如果您想了解物体检测，Error:maximumboxcoordinatevalueistoolarge的相关知识，那么本文是一篇不可错过的文章，我们将对物体检测名词解释进行全面详尽的解释，并且为您提
如果您想了解物体检测，Error: maximum box coordinate value is too large的相关知识，那么本文是一篇不可错过的文章，我们将对物体检测名词解释进行全面详尽的解释，并且为您提供关于2 条线图之间的阴影区域 ValueError: ordinal must be >= 1、@JoinColumnOrFormula(formula=@JoinFormula org.hibernate.mapping.Formula cannot be cast to org.hibernate.mapping.Column、Avito Cool Challenge 2018：D. Maximum Distance、Educational Codeforces Round 59 (Rated for Div. 2)G. Vasya and Maximum Profit 区间dp的有价值的信息。
本文目录一览：- 物体检测，Error: maximum box coordinate value is too large（物体检测名词解释）
- 2 条线图之间的阴影区域 ValueError: ordinal must be >= 1
- @JoinColumnOrFormula(formula=@JoinFormula org.hibernate.mapping.Formula cannot be cast to org.hibernate.mapping.Column
- Avito Cool Challenge 2018：D. Maximum Distance
- Educational Codeforces Round 59 (Rated for Div. 2)G. Vasya and Maximum Profit 区间dp
物体检测，Error: maximum box coordinate value is too large（物体检测名词解释）
使用 ssd 目标检测，出现 error:maximum box coordinate value is larger than 1.100000: ] [1.325]
主要原因在于，用 labelImg 标记的目标太小，以及标记工具出现问题
1 deleted all boxes that are less than 1/16 th of the image size and the training works fine.，https://github.com/EdjeElectronics/TensorFlow-Object-Detection-API-Tutorial-Train-Multiple-Objects-Windows-10/blob/master/sizeChecker.py
if x_value < 33 or y_value < 33: 删除，shutil.move (file, tree + ''/wrong_data/'')
2. 忽视结果
Actually, I just ignore this error, change the ./core/box_list_ops.py
`max_assert = tf.Assert(tf.greater_equal(1.1, box_maximum), [''maximum box coordinate value is larger '' ''than 1.1: '', box_maximum])`
1.01 to 1.1, as my error max value is 1.014.
By this way to make the code keep on. Seems the result is not affected.
参考链接：https://github.com/tensorflow/models/issues/1754
2 条线图之间的阴影区域 ValueError: ordinal must be >= 1
如何解决2 条线图之间的阴影区域 ValueError: ordinal must be >= 1？
我有以下代码可以正确呈现：
plt.xlabel(''Date'')
plt.ylabel(''Temp'')
plt.title(''Min and Max temperature 2005-2014'')
# Plotting on the first y-axis
minimum=new_df[''min'']
maximum=new_df[''max'']
plt.plot(new_df[''Date''],new_df[''min''],color=''orange'',label=''Min'')
plt.plot(new_df[''Date''],new_df[''max''],color=''olive'',label=''Max'')
现在我需要为两条线之间的区域着色：
我试过了：
plt.fill_between(minimum,maximum,color=''#539ecd'')
但后来我收到此错误：
ValueError Traceback (most recent call last)
/opt/conda/lib/python3.6/site-packages/IPython/core/formatters.py in __call__(self,obj)
305 pass
306 else:
--> 307 return printer(obj)
308 # Finally look for special method names
309 method = get_real_method(obj,self.print_method)
/opt/conda/lib/python3.6/site-packages/IPython/core/pylabtools.py in <lambda>(fig)
225
226 if ''png'' in formats:
--> 227 png_formatter.for_type(figure,lambda fig: print_figure(fig,''png'',**kwargs))
228 if ''retina'' in formats or ''png2x'' in formats:
229 png_formatter.for_type(figure,lambda fig: retina_figure(fig,**kwargs))
/opt/conda/lib/python3.6/site-packages/IPython/core/pylabtools.py in print_figure(fig,fmt,bBox_inches,**kwargs)
117
118 bytes_io = BytesIO()
--> 119 fig.canvas.print_figure(bytes_io,**kw)
120 data = bytes_io.getvalue()
121 if fmt == ''svg'':
/opt/conda/lib/python3.6/site-packages/matplotlib/backend_bases.py in print_figure(self,filename,dpi,facecolor,edgecolor,orientation,format,**kwargs)
2190 orientation=orientation,2191 dryrun=True,-> 2192 **kwargs)
2193 renderer = self.figure._cachedRenderer
2194 bBox_inches = self.figure.get_tightbBox(renderer)
/opt/conda/lib/python3.6/site-packages/matplotlib/backends/backend_agg.py in print_png(self,filename_or_obj,*args,**kwargs)
543
544 def print_png(self,**kwargs):
--> 545 figureCanvasAgg.draw(self)
546 renderer = self.get_renderer()
547 original_dpi = renderer.dpi
/opt/conda/lib/python3.6/site-packages/matplotlib/backends/backend_agg.py in draw(self)
462
463 try:
--> 464 self.figure.draw(self.renderer)
465 finally:
466 RendererAgg.lock.release()
/opt/conda/lib/python3.6/site-packages/matplotlib/artist.py in draw_wrapper(artist,renderer,**kwargs)
61 def draw_wrapper(artist,**kwargs):
62 before(artist,renderer)
---> 63 draw(artist,**kwargs)
64 after(artist,renderer)
65
/opt/conda/lib/python3.6/site-packages/matplotlib/figure.py in draw(self,renderer)
1141
1142 mimage._draw_list_compositing_images(
-> 1143 renderer,self,dsu,self.suppressComposite)
1144
1145 renderer.close_group(''figure'')
/opt/conda/lib/python3.6/site-packages/matplotlib/image.py in _draw_list_compositing_images(renderer,parent,suppress_composite)
137 if not_composite or not has_images:
138 for zorder,a in dsu:
--> 139 a.draw(renderer)
140 else:
141 # Composite any adjacent images together
/opt/conda/lib/python3.6/site-packages/matplotlib/artist.py in draw_wrapper(artist,renderer)
65
/opt/conda/lib/python3.6/site-packages/matplotlib/axes/_base.py in draw(self,inframe)
2407 renderer.stop_rasterizing()
2408
-> 2409 mimage._draw_list_compositing_images(renderer,dsu)
2410
2411 renderer.close_group(''axes'')
/opt/conda/lib/python3.6/site-packages/matplotlib/image.py in _draw_list_compositing_images(renderer,renderer)
65
/opt/conda/lib/python3.6/site-packages/matplotlib/axis.py in draw(self,**kwargs)
1134 renderer.open_group(__name__)
1135
-> 1136 ticks_to_draw = self._update_ticks(renderer)
1137 ticklabelBoxes,ticklabelBoxes2 = self._get_tick_bBoxes(ticks_to_draw,1138 renderer)
/opt/conda/lib/python3.6/site-packages/matplotlib/axis.py in _update_ticks(self,renderer)
967
968 interval = self.get_view_interval()
--> 969 tick_tups = [t for t in self.iter_ticks()]
970 if self._smart_bounds:
971 # handle inverted limits
/opt/conda/lib/python3.6/site-packages/matplotlib/axis.py in <listcomp>(.0)
967
968 interval = self.get_view_interval()
--> 969 tick_tups = [t for t in self.iter_ticks()]
970 if self._smart_bounds:
971 # handle inverted limits
/opt/conda/lib/python3.6/site-packages/matplotlib/axis.py in iter_ticks(self)
910 Iterate through all of the major and minor ticks.
911 """
--> 912 majorLocs = self.major.locator()
913 majorTicks = self.get_major_ticks(len(majorLocs))
914 self.major.formatter.set_locs(majorLocs)
/opt/conda/lib/python3.6/site-packages/matplotlib/dates.py in __call__(self)
981 def __call__(self):
982 ''Return the locations of the ticks''
--> 983 self.refresh()
984 return self._locator()
985
/opt/conda/lib/python3.6/site-packages/matplotlib/dates.py in refresh(self)
1001 def refresh(self):
1002 ''Refresh internal information based on current limits.''
