Question

当我运行以下函数时，我只得到一个值的输出。有消息， “条件长度> 1，只使用第一个元素”

我需要做些什么更改才能获得所有值？

ownsquare <- function(n)
{
  for (i in seq(from = 1, to = 506, by = 0.001))
  {
    r <- i * i
    if (r <= n) x=i
  }
 x 
}

Answer 1

您正在尝试创建一个可以对多个值执行相同操作的函数，返回与其参数长度相等的向量。在R的说法中，函数的行为被称为“向量化”。并且mapply函数周围有一个名为Vectorize的包装函数。这说明它与您的函数一起使用，它可以在单个元素上正常工作：

 Vownsquare <- Vectorize(ownsquare)
 Vownsquare( 1:10 )
 #-------
 [1] 1.000 1.414 1.732 2.000 2.236 2.449 2.645 2.828 3.000 3.162

这真的很慢，但这是实施效率低下的结果。学习使用break可能会提高效率。这避免了在达到条件r> n之后的不必要的迭代。

 ownsquare <- function(n)
 {
   for (i in seq(from = 1, to = 506, by = 0.001))
   {
     r <- i * i
     if (r > n) { x=i; break() }
   }
  x 
 }
 Vownsquare <- Vectorize(ownsquare)
 Vownsquare( 1:10 )
[1] 1.001 1.415 1.733 2.001 2.237 2.450 2.646 2.829 3.001 3.163

请注意Lundberg实现的比较，该实现计算每个参数的完整506,000长度向量：

 system.time(ownsquare2(1:10))   # makes full vectors
   user  system elapsed 
  7.311   0.069   7.292 

 system.time(ownsquare(1:10))    # uses break
   user  system elapsed 
  0.008   0.000   0.008 

 system.time(Vownsquareoriginal( 1:10 ))  # full vectors with if()
   user  system elapsed 
  3.746   0.040   3.729

Answer 2

向量化此函数的另一种方法是使用import netCDF4 as nc import pandas as pd import numpy as np import datetime as datetime import matplotlib.pyplot as plt from mpl_toolkits.basemap import Basemap ## Read in netCDF file. NOTE: Change format if it does not read. NH_Ice=nc.Dataset('Oc-Tech/Final_Project/ncFiles/NH_SeaIce.nc', 'r') ## Save netCDF parameters into arrays. lons = NH_Ice.variables['longitude'][:] lats = NH_Ice.variables['latitude'][:] time = NH_Ice.variables['time'][:] time_units = NH_Ice.variables['time'].units ## Calculate Date from given data file dbase= '2001-01-01 00:00:00' db=datetime.datetime.strptime(dbase, "%Y-%m-%d %H:%M:%S") start=db+datetime.timedelta(days=time[0]) end=db+datetime.timedelta(days=time[-1]) #print fh.time_strlen si = NH_Ice.variables['seaice_conc'][:] NH_Ice.close() m = Basemap(projection='npstere',boundinglat=45,lon_0=-156,resolution='c') #total time steps tts=1980 for ti in range(0,tts): fig = None lat = None lon = None xi = None yi = None sip = None fig=plt.figure(figsize=(10,10)) m.drawcoastlines(color='black') m.fillcontinents(color='limegreen',lake_color='blue') m.drawparallels(np.arange(-80.,81.,20.)) m.drawmeridians(np.arange(-180.,181.,20.)) m.drawmapboundary(fill_color='darkblue') lon, lat = np.meshgrid(lons, lats) #sea ice xi, yi = m(lon, lat) sip=(si[ti]) ## Plot u data from a particular day m.pcolor(xi,yi,np.squeeze(sip)) plt.savefig("fig"+str(ti)+".png") #plt.close(fig)而不是ifelse：

if

此处，ownsquare2 <- function(n) { for (i in seq(from = 1, to = 506, by = 0.001)) { r <- i * i x <- ifelse(r <= n, i, x) } x } ownsquare2(1:10) ## [1] 1.000 1.414 1.732 2.000 2.236 2.449 2.645 2.828 3.000 3.162被视为函数中的向量，x表达式中使用的替换使用if进行矢量化。

从ifelse到if的转换相当简单，但对于链接的if-else代码变得不可读（IMO）。

条件的长度> 1，只使用第一个元素。要做出哪些改变？

2 个答案: