Question

这可能是一个重复的问题，但我已经查看了重复的问题，我没有看到python的答案。

我正在尝试创建一个简单的程序，我可以对整数列表进行硬编码并过滤掉所有具有整数平方根的数字。这是我到目前为止所做的：

# Python Program to display the number of integer square roots in a list
import math

# Change this list for testing
terms = [1,2,3,4,5,6,7,9,10,11,12,13,16,24,36]

all_roots = []  #list to hold all square roots from terms list

for i in terms:
    all_roots.append(math.sqrt(i))

#integer_roots = list(filter(lambda x: [????], all_roots)) #not sure what I should put in the "[????]" part

print(all_roots) #prints integer and floating square roots
print(integer_roots) #supposed to print only square roots that are integers

为了澄清，print(all_roots)目前显示：

[1.0, 1.4142135623730951, 1.7320508075688772, 2.0, 2.23606797749979, 2.449489742783178, 2.6457513110645907, 3.0, 3.1622776601683795, 3.3166247903554, 3.4641016151377544, 3.605551275463989, 4.0, 4.898979485566356, 6.0]

但我希望print(integer_roots)显示

[1, 2, 3, 4, 6]

Answer 1

列表解析有助于在内部生成表达式以防止重复的函数调用：

>>> [int(i) for i in (math.sqrt(i) for i in terms) if i.is_integer()]
[1, 2, 3, 4, 6]

或者，如果您想要一种纯粹的功能性方法，因为您建议在帖子中使用filter()：

>>> list(map(int, filter(float.is_integer, map(math.sqrt, terms))))
[1, 2, 3, 4, 6]

当然这可以用常规for循环写出来，这个循环性能较差但完全可以接受，特别是考虑到OP似乎是Python的新东西：

lst = []
for i in terms:
    s = math.sqrt(i)
    if s.is_integer():
         lst.append(int(s))

Answer 2

import math
terms = [1,2,3,4,5,6,7,9,10,11,12,13,16,24,36]

integer_roots = [] 

for i in terms:
    tmp = math.sqrt(i)
    if (tmp.is_integer()):
         integer_roots.append(int(tmp))

print(integer_roots)

整数平方根python列表

2 个答案: