CN102243705B - Method for positioning license plate based on edge detection - Google Patents

Abstract

The invention provides a method for positioning a license plate by utilizing the edge information of a traffic image. The method comprises the following steps of: performing edge detection on a traffic image by using a Sobel operator, and binarizing the obtained gray-level edge image to obtain a binary edge image; then horizontally projecting the binary edge image in a vertical direction to obtain a vertical vector, binarizing the vertical vector, and scanning the vertical vector from top to bottom to obtain an initial row coordinate and an end row coordinate of a license plate region to realize the vertical positioning of the license plate; then obtaining the vertically positioned binary edge image by utilizing the initial row coordinate and the end row coordinate, vertically projecting the binary edge image in a horizontal direction to obtain a horizontal vector, binarizing the horizontal vector, and scanning the horizontal vector from left to right to obtain an initial line coordinate and an end line coordinate of the license plate region to realize the horizontal positioning of the license plate; and finally, positioning the license plate by using the initial row coordinate, the end row coordinate, the initial line coordinate and the end line coordinate. The method provided by the invention has higher precision for positioning the license plate and can also be used for positioning the license plate more accurately under the condition of low contrast (such as insufficient light at night).

Description

License plate locating method based on rim detection

Technical field

The present invention relates to a kind of license plate locating method based on rim detection, its purpose is the license plate area in the traffic image is positioned, and belongs to the traffic monitoring technical field.

Background technology

Along with the development of economy and carrying out fast of urbanization process, China's motor vehicle is possessed quantity sharply to be increased, and the problem of traffic administration is more and more outstanding.Under many circumstances, often need vehicle is identified in order to carry out better traffic administration.Vehicle License Plate Recognition System is utilized the technology such as image processing, pattern-recognition automatically location and identification car plate, can improve widely the efficient of vehicle identification.It can be used for the fields such as parking lot management, traffic information collection, traffic police's inspection, customs's logistics monitoring, is with a wide range of applications.

Vehicle License Plate Recognition System comprises car plate location, Character segmentation, three modules of character recognition.In engineering is used, image acquisition is subjected to the impact of environmental factor larger, and under some specific environment (such as the insufficient light at night), the contrast of license plate image is very low, may cause inaccurate even location, car plate location less than car plate, thereby reduce the accuracy of car plate identification.In order to improve the performance of Vehicle License Plate Recognition System, must improve the accuracy of the lower car plate of low-contrast circumstances (such as the insufficient light at night) location.

By the retrieval of prior art document being found existing car plate location mainly contains the intensity-based saltus step and based on these two kinds of methods of texture analysis.In the situation that night insufficient light, gray scale and the texture difference of license plate area and non-license plate area are all less, these two kinds of methods all can not position car plate well.The invention provides a kind of license plate locating method that under low-contrast circumstances, still has high accuracy.Under low-contrast circumstances, the method is compared the intensity-based saltus step and is had higher accurate positioning based on the localization method of texture analysis.

Summary of the invention

Technical matters

The invention provides a kind of license plate locating method license plate locating method based on rim detection that in low contrast (such as the insufficient light at night) situation, still has high accuracy, the method have bearing accuracy high, low-contrast circumstances is had adaptive advantage.

Technical scheme

A kind of license plate locating method based on rim detection is characterized in that:

1. utilize the Sobel operator to carry out rim detection

If need to carry out the coloured image of car plate location is F ₀, it being carried out gray processing obtain gray level image F, its line number, columns are respectively m and n, utilize horizontal Sobel operator S ₁With vertical Sobel operator S ₂Image F is carried out rim detection, obtain gradient image G ₀,

G ₀＝|S ₁*F|+|S ₂*F|，

In the formula,

* represent convolution algorithm,

Then, utilize according to threshold value T ₀To gradient image G ₀Carry out binaryzation, obtain binary edge map G ₁,

$G_1(x,y)=\left\{\begin{array}{cc}1& \mathrm{if}{G}_0(x,y)>T_0,\\ 0& \mathrm{else}\end{array}\right.$

In the formula, threshold value T ₀Be 15-20, G ₀(x, y) is gradient image G ₀Middle coordinate is the gray scale of the pixel of (x, y), G ₁(x, y) is binary edge map G ₁Middle coordinate is the value of the pixel of (x, y), if 1 this pixel of expression belongs to the edge, if 0 this pixel of expression does not belong to the edge;

2. the location of car plate vertical direction

To binary edge map G ₁Carry out horizontal projection, obtain vertical vector G _1x,

$G_{1x}\left(i_1\right)=\underset{j=1}{\overset{n}{\mathrm{\Σ}}}{G}_1(i_1,j_1),$ i ₁＝1，2，3......m，

Utilize threshold value T ₁To vertical vector G _1xCarry out binaryzation, obtain two-value vertical vector G ' _1x,

$G_{1x}^{\′}\left(i_1\right)=\left\{\begin{array}{cc}1& \mathrm{if}{G}_{1x}\left(i_1\right)>T_1\\ 0& \mathrm{else}\end{array}\right.,$ i ₁＝1，2，3......m，

In the formula, threshold value T ₁Be generally 3-5, G ' _1x(i ₁) be 1 to be illustrated in binary edge map G ₁I ₁The row place exists characters on license plate, G ' _1x(i ₁) be 0 to be illustrated in binary edge map G ₁I ₁There is not characters on license plate in the row place,

Afterwards, to two-value vertical vector G ' _1xScan from the top down, if detect license plate area P by scanning, then to the initial row coordinate P of license plate area P _X1With end line coordinate P _X2Position, scanning step is as follows:

Step 1: establish two-value vertical vector G ' _1xIn first numerical value be that the row-coordinate of 1 pixel is i ₂, with row-coordinate i ₂Give variable xbeg, and make variable xend=xbeg,

Step 2: i ₂=i ₂+ 1, if i ₂=m+1 then forwards step 3 to, if i ₂＜m+1 then forwards step 4 to,

Step 3: if xend-xbeg＞25 then show in gray level image F and detected car plate make P _X1=xbeg, P _X2=xend finishes scanning; If scanning is finished in xend-xbeg≤25 then show do not detect car plate in gray level image F,

Step 4: if i ₂-xend≤6 and i ₂-xbeg＞25 then make P _X1=xbeg, P _X2=xend finishes scanning, if i ₂-xend≤6 and i ₂-xbeg≤25 then make xend=i ₂, return step 2; If i ₂-xend＞6 and xend-xbeg＞25 item make P _X1=xbeg, P _X2=xend finishes scanning, if i ₂-xend＞6 and xend-xbeg≤25 item make xbeg=i ₂, xend=xbeg returns step 2,

Behind the end of scan, if detect car plate in the scanning process, then the initial row coordinate of license plate area P is positioned as P _X1, its end line coordinate is positioned as P _X2, obtain binary edge map G behind the perpendicular positioning according to positioning result ₂,

G ₂(i ₂，j ₂)＝G ₁(i ₂+P _x1-1，j ₂)，

I wherein ₂=1,2,3...... (P _X2-P _X1+ 1), j ₂=1,2,3......n,

3. the location of car plate horizontal direction

To the binary edge map G behind the perpendicular positioning ₂Carry out vertical projection, obtain horizontal vector G _2y,

$G_{2y}\left(i_3\right)=\underset{j_3=1}{\overset{P_{x2}-P_{x1}+1}{\mathrm{\Σ}}}{G}_1(j_3,i_3),$ i ₃＝1，2，3......n，

Utilize threshold value T ₂To horizontal vector G _2yCarry out binaryzation, obtain binarizing level vector G ' _2y,

${G^{\′}}_{2y}\left(i_3\right)=\left\{\begin{array}{cc}1& \mathrm{if}{G}_{2y}\left(i_3\right)>T_1\\ 0& \mathrm{else}\end{array}\right.,$ i ₃＝1，2，3......n，

In the formula, threshold value T ₂Be 15-20, G ' _2y(i ₃) be 1 to be illustrated in binary edge map G ₂I ₃There is characters on license plate in the row place, G ' _2y(i ₃) be 0 to be illustrated in binary edge map G ₂I ₃There is not characters on license plate in the row place,

Afterwards, to binarizing level vector G ' _2yScan from left to right, if detect license plate area P by scanning, then to the initial row coordinate P of license plate area P _Y1With end line coordinate P _Y2Position, scanning step is as follows:

Step 1: establish binarizing level vector G ' _2yIn first numerical value be that the row coordinate of 1 pixel is i ₄, with row coordinate i ₄Give variable ybeg, and make variable yend=ybeg,

Step 2: i ₄=i ₄+ 1, if i ₄=m+1 then forwards step 3 to, if i ₄＜m+1 then forwards step 4 to,

Step 3: if yend-ybeg＞70 then show in gray level image F and detected car plate make P _Y1=ybeg, P _Y2=yend finishes scanning; If scanning is finished in yend-ybeg≤70 then show do not detect car plate in gray level image F,

Step 4: if i ₄-yend≤15 and i ₄-ybeg＞70 then make P _Y1=ybeg, P _Y2=yend finishes scanning, if i ₄-yend≤15 and i ₄-ybeg≤70 then make yend=i ₄, return step 2; If i ₄-yend＞15 and yend-ybeg＞70 then make P _Y1=ybeg, P _Y2=yend finishes scanning, if _I4-yend＞15 and yend-ybeg≤70 then make ybeg=i ₄, yend=ybeg returns step 2,

Behind the end of scan, if detect car plate in the scanning process, then the initial row coordinate of license plate area P is positioned as P _Y1, its terminal point row coordinate is positioned as P _Y2, in conjunction with initial row coordinate P _X1With end line coordinate P _X2License plate area P after can obtaining locating is:

P(i ₄，j ₄)＝F ₀(i ₄+P _x1-1，j ₄+P _y1-1)，

I wherein ₄=1,2,3...... (P _X2-P _X1+ 1), j ₄=1,2,3...... (P _Y2-P _Y1+ 1).

Beneficial effect

The car plate accurate positioning is high, and low contrast (such as the insufficient light at night) situation is had adaptability.License plate area has comprised regular numeral, letter and Chinese character.Under different shooting environmental, the marginal information of these character pictures compares its gray scale and texture information is more stable.The present invention utilizes marginal information that car plate is positioned, and has higher accurate positioning.Particularly under low-contrast circumstances, also can accurately locate car plate, compare the intensity-based saltus step and improved accurate positioning based on the localization method of texture analysis.

Description of drawings

Fig. 1 is the process flow diagram of car plate location.

Fig. 2 is the coloured image F that need to carry out the car plate location ₀

Fig. 3 is gray level image F.

Fig. 4 is gradient image G ₀

Fig. 5 is binary edge map G ₁

Fig. 6 is the binary edge map G behind the perpendicular positioning ₂

Fig. 7 is the license plate area P that the location obtains.

Fig. 8 is the binary edge map G behind the perpendicular positioning ₂

Fig. 9 is to the binary edge map G behind the perpendicular positioning ₂Carry out the perspective view that vertical projection obtains.

Figure 10 is the license plate area P that final location obtains.

Embodiment

Instantiation of the present invention is described as follows in conjunction with Fig. 2-7:

1. utilize the Sobel operator to carry out rim detection

If need to carry out the coloured image of car plate location is F ₀, as shown in Figure 2, it is carried out gray processing obtain gray level image F, as shown in Figure 3, its line number, columns are respectively 768 and 576, utilize horizontal Sobel operator S ₁With vertical Sobel operator S ₂Image F is carried out rim detection, obtain gradient image G ₀, as shown in Figure 4,

G ₀＝|S ₁*F|+|S ₂*F|，

In the formula, $S_1=\left[\begin{array}{ccc}-1& 0& 1\\ -2& 0& 2\\ -1& 0& 1\end{array}\right],$ $S_2=\left[\begin{array}{ccc}-1& -2& -1\\ 0& 0& 0\\ 1& 2& 1\end{array}\right],$ * represent convolution algorithm,

Then, utilize according to threshold value T ₀To gradient image G ₀Carry out binaryzation, obtain binary edge map G ₁, as shown in Figure 5,

$G_1(x,y)=\left\{\begin{array}{cc}1& \mathrm{if}{G}_0(x,y)>T_0\\ 0& \mathrm{else}\end{array}\right.,$

In the formula, threshold value T ₀Be 15, G ₀(x, y) is gradient image G ₀Middle coordinate is the gray scale of the pixel of (x, y), G ₁(x, y) is binary edge map G ₁Middle coordinate is the value of the pixel of (x, y), if 1 this pixel of expression belongs to the edge, if 0 this pixel of expression does not belong to the edge;

