在Python中使用频率列表创建列表

我们如何创建一个新列表，根据另一个列表多次添加一个列表中的项目

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感谢使用

itertools

和

zip

Ex:

age = [19, 20, 21, 22, 23, 24, 25]
frequency = [2, 1, 1, 3, 2, 1, 1]
output_age = [19, 19, 20, 21, 22, 22, 22, 23, 23, 24, 25]

from itertools import chain
age = [19, 20, 21, 22, 23, 24, 25]
frequency = [2, 1, 1, 3, 2, 1, 1]

print( list(chain.from_iterable([[i] * v for i,v in zip(age, frequency)])) )

输出：

age = [19, 20, 21, 22, 23, 24, 25]
frequency = [2, 1, 1, 3, 2, 1, 1]
output_age = [19, 19, 20, 21, 22, 22, 22, 23, 23, 24, 25]

from itertools import chain
age = [19, 20, 21, 22, 23, 24, 25]
frequency = [2, 1, 1, 3, 2, 1, 1]

print( list(chain.from_iterable([[i] * v for i,v in zip(age, frequency)])) )

• 注意：

  链。从\u iterable

  展开列表

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使用列表理解：

[19, 19, 20, 21, 22, 22, 22, 23, 23, 24, 25]

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为什么？

我们首先将

zip

和

age

和

frequency

列表放在一起，这样我们就可以一致地对它们进行迭代。因此：

output_age = [i for l in ([a]*f for a, f in zip(age, frequency)) for i in l]
#[19, 19, 20, 21, 22, 22, 22, 23, 23, 24, 25]

给出：

for a, f in zip(age, frequency):
    print(a, f)

然后我们要重复每个元素，

a

，重复次数与

f

确定的次数相同。这可以通过创建一个列表并将其相乘来实现。就像：

19 2
20 1
21 1
22 3
23 2
24 1
25 1

然后，我们需要解压这些值，以便将该表达式包装在生成器中（用括号表示），并对其进行迭代。这将使列表变得平坦。请注意，实现这一点有多种方法（例如使用

itertools.chain.from\u iterable

）

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另一种方法是通过在

范围

对象上迭代，而不是乘以列表来获得重复，从而重复编号

a

此方法类似于：

[4] * 3
#[4, 4, 4]

你需要：

output_age = [a for a, f in zip(age, frequency) for _ in range(f)]
#[19, 19, 20, 21, 22, 22, 22, 23, 23, 24, 25]

输出：

age = [19, 20, 21, 22, 23, 24, 25]
frequency = [2, 1, 1, 3, 2, 1, 1]
output_age = [19, 19, 20, 21, 22, 22, 22, 23, 23, 24, 25]

from itertools import chain
age = [19, 20, 21, 22, 23, 24, 25]
frequency = [2, 1, 1, 3, 2, 1, 1]

print( list(chain.from_iterable([[i] * v for i,v in zip(age, frequency)])) )

---

[19,19,20,21,22,22,22,23,23,24,25]

最简单、最容易理解的方法

age = [19, 20, 21, 22, 23, 24, 25]
frequency = [2, 1, 1, 3, 2, 1, 1]
new_list = zip(age, frequency)
output_age=[]
for x,y in new_list:
  for i in range(y):
    output_age.append(x)

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下面是一个使用

zip

和

range

age = [19, 20, 21, 22, 23, 24, 25]
frequency = [2, 1, 1, 3, 2, 1, 1]

output_age = []
for age, freq in zip(age, frequency):
    for _ in range(freq):
        output_age.append(age)

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您也可以使用

sum

功能执行此操作，但不建议对生产代码执行此操作：

>>> age = [19, 20, 21, 22, 23, 24, 25]
>>> frequency = [2, 1, 1, 3, 2, 1, 1]
>>> [a for a,f in zip(age, frequency) for _ in range(f)]
[19, 19, 20, 21, 22, 22, 22, 23, 23, 24, 25]

输出：

age = [19, 20, 21, 22, 23, 24, 25]
frequency = [2, 1, 1, 3, 2, 1, 1]
output_age = sum([[age[i]] * frequency[i] for i in range(len(age))],[])
print(output_age)

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“依赖于另一个列表的次数”你到底是什么意思？项目数量或其值或？谢谢，但这不太正确，这导致输出年龄=[[19,19]，[20]，[21]，[22,22]，[23,23]，[24]，[25]。@jr100再次查看，我更正了它