Question

初学者编码器！我正在寻找一个循环来检查名为＆＃34; budget＆＃34;看它是否为负，如果是，它将使用＆＃34;年＆＃34; list将年份添加到名为＆＃34; nodef＆＃34;的列表中。我的最终结果是列表＆＃34; nodef＆＃34;包含所有没有赤字的年份。

budget = [-1075,1225,4239,6084,1489,4031,1846,600,-6409,-19262,-14011,-12969,-9220,-10453,-10315,-3500,-1500,600,600,900]

years = [2000,2001,2002,2003,2004,2005,2006,2007,2008,2009,2010,2011,2012,2013,2014,2015,2016,2017,2018,2019]

nodef = []

i = 0

if i <21:

 if budget[i]<0:

   nodef.append(years[i])
   i += 1
else:

print(nodef)

这就是我现在所拥有的。请告诉我如何批准此事。这完全是出于我自己的好奇心。

Answer 1

budgets = [-1075,1225,4239,6084,1489,4031,1846,600,-6409,-19262,-14011,-12969,-9220,-10453,-10315,-3500,-1500,600,600,900]

years = [2000,2001,2002,2003,2004,2005,2006,2007,2008,2009,2010,2011,2012,2013,2014,2015,2016,2017,2018,2019]

nodef = []

for budget, year in zip(budgets, years):
    if budget < 0:
        nodef.append(year)

print(nodef)

Answer 2

如果要将输出限制为21，可以这样做：

nodef = [year for budget, year in zip(budgets, years) if budget < 0][:21]

如果你真的不需要它，请忽略[:21]。

另一种方法：

nodef = [years[i] for i, budget in enumerate(budgets) if budget < 0]

Answer 3

另一种方法是：

nodef = list(map(lambda k: years[k], filter(lambda i: budget[i]<0, range(len(budget)))))
print(nodef)

它产生以下输出：

[2000, 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016]

添加到循环中的列表/数组

3 个答案: