venice(VGAM)
venice()所属R语言包：VGAM
Venice Maximum Sea Levels Data
威尼斯最大海平面数据
译者：生物统计家园网 机器人LoveR
描述----------Description----------
Some sea levels data sets recorded at Venice, Italy.
在威尼斯，意大利的一些海平面记录的数据集。
用法----------Usage----------
data(venice)
data(venice90)
格式----------Format----------
venice is a data frame with 51 observations on the following 11 variables. It concerns the maximum heights of sea levels between 1931 and 1981.
venice是一个数据框，51以下11个变量的观察。在1931年和1981年之间，它涉及到的海平面上升的最大高度。
year a numeric vector.
一年一个数值向量。
r1,r2,r3,r4,r5,r6,r7,r8,r9,r10 numeric vectors; r1 is the highest recorded value, r2 is the second highest recorded value, etc.
R1，R2，R3，R4，R5，R6，R7，R8，R9，R10数值向量;r1是最高的记录值，r2是第二个记录的最高值，等
venice90 is a data frame with 455 observations on the following 7 variables.
venice90 455观察以下7个变量是一个数据框。
year, month, day, hour numeric vectors;
年，月，日，小时数值向量;
sealevel numeric; sea level.
海平面数字;海平面。
ohour numeric; number of hours since the midnight of 31 Dec 1939 and 1 Jan 1940.
ohour数字，1939年12月31日和1940年1月1日午夜以来的小时数。
Year numeric vector; approximate year as a real number. The formula is start.year + ohour / (365.26 * 24) where start.year is 1940. One can treat Year as continuous whereas year can be treated as both continuous and discrete.
年份数值向量;近似作为一个实数年。其计算公式为start.year + ohour / (365.26 * 24)其中start.year为1940。可以治疗Year持续，而year可以看作是连续和离散。
Details
详细信息----------Details----------
Sea levels are in cm. For venice90, the value 0 corresponds to a fixed reference point (e.g., the mean sea level in 1897 at an old palace of Venice). Clearly since the relative (perceived) mean sea level has been increasing in trend over time (more than an overall 0.4 m increase by 2010), therefore the value 0 is (now) a very low and unusual measurement.
海平面厘米。对于venice90，0对应一个固定的参考点（例如，在1897年的平均海平面在威尼斯的古老宫殿的）。显然，由于相对（知觉）平均海平面不断上升的趋势，随着时间的推移更不是一个整体的0.4米2010年增加，因此值为0时（现在）非常低的和不寻常的测量。
For venice, in 1935 only the top six values were recorded.
对于venice，仅在1935年的前六位值记录。
For venice90, this is a subset of a data set provided by Paolo Pirazzoli consisting of hourly sea levels from 1940 to 2009. Values greater than 90 cm were extracted, and then declustered (each cluster provides no more than one value, and each value is at least 24 hours apart). Thus the values are more likely to be independent. Of the original (2009-1940+1)*365.26*24 values about 7 percent of these comprise venice90.
对于venice90，这是一个子集，数据集提供每小时海平面1940至2009年由保罗·皮拉左里。超过90厘米的值更大的值被提取，然后declustered（每个聚类不提供一个以上的值，并且每个值是相隔至少24小时）。因此，这些值是更可能是独立的。原(2009-1940+1)*365.26*24值的7％左右，这些包括venice90。
Yet to do: check for consistency between the data sets. Some external data sets elsewhere have some extremes recorded at times not exactly on the hour.
然而，要做到：检查数据集之间的一致性。一些外部的数据集，在其他地方有一些极端的不完全记录在一小时。
源----------Source----------
Pirazzoli, P. (1982) Maree estreme a Venezia (periodo 1872–1981). Acqua Aria, 10, 1023–1039.
皮拉左里，P.（1982）MAREE埃斯特雷姆：威尼斯（periodo 1872年至1981年）。夸晓月，10日，1023年至1039年。
Thanks to Paolo Pirazzoli and Alberto Tomasin for the venice90 data.
感谢保罗·皮拉左里和Alberto Tomasinvenice90数据。
参考文献----------References----------
Extreme value theory based on the r largest annual events. Journal of Hydrology, 86, 27–43.
La serie storica delle maree a Venezia, 1872–2004 (in Italian), Comune di Venezia. Istituzione Centro Previsione e Segnalazioni Maree.
参见----------See Also----------
guplot, gev, gpd.
guplot，gev，gpd。
实例----------Examples----------
## Not run: matplot(venice[["year"]], venice[, -1], xlab = "Year",[＃不运行：matplot（威尼斯[“一年”]，威尼斯[-1]，xlab =“年”，]
ylab = "Sea level (cm)", type = "l")
## End(Not run)[＃（不执行）]
ymat = as.matrix(venice[, paste("r", 1:10, sep = "")])
fit1 = vgam(ymat ~ s(year, df = 3), gumbel(R = 365, mpv = TRUE),
data = venice, trace = TRUE, na.action = na.pass)
head(fitted(fit1))
## Not run: par(mfrow = c(2, 1), xpd = TRUE)[＃不跑杆（mfrow = C（2,1），XPD = TRUE）]
plot(fit1, se = TRUE, lcol = "blue", llwd = 2, slty = "dashed")
par(mfrow = c(1,1), bty = "l", xpd = TRUE, las = 1)
qtplot(fit1, mpv = TRUE, lcol = c(1, 2, 5), tcol = c(1, 2, 5),
llwd = 2, pcol = "blue", tadj = 0.1)
## End(Not run)[＃（不执行）]
plot(sealevel ~ Year, data = venice90, type = "h", col = "blue")
summary(venice90)
dim(venice90)
round(100 * nrow(venice90) / ((2009-1940+1)*365.26*24), dig = 3)
转载请注明:出自 生物统计家园网(http://www.biostatistic.net)。
注：
注1：为了方便大家学习，本文档为生物统计家园网机器人LoveR翻译而成，仅供个人R语言学习参考使用，生物统计家园保留版权。
注2：由于是机器人自动翻译，难免有不准确之处，使用时仔细对照中、英文内容进行反复理解，可以帮助R语言的学习。
注3：如遇到不准确之处，请在本贴的后面进行回帖，我们会逐渐进行修订。
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发表于 2012-10-1 15:56:18
