我想在导航栏标题字段部分添加自定义徽标。我尝试添加一个 html 脚本并且它起作用了,但是徽标图像不适合导航栏 - 徽标被边距推开,菜单文本与徽标重叠,等等。
name: "my-website"
output_dir: .
navbar:
title: "I want to add custom logo here"
left:
- text: "Home"
Run Code Online (Sandbox Code Playgroud) 我目前在rmarkdown中使用cosmo主题.如果我想更改导航菜单的颜色,或者更具体地说是菜单突出显示,我应该编辑哪个文件?
菜鸟问题。谢谢你对我的包容。
我的计算机上安装了 Tensorflow 2.4.1,但我需要 2.4.0 版本才能与特定项目兼容。
我尝试了以下操作但没有成功:
conda install tensorflow==2.4.0
conda install -c conda-forge tensorflow==2.4.0
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它会这样做一段时间,直到 jupyter Notebook 告诉我它找不到我需要的东西
Collecting package metadata (current_repodata.json): done
Solving environment: failed with initial frozen solve. Retrying with flexible solve.
Collecting package metadata (repodata.json): done
Solving environment: failed with initial frozen solve. Retrying with flexible solve.
Solving environment:
Found conflicts! Looking for incompatible packages.
This can take several minutes. Press CTRL-C to abort.
failed
UnsatisfiableError: The following specifications were found
to be incompatible with the …Run Code Online (Sandbox Code Playgroud) 我使用的实际数据更长并且有更多变量,但我有一个看起来像这样的数据:
country <- c("US", "US", "US", "Korea", "Korea", "Korea")
cause <- c("sharp", "suicide", "others")
value <- c(30, 20, 50, 40, 40, 20)
numbers <- cbind(country, cause, value)
country cause value
[1,] "US" "sharp" "30"
[2,] "US" "suicide" "20"
[3,] "US" "others" "50"
[4,] "Korea" "sharp" "40"
[5,] "Korea" "suicide" "40"
[6,] "Korea" "others" "20"
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我希望它看起来像这样:
country sharp suicide others
[1,] "US" "30" "20" "50"
[2,] "Korea" "40" "40" "20"
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我已经尝试过 transpose 命令,但是 R 会转置列中的所有内容,并且国家/地区的名称会重复多次。如何将原因移动为列名称并在其下方分配适当的值?
这是我所拥有的:
import pandas as pd
df = pd.DataFrame()
df['date'] = ['2020-01-01', '2020-01-01','2020-01-01', '2020-01-02', '2020-01-02', '2020-01-03', '2020-01-03']
df['value'] = ['A', 'A', 'A', 'A', 'B', 'A', 'C']
df
date value
0 2020-01-01 A
1 2020-01-01 A
2 2020-01-01 A
3 2020-01-02 A
4 2020-01-02 B
5 2020-01-03 A
6 2020-01-03 C
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我想像这样随着时间的推移聚合唯一值:
date value
0 2020-01-01 1
3 2020-01-02 2
5 2020-01-03 3
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我不是在寻找这个作为答案:
date value
0 2020-01-01 1
3 2020-01-02 2
5 2020-01-03 2
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我需要2020-01-03是3因为有三个唯一值(A、B、C)。
我有类似于以下内容的数据:
q = (0,0,0)
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我想将数字元组转换为字符串。互联网上很多人提供了一个看起来像这样的解决方案:
user = 45
permission = (43, 12, 65, 34, 67)
final = "{},'{}'".format(user, ','.join(str(p) for p in permission))
print(final)
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来源:尝试将数字元组转换为字符串。
不幸的是,我不断收到一条错误消息,内容为:TypeError: 'int' object is not callable. 所以,我尝试了其他类似的解决方案','.join(map(str, q))。徒劳无功。
我的申请有问题吗?请帮忙。预期输出:
q
"0" "," "0" "," "0"
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或以下字符串:
0,0,0
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