2. the location of car plate vertical direction

To binary edge map G ₁Carry out horizontal projection, as shown in Figure 6, obtain vertical vector G _1x,

$G_{1x}\left(i_1\right)=\underset{j=1}{\overset{n}{\mathrm{\Σ}}}{G}_1(i_1,j_1),$ i ₁＝1，2，3......m，

Utilize threshold value T ₁To vertical vector G _1xCarry out binaryzation, obtain two-value vertical vector G ' _1x,

$G_{1x}^{\′}\left(i_1\right)=\left\{\begin{array}{cc}1& \mathrm{if}{G}_{1x}\left(i_1\right)>T_1\\ 0& \mathrm{else}\end{array}\right.,$ i ₁＝1，2，3......m，

In the formula, threshold value T ₁Be 5, G ' _1x(i ₁) be 1 to be illustrated in binary edge map G ₁I ₁The row place exists characters on license plate, G ' _1x(i ₁) be 0 to be illustrated in binary edge map G ₁I ₁There is not characters on license plate in the row place,

Afterwards, to two-value vertical vector G ' _1xScan from the top down, if detect license plate area P by scanning, then to the initial row coordinate P of license plate area P _X1With end line coordinate P _X2Position, as shown in Figure 7, scanning step is as follows:

Step 1: establish two-value vertical vector G ' _1xIn first numerical value be that the row-coordinate of 1 pixel is i ₂, with row-coordinate i ₂Give variable xbeg, and make variable xend=xbeg,

Step 2: i ₂=i ₂+ 1, if i ₂=m+1 then forwards step 3 to, if i ₂＜m+1 then forwards step 4 to,

Step 3: if xend-xbeg＞25 then show in gray level image F and detected car plate make P _X1=xbeg, P _X2=xend finishes scanning; If scanning is finished in xend-xbeg≤25 then show do not detect car plate in gray level image F,

Step 4: if i ₂-xend≤6 and i ₂-xbeg＞25 then make P _X1=xbeg, P _X2=xend finishes scanning, if i ₂-xend≤6 and i ₂-xbeg≤25 then make xend=i ₂, return step 2; If i ₂-xend＞6 and xend-xbeg＞25 item make P _X1=xbeg, P _X2=xend finishes scanning, if i ₂-xend＞6 and xend-xbeg≤25 item make xbeg=i ₂, xend=xbeg returns step 2,

In scanning process, detect car plate, then the initial row coordinate P of license plate area P _X1Be positioned as 387, its end line coordinate P _X2Be positioned as 409, obtain binary edge map G behind the perpendicular positioning according to positioning result ₂, as shown in Figure 8,

G ₂(i ₂，j ₂)＝G ₁(i ₂+386，j ₂)，

I wherein ₂=1,2,3......23, j ₂=1,2,3......n,

3. the location of car plate horizontal direction

To the binary edge map G behind the perpendicular positioning ₂Carry out vertical projection, as shown in Figure 9, obtain horizontal vector G _2y,

$G_{2y}\left(i_3\right)=\underset{j_3=1}{\overset{P_{x2}-P_{x1}+1}{\mathrm{\Σ}}}{G}_1(j_3,i_3),$ i ₃＝1，2，3......n，

Utilize threshold value T ₂To horizontal vector G _2yCarry out binaryzation, obtain binarizing level vector G ' _2y,

$G_{2y}^{\′}\left(i_3\right)=\left\{\begin{array}{cc}1& \mathrm{if}{G}_{2y}\left(i_3\right)>T_1\\ 0& \mathrm{else}\end{array}\right.,$ i ₃＝1，2，3......n，

In the formula, threshold value T ₂Be 15, G ' _2y(i ₃) be 1 to be illustrated in binary edge map G ₂I ₃There is characters on license plate in the row place, G ' _2y(i ₃) be 0 to be illustrated in binary edge map G ₂I ₃There is not characters on license plate in the row place,

Afterwards, to binarizing level vector G ' _2yScan from left to right, if detect license plate area P by scanning, then to the initial row coordinate P of license plate area P _Y1With end line coordinate P _Y2Position, as shown in Figure 7, scanning step is as follows:

