我正在尝试使用ggtexttable函数从公公创建一个可发布的表。我有一个数据框架:
dput(df)
structure(list(feature = list("start_codon", "stop_codon", "intergenic",
"3UTR", "5UTR", "exon", "intron", "ncRNA", "pseudogene"),
observed = list(structure(1L, .Names = "start_codon"), structure(1L, .Names = "stop_codon"),
structure(418L, .Names = "intergenic"), structure(48L, .Names = "3UTR"),
structure(28L, .Names = "5UTR"), structure(223L, .Names = "exon"),
structure(578L, .Names = "intron"), structure(20L, .Names = "ncRNA"),
structure(1L, .Names = "pseudogene")), expected = list(
0.286, 0.286, 369.02, 72.461, 33.165, 257.869, 631.189,
48.491, 3.172), fc = list(3.5, 3.5, 1.1, 0.7, 0.8, 0.9,
0.9, 0.4, 0.3), test = list("enrichment", "enrichment",
"enrichment", "depletion", "depletion", "depletion",
"depletion", "depletion", "depletion"), sig = list("F",
"F", "T", "T", "F", "T", "T", "T", "F"), p_val = list(
"0.249", "0.249", "0.00186", "0.00116", "0.209", "0.00814",
"0.00237", "<1e-04", "0.175")), class = "data.frame", row.names = c(NA,
-9L), .Names = c("feature", "observed", "expected", "fc", "test",
"sig", "p_val"))当我试图把它变成一张桌子时
ggtexttable(df)
我知道错误:
错误在(函数(标签,解析=假,col = "black",fontsize = 12,未使用的参数(label.feature = dots[5][1],label.observed = dots[6][1],label.expected = dots[7][1],label.fc = dots[8][1],label.test = dots[9][1],label.sig_val =dots[10][1],label.sig_val=dots[10]],dots[6][1],label.expected=dots[7][1],label.fc=dots[8][1],label.test=dots[9][1],label.sig_val=dots[10]],label.p_val = dots[11][1])
有人知道是什么导致了这一切吗?
这样做很好:
df <- head(iris)
ggtexttable(df)

发布于 2017-08-02 12:04:59
我已经找到了对你有用的问题和解决办法。首先,您的数据没有正确的格式(嵌套列表),这就是为什么您在试图显示数据时会遇到这个错误。通过在控制台中粘贴:str(data),您可以轻松地检查数据集的格式。
下面是将数据转换为data.frame的解决方案
first.step <- lapply(data, unlist)
second.step <- as.data.frame(first.step, stringsAsFactors = F) 然后,您可以很容易地使用函数ggtexttable(second.step),它用数据显示表。
https://stackoverflow.com/questions/45455433
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