Python 返回一个列表中的索引范围,以从另一个列表中获取值
我有两个大小相同的列表,一个具有数据类型(连续出现): 第二个列表是关于数据的:Python 返回一个列表中的索引范围,以从另一个列表中获取值,python,list,indices,Python,List,Indices,我有两个大小相同的列表,一个具有数据类型(连续出现): 第二个列表是关于数据的: data = ["xx","tr","kl","ut","ew","uy","lp","eq","aq"] 从数据列表中,我知道“xx”、“tr”是“vigi”和“kl”、“ut”、“ew”是“水果”等等 我每次需要的是将数据分成两个数
data = ["xx","tr","kl","ut","ew","uy","lp","eq","aq"]
从数据列表中,我知道“xx”、“tr”
是“vigi”
和“kl”、“ut”、“ew”
是“水果”
等等
我每次需要的是将数据
分成两个数据集,其中包括:
data1 = data[indices for type "vigi"]
data2 = data[indices for the remaining (i.e. data for "fruits" and "nothing")]
第二次将有:
data1 = data[indices for type "fruits"]
data2 = data[indices for the remaining (i.e. data for "vigi" and "nothing")]
等等
请提供任何帮助。您可以使用zip()
函数:
types = ["vigi","vigi","fruits","fruits","fruits","nothing","nothing","nothing","nothing"]
data = ["xx","tr","kl","ut","ew","uy","lp","eq","aq"]
data1 = [d for t, d in zip(types, data) if t == 'vigi']
data2 = [d for t, d in zip(types, data) if t != 'vigi']
print(data1)
print(data2)
印刷品:
['xx', 'tr']
['kl', 'ut', 'ew', 'uy', 'lp', 'eq', 'aq']
其他版本(仅在列表上迭代一次):
以下是如何使用
zip()
:
输出:
['xx', 'tr']
['kl', 'ut', 'ew', 'uy', 'lp', 'eq', 'aq']
['xx', 'tr']
['kl', 'ut', 'ew', 'uy', 'lp', 'eq', 'aq']
另一种方法是使用
enumerate()
:
输出:
['xx', 'tr']
['kl', 'ut', 'ew', 'uy', 'lp', 'eq', 'aq']
['xx', 'tr']
['kl', 'ut', 'ew', 'uy', 'lp', 'eq', 'aq']
但是当只需要一个时,它会在数据列表上迭代两次。相反,要提高时间复杂性,请使用for循环:
types = ["vigi","vigi","fruits","fruits","fruits","nothing","nothing","nothing","nothing"]
data = ["xx","tr","kl","ut","ew","uy","lp","eq","aq"]
data1 = []
data2 = []
for i,d in enumerate(data):
if types[i] == 'vigi':
data1.append(types[i])
elif types[i] in ['fruits','nothing']:
data2.append(types[i])
print(data1)
print(data2)
zip()
选项也是如此(参见@AndrejKesely对for循环的回答)。单循环解决方案的变体(参见其他答案)。依赖于这样一个事实,即布尔值在转换为int
时将变为0
或1
。可能被认为是黑客,但包括娱乐
data1, data2 = [], []
out = (data2, data1)
for t, d in zip(types, data):
out[t == 'vigi'].append(d)
print(data1)
print(data2)
可能不完全是您想要的,但如果您将来有更复杂的要求,发布此内容可能会有所帮助:
>>> from itertools import groupby, count
>>> index = count()
>>> database = {key: [*group] for key, group in groupby(data, lambda x:types[next(index)])}
>>> database
{'vigi': ['xx', 'tr'],
'fruits': ['kl', 'ut', 'ew'],
'nothing': ['uy', 'lp', 'eq', 'aq']}
参考资料:
可能只有我一个人,但这需要压缩和迭代两次,这让我非常恼火。可能使用“普通”循环,并根据需要附加到每个列表中
data1, data2 = [], []
out = (data2, data1)
for t, d in zip(types, data):
out[t == 'vigi'].append(d)
print(data1)
print(data2)
>>> from itertools import groupby, count
>>> index = count()
>>> database = {key: [*group] for key, group in groupby(data, lambda x:types[next(index)])}
>>> database
{'vigi': ['xx', 'tr'],
'fruits': ['kl', 'ut', 'ew'],
'nothing': ['uy', 'lp', 'eq', 'aq']}