-> 1003 dmin,dmax = self.viewlim_to_dt()
1004 self._locator = self.get_locator(dmin,dmax)
1005
/opt/conda/lib/python3.6/site-packages/matplotlib/dates.py in viewlim_to_dt(self)
758 vmin,vmax = vmax,vmin
759
--> 760 return num2date(vmin,self.tz),num2date(vmax,self.tz)
761
762 def _get_unit(self):
/opt/conda/lib/python3.6/site-packages/matplotlib/dates.py in num2date(x,tz)
399 tz = _get_rc_timezone()
400 if not cbook.iterable(x):
--> 401 return _from_ordinalf(x,tz)
402 else:
403 x = np.asarray(x)
/opt/conda/lib/python3.6/site-packages/matplotlib/dates.py in _from_ordinalf(x,tz)
252
253 ix = int(x)
--> 254 dt = datetime.datetime.fromordinal(ix).replace(tzinfo=UTC)
255
256 remainder = float(x) - ix
ValueError: ordinal must be >= 1
<matplotlib.figure.figure at 0x7fb311ea1cf8>
编辑：
数据框如下所示：
Date min max min2015 max2015
0 2014-01-01 -160 156 -133 11
1 2014-01-02 -267 139 -122 39
2 2014-01-03 -267 133 -67 39
3 2014-01-04 -261 106 -88 44
4 2014-01-05 -150 128 -155 28
我将日期转换为日期时间类型，如下所示：
new_df[''Date''] = pd.to_datetime(new_df[''Date''],infer_datetime_format=True)
编辑：
当我运行时： plt.fill_between(new_df[''Date''],最小值,最大值)
我收到此错误
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-23-0f6fcbb48fdc> in <module>()
59
60
---> 61 leaflet_plot_stations(400,''fb441e62df2d58994928907a91895ec62c2c42e6cd075c2700843b89'')
<ipython-input-23-0f6fcbb48fdc> in leaflet_plot_stations(binsize,hashid)
44 minimum = new_df[''min'']
45 maximum = new_df[''max'']
---> 46 plt.fill_between(new_df[''Date''],minimum,maximum)
47 #plt.scatter(new_df[''Date''],new_df[''min2015''],''o'')
48 #plt.scatter(new_df[''Date''],new_df[''max2015''],''o'')
/opt/conda/lib/python3.6/site-packages/matplotlib/pyplot.py in fill_between(x,y1,y2,where,interpolate,step,hold,data,**kwargs)
2999 ret = ax.fill_between(x,y2=y2,where=where,3000 interpolate=interpolate,step=step,data=data,-> 3001 **kwargs)
3002 finally:
3003 ax._hold = washold
/opt/conda/lib/python3.6/site-packages/matplotlib/__init__.py in inner(ax,**kwargs)
1890 warnings.warn(msg % (label_namer,func.__name__),1891 RuntimeWarning,stacklevel=2)
-> 1892 return func(ax,**kwargs)
1893 pre_doc = inner.__doc__
1894 if pre_doc is None:
/opt/conda/lib/python3.6/site-packages/matplotlib/axes/_axes.py in fill_between(self,x,**kwargs)
4770
4771 # Convert the arrays so we can work with them
-> 4772 x = ma.masked_invalid(self.convert_xunits(x))
4773 y1 = ma.masked_invalid(self.convert_yunits(y1))
4774 y2 = ma.masked_invalid(self.convert_yunits(y2))
/opt/conda/lib/python3.6/site-packages/numpy/ma/core.py in masked_invalid(a,copy)
2343 cls = type(a)
2344 else:
-> 2345 condition = ~(np.isfinite(a))
2346 cls = MaskedArray
2347 result = a.view(cls)
TypeError: ufunc ''isfinite'' not supported for the input types,and the inputs Could not be safely coerced to any supported types according to the casting rule ''''safe''''
解决方法
看起来是在时间轴上水平尝试填充，minimum 和 maximum 包含不是日期的值。我查过documentation：
matplotlib.pyplot.fill_between(x,y1,y2=0,where=None,interpolate=False,step=None,*,data=None,**kwargs)
这是第一个错误。
