had to use a pip install after I had activated the virtualenv because Natasha (as a pip install) couldn't be found by the default channels, I had to pip install directly into the virtual environment.

Literalomicus0
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