Question

我想快速访问np.array元素，例如从0-6加10到最后的索引。到目前为止，我已经尝试过：

Fold_5 = len(predictorX)/5
trainX = np.concatenate(predictorX[:3*int(Fold_5)],predictorX[4*int(Fold_5)])

或

[[0.1,0.4,0.6,0.2],[..]....]

两个都给了我错误，第二个给了我：“TypeError：只有整数标量数组可以转换为标量索引”

编辑：concatenate仍然给我带来问题，所以我将在这里发布我的完整代码：

set.seed(123)

names <- c("acquisitionsmergers", "analystratings", "assets", "balanceofpayments", 
"bankruptcy", "civilunrest", "corporateresponsibility", "credit", 
"creditratings", "crime", "dividends", "earnings", "equityactions", 
"exploration", "government", "indexes", "industrialaccidents", 
"insidertrading", "investorrelations", "laborissues", "legal", 
"marketing", "orderimbalances", "partnerships", "pricetargets", 
"productsservices", "publicopinion", "regulatory", "revenues", 
"security", "stockprices", "taxes", "technicalanalysis", "transportation", 
"warconflict")

mylabels <- c("acquisitionsmergers" = "Acquisitions/Mergers",
                   "analystratings" = "Analyst Ratings",
                   "assets" = "Assets",
                   "balanceofpayments" = "Balance of Payments",
                   "bankruptcy" = "Bankruptcy",
                   "civilunrest" = "Civil Unrest",
                   "corporateresponsibility" = "Corporate Responsibility",
                   "credit" = "Credit",
                   "creditratings" = "Credit Ratings",
                   "crime" = "Crime",
                   "dividends" = "Dividends",
                   "earnings" = "Earnings",
                   "equityactions" = "Equity Actions",
                   "exploration" = "Exploration",
                   "government" = "Government",
                   "indexes" = "Indexes",
                   "industrialaccidents" = "Industrial Accidents",
                   "insidertrading" = "Insider Trading",
                   "investorrelations" = "Investor Relations",
                   "laborissues" = "Labor Issues",
                   "legal" = "Legal",
                   "marketing" = "Marketing",
                   "orderimbalances" = "Order Imbalances",
                   "partnerships" = "Partnerships",
                   "pricetargets" = "Price Targets",
                   "productsservices" = "Product Services",
                   "publicopinion" = "Public Opinion",
                   "regulatory" = "Regulatory",
                   "revenues" = "Revenues",
                   "security" = "Security",
                   "stockprices" = "Stockprices",
                   "taxes" = "Taxes",
                   "technicalanalysis" = "Technical Analysis",
                   "transportation" = "Transportation",
                   "warconflict" = "War Conflict")

df <- data.frame(item = rep(names, each=5), value=rnorm(5*35,5,2), date = rep(seq(as.Date("2000/1/1"), by = "month", length.out = 5),35))

预测变量X是一个值为

的数组

library(ggplot2)
ggplot(df, aes(x=date, y=value, color=item)) +
  geom_line() + 
  facet_wrap( ~ item, ncol=4, scales="free_y", labeller = mylabels)

Answer 1

以下是获得所需子阵列的两种简短方法：

np.delete(a, np.s_[6:10])

和

np.r_[a[:6], a[10:]]

Answer 2

在：

a[0:6,10:]

0：6选择行，10：选择列。如果a不是2d或足够大会导致错误。

在

np.concatenate(a[0:6],a[10:])

问题是参数的数量;它需要一个数组列表。第二个，如果给出被理解为axis，它应该是一个整数（因此你的错误）。

np.concatenate([a[0:6],a[10:]])

应该有用。

另一个选择是使用列表

进行索引

a[0,1,2,3,4,5,10,11,...]]

np.r_是构建此类列表的便捷小工具：

In [73]: np.r_[0:6, 10:15]
Out[73]: array([ 0,  1,  2,  3,  4,  5, 10, 11, 12, 13, 14])

它实际上是np.concatenate([np.arange(0,6),np.arange(10,15)])。

无论是首先索引还是连接，或者先连接索引然后索引，都无关紧要。效率大致相同。 np.delete选择多种方法，包括这些方法，具体取决于＆＃39;删除＆＃39;的大小和类型。区域。

在trainX表达式中，将[]添加到连接调用应该有效。但是，predictorX[4*Fold_5]可能是个问题。你错过了:（例如10：例子）吗？如果您只想要一个值，那么您需要将其转换为1d，例如predictorX[[4*Fold_5]]

Fold_5 = len(predictorX)//5   # integer division in py3
trainX = np.concatenate([predictorX[:3*Fold_5], predictorX[4*Fold_5:]])

Answer 3

np.concatenate采用一系列数组。试试

np.concatenate([a[0:6],a[10:]])

或

np.concatenate((a[0:6],a[10:]))

访问np.array的部分

3 个答案: