Python 重复列表的元素 n 次
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Repeating elements of a list n times
提问by user3720101
How do I repeat each element of a list ntimes and form a new list? For example:
如何重复列表中的每个元素n并形成一个新列表？例如：
x=[1,2,3,4]
n=3
x1=[1,1,1,2,2,2,3,3,3,4,4,4]
x*ndoesn't work
x*n不起作用
for i in x[i]
x1=n*x[i]
There must be a simple and smart way.
必须有一个简单而聪明的方法。
采纳答案by m.wasowski
In case you really want result as list, and generator is not sufficient:
如果你真的想要结果作为列表，而生成器是不够的：
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]
回答by mgilson
A nested list-comp works here:
一个嵌套的 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]
Or to use your example:
或者使用您的示例：
>>> 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]
回答by Dair
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)
Gives:
给出：
[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
Explanation:
解释：
Doing, [3]*3gives the result of [3,3,3], replacing this with nwe get [3,3,3,...3] (n times)Using a list comprehension we can go through each elem of the list and perform this operation, finally we need to flatten the list, which we can do by list(itertools.chain.from_iterable(lst))
做，[3]*3给出的结果[3,3,3]，用n我们得到的替换它[3,3,3,...3] (n times)使用列表推导式，我们可以遍历列表的每个元素并执行此操作，最后我们需要展平列表，我们可以通过list(itertools.chain.from_iterable(lst))
回答by A.J. Uppal
You can use list comprehension:
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)]
#[1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
回答by CT Zhu
The ideal way is probably numpy.repeat:
理想的方式可能是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])
回答by Artyer
If you want to modify the list in-place, the best way is to iterate from the back and assign a slice of what was previously one item to a list of that item ntimes.
如果您想就地修改列表，最好的方法是从后面迭代并将之前一个项目的一部分分配给该项目n时间的列表。
This works because of slice assignment:
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
Demo usage:
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]
(If you don't want to modify it in-place, you can copy the list and return the copy, which won't modify the original. This would also work for other sequences, like tuples, but is not lazy like the itertools.chain.from_iterableand itertools.repeatmethod)
（如果你不想就地修改它，你可以复制列表并返回副本，这不会修改原始的。这也适用于其他序列，如tuples，但不像itertools.chain.from_iterableand那样懒惰itertools.repeat方法）
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
回答by Sohaib Farooqi
A simpler way to achieve this to multiply the list xwith nand sort the resulting list. e.g.
实现此目的的一种更简单的方法是将列表x与n结果列表相乘并对结果列表进行排序。例如
>>> x = [1,2,3,4]
>>> n = 3
>>> a = sorted(x*n)
>>> a
>>> [1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
回答by Peyman Obeidy
zAxe=[]
for i in range(5):
zAxe0 =[i] * 3
zAxe +=(zAxe0) # append allows accimulation of data
回答by Ross
For base Python 2.7:
对于基础 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])