然后使用比 1.17.0 更新的 numpy，你可以这样做：
size = 500
minimum = np.random.normal(0,100,size)
minimum.sort()
minimum = np.random.randint(250,300,size) - np.abs(minimum)
df = pd.DataFrame(minimum,pd.date_range("2005-01-01",periods=size,freq="d"),columns=[''min''],)
df[''max''] = df[''min''] + np.random.randint(200,250,size)
fig = df.plot()
fig.fill_between(df.index,df[''min''],df[''max''],color=''#539ecd'')
但是从第二个回溯中，我们可以看到 fill_between 正在所有轴上尝试 ~(np.isfinite(a))。 Which isn''t supported on older numpy.datetime64，x 轴的类型。
因此我们将不得不使用数字 x 轴，然后更改标签。
df = pd.DataFrame(minimum,color=''#539ecd'')
# We take the original datetime axis
date_axis = pd.date_range("2005-01-01",freq="d")
# and map a function from (axis,tick) -> wanted string
def label(axis,tick):
tick = int(tick)
if tick == len(axis):
tick -= 1
if 0 <= tick < len(axis):
return f"{axis[tick].year}-{axis[tick].month}"
else:
return '' ''
fig.set_xticks(fig.get_xticks()) #silence a warning
fig.set_xticklabels(
[label(date_axis,tick) for tick in fig.get_xticks()]
)
fill_between 将 x 坐标作为第一个参数。以下应该有效：
plt.fill_between(new_df["Date"],minimum,maximum,color="lemonchiffon")
请注意，使用 matplotlib 3.4 我无法重现该错误。无论 Date 中的值是转换为日期还是保留为字符串，fill_between(minimum,maximum) 都不会引发错误，但会产生意外的图。
编辑
使用 numpy 1.11.3 和 matplotlib 2.0.2，我已经能够通过 TypeError 重现 np.isfinite 加注（有关详细信息，请参阅 this post）。实际上，此函数仅从 1.17 版开始支持 datetime64。 我强烈建议您更新 matplotlib 和 numpy 的版本。但是，使用本段中描述的早期版本，通过显式转换日期来绕过错误：
plt.fill_between(np.array(new_df["Date"]),color=''lemonchiffon'')
@JoinColumnOrFormula(formula=@JoinFormula org.hibernate.mapping.Formula cannot be cast to org.hibernate.mapping.Column
@JoinColumnOrFormula (formula=@JoinFormula 在 hibernate4.3.11 中报错:
Caused by: java.lang.ClassCastException: org.hibernate.mapping.Formula cannot be cast to org.hibernate.mapping.Column
at org.hibernate.cfg.annotations.TableBinder.bindFk(TableBinder.java:433)
at org.hibernate.cfg.annotations.CollectionBinder.bindCollectionSecondPass(CollectionBinder.java:1467)
at org.hibernate.cfg.annotations.CollectionBinder.bindOneToManySecondPass(CollectionBinder.java:865)
at org.hibernate.cfg.annotations.CollectionBinder.bindStarToManySecondPass(CollectionBinder.java:780)
at org.hibernate.cfg.annotations.CollectionBinder$1.secondPass(CollectionBinder.java:729)
at org.hibernate.cfg.CollectionSecondPass.doSecondPass(CollectionSecondPass.java:70)
at org.hibernate.cfg.Configuration.originalSecondPassCompile(Configuration.java:1697)
at org.hibernate.cfg.Configuration.secondPassCompile(Configuration.java:1426)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1846)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1930)
at org.springframework.orm.hibernate4.LocalSessionFactoryBuilder.buildSessionFactory(LocalSessionFactoryBuilder.java:372)
at org.springframework.orm.hibernate4.LocalSessionFactoryBean.buildSessionFactory(LocalSessionFactoryBean.java:454)
at org.springframework.orm.hibernate4.LocalSessionFactoryBean.afterPropertiesSet(LocalSessionFactoryBean.java:439)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1631)
