小编san*_*epp的帖子

存在具有时间复杂度的算法

    T(n)=T(n-1)+1/n if n>1
        =1          otherwise

我正在解决它的渐近复杂性,并将命令作为'n',但给出的答案是'log n'.这是对的吗？如果是log n,那么为什么呢？

我有看起来像这样的数据集`

"Name of Countries","2001","2002","2003","2004","2005","2006","2007","2008","2009","2010"
"BANGLADESH","431312","435867","454611","477446","456371","484401","480240","541884","468899","431962" 
"SRILANKA","112813","108008","109098","128711","136400","154813","204084","218805","239995","266515" 
"UK","405472","387846","430917","555907","651803","734240","796191","776530","769251","759494" 
"USA","329147","348182","410803","526120","611165","696739","799062","804933","827140","931292"

我想用y轴作为值绘制行，而x轴是年份。我尝试过美国

t=df[df['Name of Countries']=='USA']
x=pd.DataFrame([t.iloc[0].index,t.iloc[0].values]).T
x.plot()
plt.show() 

看起来很丑陋的代码。我得到的是

我想要-USA在图例和X轴上作为列的名称[2001,2002 ... 2010]，并且可以以更好的方式完成它，而不必像我正在做的那样遍历单个行。`

我努力了

re.findall(r'(\d\*\*\d)','3*2**3**2*5**4**')

输出是['2**3', '5**4']. 我想要的输出是['2**3','3**2', '5**4']. 重新需要什么改变？