Step 1: establish binarizing level vector G ' _2yIn first numerical value be that the row coordinate of 1 pixel is i ₄, with row coordinate i ₄Give variable ybeg, and make variable yend=ybeg,

Step 2: i ₄=i ₄+ 1, if i ₄=m+1 then forwards step 3 to, if i ₄＜m+1 then forwards step 4 to,

Step 3: if yend-ybeg＞70 then show in gray level image F and detected car plate make P _Y1=ybeg, P _Y2=yend finishes scanning; If scanning is finished in yend-ybeg≤70 then show do not detect car plate in gray level image F,

Step 4: if i ₄-yend≤15 and i ₄-ybeg＞70 then make P _Y1=ybeg, P _Y2=yend finishes scanning, if i ₄-yend≤15 and i ₄-ybeg≤70 then make yend=i ₄, return step 2; If i ₄-yend＞15 and yend-ybeg＞70 then make P _Y1=ybeg, P _Y2=yend finishes scanning, if i ₄-yend＞15 and yend-ybeg≤70 then make ybeg=i ₄, yend=ybeg returns step 2,

In scanning process, detect car plate, then the initial row coordinate P of license plate area P _Y1Be positioned as 85, its terminal point row coordinate P _Y2Be positioned as 210, in conjunction with initial row coordinate P _X1With end line coordinate P _X2License plate area P after can obtaining locating is:

P(i ₄，j ₄)＝F ₀(i ₄+386，j ₄+84)，

I wherein ₄=1,2,3......23, j ₄=1,2,3......117,

The license plate area P that final location obtains as shown in figure 10.

Claims (1)

1. license plate locating method based on rim detection is characterized in that:

1. utilize the Sobel operator to carry out rim detection

If need to carry out the coloured image of car plate location is F ₀, it being carried out gray processing obtain gray level image F, its line number, columns are respectively m and n, utilize horizontal Sobel operator S ₁With vertical Sobel operator S ₂Image F is carried out rim detection, obtain gradient image G ₀,

G ₀=|S ₁*F|+|S ₂*F|，

In the formula, $S_1=\left[\begin{array}{ccc}-1& 0& 1\\ -2& 0& 2\\ -1& 0& 1\end{array}\right],$ $S_2=\left[\begin{array}{ccc}-1& -2& -1\\ 0& 0& 0\\ 1& 2& 1\end{array}\right],$ * represent convolution algorithm,

Then, utilize threshold value T ₀To gradient image G ₀Carry out binaryzation, obtain binary edge map G ₁,

$G_1(x,y)=\left\{\begin{array}{cc}1& \mathrm{if}{G}_0(x,y)>T_0\\ 0& \mathrm{else}\end{array}\right.,$

In the formula, threshold value T ₀Be 15-20, G ₀(x, y) is gradient image G ₀Middle coordinate is the gray scale of the pixel of (x, y), G ₁(x, y) is binary edge map G ₁Middle coordinate is the value of the pixel of (x, y), if 1 this pixel of expression belongs to the edge, if 0 this pixel of expression does not belong to the edge;

2. the location of car plate vertical direction

To binary edge map G ₁Carry out horizontal projection, obtain vertical vector G _1x,

$G_{1x}\left(i_1\right)=\underset{j=1}{\overset{n}{\mathrm{\Σ}}}{G}_1(i_1,j_1),i_1=\mathrm{1,2,3}......m,$

Utilize threshold value G ₁To vertical vector G _1xCarry out binaryzation, obtain the two-value vertical vector

$G_{1x}^{\′}\left(i_1\right)=\left\{\begin{array}{cc}1& \mathrm{if}{G}_{1x}\left(i_1\right)>T_1\\ 0& \mathrm{else}\end{array}\right.,i_1=\mathrm{1,2,3}......m,$

In the formula, threshold value T ₁Be 3-5,

Be 1 to be illustrated in binary edge map G ₁I ₁There is characters on license plate in the row place,

Be 0 to be illustrated in binary edge map G ₁I ₁There is not characters on license plate in the row place,

