Looking at the Particle Data Group tables of the $\Lambda$ baryon, I find that the rate of the hadronic decay $\Lambda \to p \pi^-$ is 64% while the semileptonic decay $\Lambda \to p e^-\nu_e$ has the rate of $8 \times 10^{-4}$.
I cannot explain what is the reason for such large suppression of the semileptonic decay of the $\Lambda$. Both $\Lambda \to p \pi^-$ and $\Lambda \to p e^-\nu_e$ decays originate from the same Feynman diagram, $s \to u W^-$ at the quark level, with $W^-$ going either to $\overline{u}d$ or $e^-\nu_e$. So, I would expect the factor 3 difference between these two diagrams, due to the color factor; which could then be diminished by the smaller phase-space in the hadronic decay compared to the semileptonic one. However, the actual difference in rate of $\Lambda \to p \pi^-$ and $\Lambda \to p e^-\nu_e$ is factor 800, so my estimation is wrong.
The assumption that a three-body decay would somehow be more suppressed than a two-body decay, also does not work, as the charged kaon, $K^-$, decays (via the same Feynman diagrams as the decays discussed above, modulo one spectator quark) to $\pi^0\pi^-$ in about 21% of times, and to $\pi^0 e^-\nu_e$ in about 5% of times (so, factor 4 difference instead of 800). There should be some suppression affecting only baryonic decays but not mesonic... but I do not see where it might come from.
In a related question, a suggestion has been made that semileptonic decays might be affected by the helicity suppression. However, under that assumption, I would expect the $\Lambda \to p \mu^-\nu_\mu$ decay to dominate over $\Lambda \to p e^-\nu_e$ (like in a pion decay), which is not the case.
What am I missing?
(This question is motivated by my previous question on hyperon lifetimes.)