Python 创建一个包含 100 个整数的列表，其值等于它们的索引
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Create a list of 100 integers whose values equal their indexes
提问by Bob
Create a list of 100 integers whose value and index are the same, e.g.
创建一个包含 100 个值和索引相同的整数的列表，例如
mylist[0] = 0, mylist[1] = 1, mylist[2] = 2, ...
Here is my code.
这是我的代码。
x_list=[]
def list_append(x_list):
for i in 100:
x_list.append(i)
return(list_append())
print(x_list)
采纳答案by Veedrac
Since nobody else realised you're using Python 3, I'll point out that you should be doing list(range(100))to get the wanted behaviour.
由于没有其他人意识到您使用的是 Python 3，我会指出您应该这样做list(range(100))以获得所需的行为。
回答by TerryA
for i in 100doesn't do what you think it does. intobjects are not iterable, so this won't work. The for-loop tries to iterate through the object given.
for i in 100不会做你认为它会做的事情。int对象不可迭代，所以这行不通。for 循环尝试遍历给定的对象。
If you want to get a list of numbers between 0-100, use range():
如果您想获取 0-100 之间的数字列表，请使用range()：
for i in range(100):
dostuff()
The answer to your question is pretty much range(100)anyway:
range(100)无论如何，您的问题的答案几乎是：
>>> range(100)[0]
0
>>> range(100)[64]
64
回答by Andreas Jung
Use range() for generating such a list
使用 range() 生成这样的列表
>>> range(10)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> range(10)[5]
5
回答by tom10
You can use range(100), but it seems that you are probably looking to make the list from scratch, so there you can use while:
您可以使用range(100)，但似乎您可能希望从头开始制作列表，因此您可以使用while：
x_list=[]
i = 0
while i<100:
x_list.append(i)
i += 1
Or you could do this recursively:
或者您可以递归地执行此操作：
def list_append(i, L):
L.append(i)
if i==99:
return L
list_append(i+1, L)
x_list = []
list_append(0, x_list)
print x_list
回答by PiMathCLanguage
If you want to import numpyyou could do something like this:
如果要导入numpy，可以执行以下操作：
import numpy as np
x_list = np.arange(0, 100).tolist()
Should work in python2.7 and python3.x
应该在 python2.7 和 python3.x 中工作
回答by codingBoy
Also can use List Comprehensions, like
也可以使用列表理解，比如
[x for x in range(100)]
回答by zyxwvutsr qpnmlkj
import random
data1=[]
def f(x):
return(random.randrange(0,1000))
for x in range (0,100):
data1.append(f(x))
data1