Afterwards, to the two-value vertical vector

Scan from the top down, if detect license plate area P by scanning, then to the initial row coordinate P of license plate area P _X1With end line coordinate P _X2Position, behind the end of scan, obtain binary edge map G behind the perpendicular positioning according to positioning result ₂,

G ₂(i ₂,j ₂)=G ₁(i ₂+P _x1-1,j ₂)，

I wherein ₂=1,2,3...... (P _X2-P _X1+ 1), j ₂=1,2,3......n,

Concrete scanning step is as follows:

Step 1: establish the two-value vertical vector

In first numerical value be that the row-coordinate of 1 pixel is i ₂, with row-coordinate i ₂Give variable xbeg, and make variable xend=xbeg,

Step 2: i ₂=i ₂+ 1, if i ₂=m+1 then forwards step 3 to, if i ₂＜m+1 then forwards step 4 to,

Step 3: if xend-xbeg〉25 then show in gray level image F and detected car plate, make P _X1=xbeg, P _X2=xend finishes scanning; If scanning is finished in xend-xbeg≤25 then show do not detect car plate in gray level image F,

Step 4: if i ₂-xend≤6 and i ₂-xbeg〉25, then make P _X1=xbeg, P _X2=xend finishes scanning, if i ₂-xend≤6 and i ₂-xbeg≤25 then make xend=i ₂, return step 2; If i ₂-xend〉6 and xend-xbeg 25 items make P _X1=xbeg, P _X2=xend finishes scanning, if i ₂-xend〉6 and xend-xbeg≤25 item make xbeg=i ₂, xend=xbeg returns step 2,

3. the location of car plate horizontal direction

To the binary edge map G behind the perpendicular positioning ₂Carry out vertical projection, obtain horizontal vector G _2y,

$G_{2y}\left(i_3\right)=\underset{j_3=1}{\overset{P_{x2}-P_{x1}+1}{\mathrm{\Σ}}}{G}_1(j_3,i_3),i_3=\mathrm{1,2,3}......n,$

Utilize threshold value T ₂To horizontal vector T _2yCarry out binaryzation, obtain the binarizing level vector

$G_{2y}^{\′}\left(i_3\right)=\left\{\begin{array}{cc}1& \mathrm{if}{G}_{2y}\left(i_3\right)>T_2\\ 0& \mathrm{else}\end{array}\right.,i_3=\mathrm{1,2,3}......n,$

In the formula, threshold value T ₂Be 15-20,

Be 1 to be illustrated in binary edge map G ₂I ₃There is characters on license plate in the row place,

Be 0 to be illustrated in binary edge map G ₂I ₃There is not characters on license plate in the row place,

Afterwards, to the binarizing level vector Scan from left to right, if detect license plate area P by scanning, then to the initial row coordinate P of license plate area P _Y1With terminal point row coordinate P _Y2Position, behind the end of scan, in conjunction with initial row coordinate P _X1With end line coordinate P _X2License plate area P after can obtaining locating is:

P(i ₄,j ₄)=F ₀(i ₄+P _x1-1,j ₄+P _y1-1)，

I wherein ₄=1,2,3...... (P _X2-P _X1+ 1), j ₄=1,2,3...... (P _Y2-P _Y1+ 1),

Concrete scanning step is as follows:

Step 1: establish the binarizing level vector In first numerical value be that the row coordinate of 1 pixel is i ₄, with row coordinate i ₄Give variable ybeg, and make variable yend=ybeg,

Step 2: i ₄=i ₄+ 1, if i ₄=n+1 then forwards step 3 to, if i ₄＜n+1 then forwards step 4 to,

Step 3: if yend-ybeg〉70 then show in gray level image F and detected car plate, make P _Y1=ybeg, P _Y2=yend finishes scanning; If scanning is finished in yend-ybeg≤70 then show do not detect car plate in gray level image F,

Step 4: if i ₄-yend≤15 and i ₄-ybeg〉70, then make P _Y1=ybeg, P _Y2=yend finishes scanning, if i ₄-yend≤15 and i ₄-ybeg≤70 then make yend=i ₄, return step 2; If i ₄-yend〉15 and yend-ybeg 70, then make P _Y1=ybeg, P _Y2=yend finishes scanning, if i ₄-yend＞15 and yend-ybeg≤70 then make ybeg=i ₄, yend=ybeg returns step 2.

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