python浮点数可以不带小数吗_python – 对于给定的浮点数列表,有没有办法可以找到给定整个部分的最大小数部分？...

我遇到了这个问题,在那里我被绊倒了.

问题是这个.

我有一个这样的列表：

L = [2.1, 2.2, 2.3, 2.4, 2.5, 3.1, 3.2, 3.3, 3.4, 4.1, 4.2, 4.3, 4.4, 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 6.1, 6.2, 6.3, 6.4, 6.5, 7.1, 7.2, 7.3, 7.4, 7.5, 8.1, 8.2, 8.3, 8.4, 8.5, 9.1, 9.2, 9.3, 9.4, 9.5, 10.1, 10.2, 10.3, 10.4, 10.5, 11.1, 11.2, 11.3, 11.4, 11.5, 12.1, 12.2, 12.3, 12.4, 12.5]

需要：

ans = [[2,5],[3,4],[4,4],……]

即整数和最大小数部分.

我们应该怎么做？

到目前为止,我试过这个：

# Imports

import numpy as np

import math

L = [2.1, 2.2, 2.3, 2.4, 2.5, 3.1, 3.2, 3.3, 3.4, 4.1, 4.2, 4.3, 4.4, 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 6.1, 6.2, 6.3, 6.4, 6.5, 7.1, 7.2, 7.3, 7.4, 7.5, 8.1, 8.2, 8.3, 8.4, 8.5, 9.1, 9.2, 9.3, 9.4, 9.5, 10.1, 10.2, 10.3, 10.4, 10.5, 11.1, 11.2, 11.3, 11.4, 11.5, 12.1, 12.2, 12.3, 12.4, 12.5]

whole = [ int(str(x).split(".")[0]) for x in L]

frac = [ int(str(x).split(".")[-1]) for x in L]

wf = [[w,f] for w,f in zip(whole,frac)]

wset = list(set(whole))

print(whole)

print(frac)

print(wf)

print(wset)

解决方法:

这是一个numpy解决方案：

L.sort() # skip this if L is already sorted

Li = L.astype(int)

uniq = np.r_[np.where(np.diff(Li))[0], Li.size-1]

int_part = Li[uniq]

max_frac = np.round((L[uniq]-int_part) * 10).astype(int)

np.c_[int_part, max_frac].tolist()

# [[2, 5], [3, 4], [4, 4], [5, 6], [6, 5], [7, 5], [8, 5], [9, 5], [10, 5], [11, 5], [12, 5]]

一些时间：

import numpy as np

import itertools as it

from timeit import timeit

def pp(L):

Li = L.astype(int)

uniq = np.r_[np.where(np.diff(Li))[0], Li.size-1]

int_part = Li[uniq]

max_frac = np.round((L[uniq]-int_part) * 10).astype(int)

return np.c_[int_part, max_frac].tolist()

def johnchase(L):

vals = []

foo = [str(e).split('.') for e in L]

for key, group in it.groupby(foo, lambda x: (x[0])):

vals.append([int(e) for e in max(group)])

return vals

def bernie(L):

return [[k, int(round(max(i % 1 for i in g)*10,1))]

for k, g in it.groupby(L,lambda x:int(x))]

def kprabhakaran(L):

return [map(int,str(max(list(cgen))).split('.'))

for c,cgen in it.groupby(L,lambda x:int(x))]

def prep(N, k):

global L

L = np.sort(np.random.choice(np.arange(N) / 10, k, False))

N, k, n = 2 * 10**6, 10**6, 10

for func in pp, johnchase, bernie, kprabhakaran:

print('{:20s}'.format(func.__name__), '{:6.4f} secs'.format(timeit(

lambda: func(L), lambda: prep(N, k), number=n) / n))

打印：

pp 0.0379 secs

johnchase 1.7252 secs

bernie 1.4773 secs

kprabhakaran 0.7592 secs

标签：python,numpy

来源： https://codeday.me/bug/20190727/1552497.html