Question

我尝试在: seprated之间添加逗号分隔值，然后乘以整个值例如，考虑我的值为1,2,3:4,5,6 我想添加1 + 2 + 3和4 + 5 + 6然后乘以该值的结果，所以答案是6 * 15 = 90

对于我的波纹管数据，我希望结果为7.224，但此脚本会显示61.658886435 我不知道我的脚本'

中有什么问题

ar = "0.212,1.231,0.112:1.001,3.212,0.002:0.002,0.0001,1.1"
x_data = ar.split(":")
x_final = 1
x_add = 0
for i in x_data:
    x_each = i.split(",")
    for j in x_each:
        x_add = x_add + float(j)
    x_final = x_add * x_final

print x_final

有没有可能的方法来获得没有迭代循环的结果？对于上述问题

Answer 1

这个问题也可以通过功能性的方式解决：

您必须将列表中的所有值相乘 - 这是functools.reduce + operator.mul
您必须总结所有内部列表中的所有值 - 这是sum

示例：

In [5]: ar = "0.212,1.231,0.112:1.001,3.212,0.002:0.002,0.0001,1.1"

In [6]: import operator

In [7]: import functools

In [8]: functools.reduce(operator.mul, (sum(float(x) for x in s.split(',')) for s in ar.split(':')))
Out[8]: 7.223521582500001

Answer 2

我不一定会推荐这个复杂的表达式，但是您可以使用列表推导来避免for循环：

import operator

ar = "0.212,1.231,0.112:1.001,3.212,0.002:0.002,0.0001,1.1"
reduce(operator.mul, [sum([float(n) for n in e]) for e in [x.split(',') for x in ar.split(":")]], 1)

Answer 3

在每次迭代中使用错过初始化值为零（x_add = 0）。所以你的脚本添加了以前的值

ar = "0.212,1.231,0.112:1.001,3.212,0.002:0.002,0.0001,1.1"
x_data = ar.split(":")
x_final = 1
for i in x_data:
    x_each = i.split(",")
    x_add = 0             # Here you not initialize it
    for j in x_each:
        x_add = x_add + float(j)
    x_final = x_add * x_final

print x_final

!!!来自@ jpmc26和@soon评论。避免使用eval，并符合您的输入字符串格式。没有循环使用正则表达式来做到这一点

使用正则表达式解决问题而不循环。

ar = "0.212,1.231,0.112:1.001,3.212,0.002:0.002,0.0001,1.1"
import re
ar = "("+ar                      #Add the ( with your data 
ar = re.sub(r",","+",ar)         #Substitute with + instead of ,
ar = re.sub(r"(?=\:|$)",")",ar)  #look ahead for add `)` after colon
ar = re.sub(r"(?<=)\:","*(",ar)  #Replace the color with *

#NOw you data look likes(0.212+1.231+0.112)*(1.001+3.212+0.002)*(0.002+0.0001+1.1)
#Finally evaluvate the string as a expression
print eval(ar)

为每个人提供不同的结果？

3 个答案: