Idiomatic way to do basic pattern matching on multiple variables in Python

I want to achieve what this code (written in Crystal) does:

enum PieceKind
Pawn, Rook, Bishop, Knight, King, Queen
end

def piece_kind_at_init(x, y)
case y
when 2, 7 then PieceKind::Pawn
when 1, 8
case x
when 1, 8 then PieceKind::Rook
when 2, 7 then PieceKind::Bishop
when 3, 6 then PieceKind::Knight
when 4 then PieceKind::King
when 5 then PieceKind::Queen
end
end
end


This is the "nicest" version I was able to come up with in Python:

class PieceKind(Enum):
Pawn = auto()
Rook = auto()
Bishop = auto()
Knight = auto()
King = auto()
Queen = auto()

def piece_kind_at_init(x, y):
if y in [2, 7]:
return PieceKind.Pawn
elif y in [1, 8]:
if x in [1, 8]:
return PieceKind.Rook
elif x in [2, 7]:
return PieceKind.Bishop
elif x in [3, 6]:
return PieceKind.Knight
elif x == 4:
return PieceKind.King
elif x == 5:
return PieceKind.Queen


It's fine, but it's definitely less readable the version with proper pattern matching. What's the most idiomatic way to write the code?

switch

elif

list

[Piecekind.Rook, Piecekind.Knight, Piecekind.Bishop, ...]

1 Answer
1

One way to get rid of lengthy elif chain is to create a translation table.

elif

outer_row_pieces = [PieceKind.Rook,
PieceKind.Knight,
PieceKind.Bishop,
PieceKind.King,
PieceKind.Queen,
PieceKind.Bishop,
Piecekind.Knight,
PieceKind.Rook]

def piece_kind_at_init(x, y):
if y == 2 or y == 7:
return PieceKind.Pawn
if y == 1 or y == 8:
return outer_row_pieces[x]
return None


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