Fluorine is the most electronegative halogen and therefore, there is larger difference in electronegativity between the atoms of $\ce{HF}$ than any other hydrogen halide, which means the positive charge on hydrogen atom is the greatest in this compound and hence a comparatively small negative charge is needed to attract it. If so, why is it classified as least acidic hydrogen halide if the $\ce{H+}$ is easy to remove?
Answer
First of all as @chipbuster says $\ce{HF}$ in diluted solutions in water is nearly completely dissociated and therefore shouldn't be called weak. Wikipedia describes this nicely and cites several sources for this claim.
It was rather difficult to prove (spectroscopic methods were used), because hydronium ions created in dissociation are mostly bound to fluorine anions with hydrogen bonds, in what is called tight ion pairs. It prevents from detecting the true strength of $\ce{HF}$ with methods like acid–base titration - they show dissociation of ionic pair. Similar effect is present in ion exchange resins.
Strength of this acid is revealed in more concentrated solutions, where $\ce{HF}$ molecules replace hydronium ions, creating bifluoride anions, freeing them - it's homoassotiation mentioned by Wildcat.
More to the point - does it mean that $\ce{HF}$ isn't weaker acid than other hydrogen halides? The answer isn't simple, and acidity itself depends on many factors.
In terms of $\ce{K_a}$ in diluted aqueous solutions $\ce{HF}$ is probably still weaker than $\ce{HCl}$ and the rest but getting exact value is greater problem than normally (generally values of $\ce{K_a}$ for strong acids aren't precise). Decrease in acidity between hydrogen halides is described in answer to this question
For concentrated solutions one needs to use Hammett $H_0$ function which gets lower than -10 in pure $\ce{HF}$ (pure sulfuric acid has -12) - it means it's very strong, although not superacidic.
See also:
The Hammett Acidity Function $H_0$ for Hydrofluoric Acid Solutions
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