你不应该担心寻找元素的“性能”:
- 元素不超过 200 个,对于计算机来说是个小数目;
- 由于程序与人类用户交互,因此人类无论如何都会比计算机慢几个数量级。
选项 1:pandas.DataFrame
因此我建议一个简单的pandas DataFrame:
import pandas as pd
df = pd.read_csv('element.txt')
df.columns = ['Number', 'Symbol', 'Name']
def get_column_and_key(s):
s = s.strip()
try:
k = int(s)
return 'Number', k
except ValueError:
if len(s) <= 2:
return 'Symbol', s
else:
return 'Name', s
def find_element(s):
column, key = get_column_and_key(s)
return df[df[column] == key]
def play():
keep_going = True
while keep_going:
s = input('>>>> ')
if s[0] == 'q':
keep_going = False
else:
print(find_element(s))
if __name__ == '__main__':
play()
另见:
选项 2:三个冗余字典
python 最常用的数据结构之一是dict。这里我们有三个不同的可能键,所以我们将使用三个 dict。
import csv
with open('element.txt', 'r') as f:
data = csv.reader(f)
elements_by_num = {}
elements_by_symbol = {}
elements_by_name = {}
for row in data:
num, symbol, name = int(row[0]), row[1], row[2]
elements_by_num[num] = num, symbol, name
elements_by_symbol[symbol] = num, symbol, name
elements_by_name[name] = num, symbol, name
def get_dict_and_key(s):
s = s.strip()
try:
k = int(s)
return elements_by_num, k
except ValueError:
if len(s) <= 2:
return elements_by_symbol, s
else:
return elements_by_name, s
def find_element(s):
d, key = get_dict_and_key(s)
return d[key]
def play():
keep_going = True
while keep_going:
s = input('>>>> ')
if s[0] == 'q':
keep_going = False
else:
print(find_element(s))
if __name__ == '__main__':
play()