at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1568)
... 75 moreM_CIF_INTERNATIONAL_CHECK
package com.hnair.opcnet.ods.ml.model;
import java.util.Date;
import java.util.HashSet;
import java.util.Set;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.OneToMany;
import javax.persistence.Table;
import org.hibernate.annotations.Where;
import com.hnair.opcnet.common.util.DateUtils;
@Entity
@Table(name="M_CIF_INTERNATIONAL_CHECK")
public class M_CIF_INTERNATIONAL_CHECK implements java.io.Serializable {
private static final long serialVersionUID = -1;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "ID")
private Long id;
@Column(name = "SRC_ID")
private String srcId;
@OneToMany(fetch = FetchType.EAGER, mappedBy="check")
// @Where(clause="DELETED = 0")
private Set<M_CIF_INTERNATIONAL_CHECK_FILE> checkFile = new HashSet<M_CIF_INTERNATIONAL_CHECK_FILE>();
public void setId(Long id){
this.id=id;
}
public Long getId(){
return id;
}
public Set<M_CIF_INTERNATIONAL_CHECK_FILE> getCheckFile() {
return checkFile;
}
public void setCheckFile(Set<M_CIF_INTERNATIONAL_CHECK_FILE> checkFile) {
this.checkFile = checkFile;
}
}M_CIF_INTERNATIONAL_CHECK_FILE
package com.hnair.opcnet.ods.ml.model;
import java.util.Date;
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.Table;
import org.hibernate.annotations.JoinColumnOrFormula;
import org.hibernate.annotations.JoinColumnsOrFormulas;
import org.hibernate.annotations.JoinFormula;
import com.hnair.opcnet.common.util.DateUtils;
@Entity
@Table(name="M_CIF_INTERNATIONAL_CHECK_FILE")
public class M_CIF_INTERNATIONAL_CHECK_FILE implements java.io.Serializable {
private static final long serialVersionUID = -1;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "ID")
private Long id;
@Column(name = "SRC_ID")
private String srcId;
@ManyToOne(cascade = CascadeType.REFRESH, targetEntity = M_CIF_INTERNATIONAL_CHECK.class)
@JoinColumnsOrFormulas(value={
@JoinColumnOrFormula(column=@JoinColumn(name = "CHECKID", referencedColumnName = "SRC_ID", nullable = true, insertable = false, updatable = false))
,@JoinColumnOrFormula(formula=@JoinFormula(value="0", referencedColumnName = "DELETED"))
})
private M_CIF_INTERNATIONAL_CHECK check;
/** 删除标志 */
@Column(name = "DELETED")
private Integer deleted;
public void setId(Long id){
this.id=id;
}
public Long getId(){
return id;
}
/** 设置 删除标志 */
public void setDeleted(Integer deleted){
this.deleted=deleted;
}
/** 获取 删除标志 */
public Integer getDeleted(){
return deleted;
}
public M_CIF_INTERNATIONAL_CHECK getCheck() {
return check;
}
public void setCheck(M_CIF_INTERNATIONAL_CHECK check) {
this.check = check;
}
}
主表 M_CIF_INTERNATIONAL_CHECK 中使用 @Where (clause="DELETED = 0")
子表 M_CIF_INTERNATIONAL_CHECK_FILE 取消，@JoinColumnOrFormula (formula=@JoinFormula (value="0", referencedColumnName = "DELETED"))
是正常的
Avito Cool Challenge 2018：D. Maximum Distance
D. Maximum Distance
题目链接：https://codeforces.com/contest/1081/problem/D
题意：
给出一个连通图以及一些特殊点，现在定义cost(u,v)为一条从u到v的路径上面边权的最大值，然后定义dis(u,v)为从u到v所有路径上面cost的最小值。
最后求所有特殊点到其它特殊点的最大距离...
