Question

如何重复列表n次的每个元素并形成新列表？例如：

x=[1,2,3,4]
n=3

x1=[1,1,1,2,2,2,3,3,3,4,4,4]

x*n无法正常工作

for i in x[i]
    x1=n*x[i]

必须有一种简单而聪明的方式。

Answer 1

理想的方法可能是numpy.repeat：

In [16]:

x1=[1,2,3,4]
In [17]:

np.repeat(x1,3)
Out[17]:
array([1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4])

Answer 2

如果你真的想要结果作为列表，并且生成器是不够的：

import itertools
lst = range(1,5)
list(itertools.chain.from_iterable(itertools.repeat(x, 3) for x in lst))

Out[8]: [1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

Answer 3

您可以使用列表理解：

[item for item in x for i in range(n)]

>>> x=[1, 2, 3, 4]
>>> n = 3
>>> new = [item for item in x for i in range(n)]
>>> new
[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
>>>

Answer 4

嵌套的list-comp在这里工作：

>>> [i for i in range(10) for _ in xrange(3)]
[0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9]

或者使用你的例子：

>>> x = [1, 2, 3, 4]
>>> n = 3
>>> [i for i in x for _ in xrange(n)]
[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

Answer 5

 [myList[i//n] for i in range(n*len(myList))]

Answer 6

实现此目的的更简单方法是将列表x与n相乘并对结果列表进行排序。 e.g。

>>> x = [1,2,3,4]
>>> n = 3
>>> a = sorted(x*n)
>>> a
>>> [1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

Answer 7

这将解决您的问题：

x=[1,2,3,4]
n = 3
x = sorted(x * n)

Answer 8

import itertools

def expand(lst, n):
    lst = [[i]*n for i in lst]
    lst = list(itertools.chain.from_iterable(lst))
    return lst

x=[1,2,3,4]
n=3
x1 = expand(x,3)

print(x1)

给出：

[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]

说明：

执行，[3]*3给出[3,3,3]的结果，将其替换为n我们得到[3,3,3,...3] (n times)使用列表理解我们可以遍历列表的每个元素并执行这个操作，最后我们需要展平列表，我们可以通过list(itertools.chain.from_iterable(lst))

来完成

Answer 9

如果您想要就地修改列表，最好的方法是从后面进行迭代，并将之前一个项目的切片分配给该项目的列表n次。

这是因为切片分配：

>>> ls = [1, 2, 3]
>>> ls[0: 0+1]
[1]
>>> ls[0: 0+1] = [4, 5, 6]
>>> ls
>>> [4, 5, 6, 2, 3]

def repeat_elements(ls, times):
    for i in range(len(ls) - 1, -1, -1):
        ls[i: i+1] = [ls[i]] * times

演示用法：

>>> a = [1, 2, 3]
>>> b = a
>>> b
[1, 2, 3]
>>> repeat_elements(b, 3)
>>> b
[1, 1, 1, 2, 2, 2, 3, 3, 3]
>>> a
[1, 1, 1, 2, 2, 2, 3, 3, 3]

（如果你不想就地修改它，你可以复制列表并返回副本，这不会修改原文。这也适用于其他序列，如tuple s ，但不像itertools.chain.from_iterable和itertools.repeat方法那样懒惰）

def repeat_elements(ls, times):
    ls = list(ls)  # Makes a copy
    for i in range(len(ls) - 1, -1, -1):
        ls[i: i+1] = [ls[i]] * times
    return ls

Answer 10

zAxe=[]
for i in range(5):
    zAxe0 =[i] * 3
    zAxe +=(zAxe0) # append allows accimulation of data

Answer 11

方式1：

def foo():
    for j in [1, 3, 2]:
        yield from [j]*5

方式2：

from itertools import chain
l= [3, 1, 2]
chain(*zip(*[l]*3))

方式3：

sum(([i]*5 for i in [2, 1, 3]), [])

Answer 12

对于基本Python 2.7：

    from itertools import repeat
    def expandGrid(**kwargs):
        # Input is a series of lists as named arguments
        # output is a dictionary defining each combination, preserving names
        #
        # lengths of each input list
        listLens = [len(e) for e in kwargs.itervalues()] 
        # multiply all list lengths together to get total number of combinations
        nCombos = reduce((lambda x, y: x * y), listLens) 
        iDict = {}
        nTimesRepEachValue=1 #initialize as repeating only once
        for key in kwargs.keys():
            nTimesRepList=nCombos/(len(kwargs[key])*nTimesRepEachValue)
            tempVals=[] #temporary list to store repeated
            for v in range(nTimesRepList):
                indicesToAdd=reduce((lambda x,y: list(x)+list(y)),[repeat(x, nTimesRepEachValue) for x in kwargs[key]])
                tempVals=tempVals+indicesToAdd
            iDict[key] = tempVals
            # Accumulating the number of times needed to repeat each value
            nTimesRepEachValue=len(kwargs[key])*nTimesRepEachValue
        return iDict

    #Example usage:
    expandedDict=expandGrid(letters=["a","b","c","d"],nums=[1,2,3],both=["v",3])

重复列表的元素n次

12 个答案: