是否可以使用Python matplotlib代码在RStudio中绘制图形?
例如,在Python matplotlib代码下面:
import numpy as np
import matplotlib.pyplot as plt
n = 256
X = np.linspace(-np.pi,np.pi,n,endpoint=True)
Y = np.sin(2*X)
plt.plot (X, Y+1, color='blue', alpha=1.00)
plt.plot (X, Y-1, color='blue', alpha=1.00)
plt.show()
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输出图将是:
然后我需要写一个R Markdown来包含这些代码并在编织降价后自动生成图形.
我找到了一个googlemaps包来从python访问谷歌地图,但没有找到任何演示以在python代码中显示结果.
例如:
gmaps = googlemaps.Client(key='Add Your Key here')
# Geocoding an address
geocode_result = gmaps.geocode('1600 Amphitheatre Parkway, Mountain View, CA')
# Look up an address with reverse geocoding
reverse_geocode_result = gmaps.reverse_geocode((40.714224, -73.961452))
# Request directions via public transit
now = datetime.now()
directions_result = gmaps.directions("Sydney Town Hall",
"Parramatta, NSW",
mode="transit",
departure_time=now)
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我可以将directions_result打印到控制台,但不知道如何以图形模式显示地图.如何在python中将搜索结果显示为地图?
我现在正在使用emacs 25.1和gdb 7.12.但是当我运行"Mx gdb -i = mi a.out"时,然后设置断点并发出下一个命令,一个源代码窗口总是弹出以获得焦点,然后如果我想再次发出下一个命令,我必须切换到首先是gdb窗口,否则,键盘输入将直接到源代码文件.
我们应该专注于gdb窗口,而不是将焦点切换到源代码窗口.
从其他线程,我知道我们不应该在析构函数中抛出异常!但是对于下面的例子,它确实有效.这是否意味着我们只能在一个实例的析构函数中抛出异常?我们该如何理解这个代码示例!
#include <iostream>
using namespace std;
class A {
public:
~A() {
try {
printf("exception in A start\n");
throw 30;
printf("exception in A end\n");
}catch(int e) {
printf("catch in A %d\n",e);
}
}
};
class B{
public:
~B() {
printf("exception in B start\n");
throw 20;
printf("exception in B end\n");
}
};
int main(void) {
try {
A a;
B b;
}catch(int e) {
printf("catch in main %d\n",e);
}
return 0;
}
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输出是:
exception in B start
exception in A start …Run Code Online (Sandbox Code Playgroud) 我在这里找到了一个很好的活动图,但没有源代码.如下所示:

我尝试用plantuml绘制它,下面是我的代码:
@startuml
skinparam linetype ortho
(*) -down-> "clone repository"
note right
A:git clone url/to/repository
scp -p url/to/hooks/commit_msg repository/.git/hooks
end note
-down->[get an assignment] "coding"
-down-> "commit locally"
note right
B:git add xxx
git commit
end note
-down-> "review"
note right
C:git push origin refs/for/<branch>
end note
if "" then
-left->[rejected] "rework"
-up-> "recommit"
'note left
'D:git add xxx
'git commit --amend
'end note
-right-> "review"
else
--right-->[approved] "submit"
note right: E:click "submit" on on corresponding gerrit page …Run Code Online (Sandbox Code Playgroud) 当用户单击链接时,我有以下代码来调用 elisp 函数“myfun”:
#+BEGIN_SRC elisp :results output raw
(defun myfun(filepath lineno)
(if (not (get-buffer bufname))
(get-buffer-create bufname)
)
(switch-to-buffer-other-window bufname)
(global-linum-mode)
(erase-buffer)
(insert-file-contents (format "%s" filepath))
(goto-char (point-min))
(forward-line lineno)
(setq start (point))
(forward-line -1)
(setq end (point))
(let ((x (make-overlay start end)))
(overlay-put x 'face '(:background "#fef0f1")))
)
(defun createSampleFile(file-name count)
(let ((c 0))
(with-temp-buffer
(while (< c (* 2 count))
(setq c (1+ c))
(insert (format "line %d\n" c))
(write-file filename)
))))
(let ((c 0)(filename nil))
(while (< …Run Code Online (Sandbox Code Playgroud) 我正在阅读这份文件,它说:
Run Code Online (Sandbox Code Playgroud)char *strncpy(char *destination, const char *source, size_t num);从字符串复制的字符复制第一
num的人物source来destination.如果在复制字符source之前找到C字符串的末尾(由空字符表示)num,destination则用零填充,直到num写入总共字符为止.如果source长于,则不会在目标末尾隐式附加空字符
num.因此,在这种情况下,destination不应将其视为空终止的C字符串(读取它会溢出).
destination并且source不得重叠(重叠时参见memmove更安全的替代方案).
但我对此声明感到困惑:
在这种情况下,目标不应被视为空终止的C字符串(读取它会溢出)
因为if num > strlen(source),它将'\0'在最后填充,'\0'实际上是字符串中的空(终止)字符,为什么它不应被视为以null结尾的C字符串?
我写了下面的代码来验证:
char from[] = { 'h', 'e', 'l', 'l', 'o', '\0' };
char to[1024];
for (int i = 0; i < 1024; i++) {
to[i] = 'e';
}
strncpy(to, from, 1024);
printf("from …Run Code Online (Sandbox Code Playgroud) c ×1
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