题解：
这个题意似乎有点绕...
我们考虑一下最小生成树，那么点与点之间的距离就为最小生成树路径上面边权的最大值。
我们来证明一下：假设在最小生成树上面的路径cost为w1，另外在原图中还有一条路径从u到v，其cost为w2，那么必然有w2>w1的。那么我们最后的dis一定是w1。
那么我们现在的目标就是求特殊点到特殊点之间的最大距离。注意一下这里是从一个特殊点到其它所有特殊点的最大距离。
我们知道在Kruskal加边时，后加的边权一定时大于前面的边权的，既然要求最大权值，那么我们可以想加的最后一条边是否可以作为答案。
我们假设现在有两个集合，现在将其连接起来，当满足两个集合里面都有特殊点时我们就可以更新答案了，否则就不行。
所以我们合并的时候顺带维护一下集合里面特殊点的信息就可以了。
代码如下:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N =1e5+5;
struct Edge{
int u,v,w;
bool operator < (const Edge&A)const{
return w<A.w;
}
}e[N];
int n,m,k;
int a[N],f[N],val[N];
int find(int x){
if(x==f[x]) return f[x];
f[x]=find(f[x]);
return f[x];
}
int main(){
cin>>n>>m>>k;
int tot=k;
for(int i=1,t;i<=k;i++){
cin>>t;
a[t]=1;
}
for(int i=1;i<=m;i++){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
e[i]=Edge{u,v,w};
}
for(int i=1;i<=n;i++) f[i]=i;
sort(e+1,e+m+1);
int ans;
for(int i=1;i<=m;i++){
int u=e[i].u,v=e[i].v,w=e[i].w;
int fx=find(u),fy=find(v);
if(fx==fy) continue;
f[fx]=fy;
if(a[u]) val[fx]++;if(a[v]) val[fy]++;
if(val[fx]&&val[fy]) ans=w;
val[fy]+=val[fx];
}
for(int i=1;i<=k;i++) cout<<ans<<" ";
return 0;
}
Educational Codeforces Round 59 (Rated for Div. 2)G. Vasya and Maximum Profit 区间dp
题目链接：G. Vasya and Maximum Profit
与这道题相似
题意，给个长度为n的01字符串s,以及长度为n数组a.你每次可以选择任意长度L的连续子串（要求子串每个字符相同）从原串中去掉，并获得a【L】的值。问你可能获得的值最多是多少
题解:
dp[i][j][k]代表区间i,j后面跟k的与s[j]相同的字符。
状态转移
第一种去掉j-1后面的k+1字符dp[l][r][k]=max(dp[l][r][k],dp[l][r-1][0]+a[k+1]);
第二种在i,j-1的区间中找一个与s[j]相同的位置p然后去掉中间的部分[p+1,j-1]
dp[i][j][k]=max(dp[i][j][k],dp[i][p][k+1]+dp[p+1][j-1][0]);
最后答案就是dp[0][n-1][0];
#include<bits/stdc++.h> #define ws wewe #define ll long long using namespace std; const int N=105; ll a[N],dp[N][N][N]; int n; string ss; int main() { scanf("%d",&n); cin>>ss; for(int i=1;i<=n;i++)scanf("%I64d",&a[i]); for(int i=0;i<n;i++) { for(int j=0;j<n;j++) { dp[i][i][j]=a[j+1]; } } for(int s=2;s<=n;s++)//枚举区间长度 { for(int l=0;l+s-1<n;L++)//枚举区间左节点 { int r=l+s-1; for(int k=0;k<n;k++)//枚举后面有多少个s[r]相同的字符 { dp[l][r][k]=dp[l][r-1][0]+a[k+1]; for(int j=l;j<r;j++) { if(ss[j]==ss[r]) { dp[l][r][k]=max(dp[l][r][k],dp[l][j][k+1]+dp[j+1][r-1][0]); } } } } } printf("%I64d\n",dp[0][n-1][0]); return 0; }
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