feat: function now creates blueprint for integer part
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@ -26,7 +26,8 @@ C = {
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19: "十九",
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}
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D = {2: 10, 3: 100, 4: 1000, 5: 10000}
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# "Ten^x" dictonary
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T = {2: 10, 3: 100, 4: 1000, 5: 10000}
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def to_chinese_numeral(n):
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@ -36,29 +37,61 @@ def to_chinese_numeral(n):
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except KeyError:
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pass
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# Check if negative
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# Check if negative, turn positive and remeber state
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if n < 0:
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n = abs(n)
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is_negative = True
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else:
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is_negative = False
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"""# Put single digits in list
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split_number = [int(x) for x in str(n)]
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# Split number in iteger list and fractional (None if not exists)
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if type(n) is not int:
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n = str(n).split(".")
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integer = [int(x) for x in n[0]]
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fractional = [int(x) for x in n[1]]
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else:
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integer = [int(x) for x in str(n)]
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fractional = None
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arranged_number = []
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for _ in range(0, len(split_number)):
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arranged_number.append(split_number[0])
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if len(split_number) in D:
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arranged_number.append(D[len(split_number)])
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split_number.pop(0)
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if arranged_number[-1] == 0:
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arranged_number.pop(-1)
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# Blueprint list will contain building instruction on how to encode the number: 90090 -> [9, 10000, 0, 9, 10]
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blueprint = []
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if is_negative:
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arranged_number.insert(0, "-")"""
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blueprint.append(integer[0]) # First digit won't be 0 it can just be appended
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blueprint.append(T[len(integer)]) # Length of integer determines power of ten
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integer.pop(0) # Pop first digit
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# create rest of blueprint by looping rest of integer
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for _ in range(0, len(integer)):
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# 0 needs no power of ten
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if integer[0] == 0:
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blueprint.append(0)
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else:
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blueprint.append(integer[0])
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blueprint.append(T[len(integer)])
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integer.pop(0)
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# Pop last digit in blueprint if it's a 0
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for i in range(len(blueprint) - 1, 1, -1):
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if blueprint[i] == 0:
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blueprint.pop(i)
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else:
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break
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# Remove grouped zeros
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was_zero = False
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for j, i in enumerate(blueprint):
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if i == 0 and not was_zero:
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was_zero = True
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elif i == 0 and was_zero:
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blueprint.pop(j)
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elif i != 0 and was_zero:
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was_zero = False
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print(blueprint)
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print(fractional)
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return is_negative
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print(to_chinese_numeral(-90909)) # 9 10000 9
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to_chinese_numeral(-90090.358)
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to_chinese_numeral(90000)
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