JPS6319826B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a method for classifying leukocytes, and more particularly to a method for distinguishing between neutrophils, cyclokaryocytes and metamyelocytes within leukocytes.

Conventionally, when distinguishing between neutrophil cyclokaryocytes and metamyelocytes, the long axis and short axis of the nucleus are measured, and those with a ratio of 3 to 1 or more are considered neutrophil cyclokaryocytes, and those with a ratio of 3 to 1 or more are classified as metamyelocytes. I was discerning. However, the accuracy of this discrimination method was poor.

An object of the present invention is to provide a method for reliably and highly accurately distinguishing neutrophil annular cells and metamyelocytes by utilizing the fact that their nucleus opening angles are different.

Hereinafter, the present invention will be explained in detail based on one embodiment shown in the drawings. Figure 1 is a block diagram of the apparatus used to carry out the invention of this method, in which 1 is a microscope section, 2 is a
3 is a photoelectric converter, 3 is an analog/digital converter, 4 is a microcomputer, 5 is a host computer, and 6 is an output device. The microscope section 1 is
White blood cells are captured at approximately 1500x magnification from the May-Giemsa-stained specimen set on the stage.
In the optical image obtained by the microscope section 1, the background where no white blood cells are shown is bright, the nucleus of the white blood cell is dark, and the protoplasm of the white blood cell has an intermediate brightness. The photoelectric conversion unit 2 captures this optical image with a television camera and generates a video signal representing the brightness of each part of the optical image. This video signal is converted into an 8-bit (256 level) digital signal by the analog-to-digital converter 3. This digital signal converts the optical image into 4800 (60x80)
The brightness of each pixel is expressed as 8-bit binary information. Each of these digital signals is transferred to the microcomputer 4 by DMA.
stored in internal memory. The microcomputer 4 arranges the stored digital signals (pixel data). The host computer 5 processes the image data using the Fortran language,
It is determined whether the white blood cells captured by the microscope section 1 are ring cells or metamyelocytes, and the determination result is output to the output device 6.

The process performed by the host computer 5 is to extract the core part from the pixel data, thin it, and
A step of obtaining a nuclear skeleton that represents the characteristics of the nuclear morphology, and a step of removing and unifying the parts of the skeleton that cause errors when calculating the nuclear opening angle θ from this nuclear skeleton. The step of calculating the nuclear opening angle θ from the nuclear skeleton obtained and 0° < θ <
If the angle is 100°, it is determined to be a neutrophil ring nucleus, and 100°<θ
If the angle is <180°, it is determined to be a metamyelocyte.

The step of obtaining the nuclear skeleton is Hildrich's 8
This is carried out as follows according to the connected thinning method.

(1) Scan the pixel data screen in sequence from the one corresponding to the upper left to the lower right, and discover the pixels that make up the outline of the nucleus. As mentioned above, optical images representing portions of the nucleus are always dark. Therefore, when sequentially scanning the pixel data representing the brightness of each pixel in this optical image, if there is a sudden change from a bright value to a dark value, this is due to the boundary between the background and the protoplasm, or the boundary between the protoplasm and the nucleus. Since it represents a boundary, for example, if the value of pixel data that suddenly changes and becomes darker is below a predetermined level,
It can be seen that the darkened pixel data corresponds to the outline of the nucleus.

(2) (a) If the connection of the pixels that make up the nucleus's outline among the eight pixels around this pixel is not lost by deleting the pixel that makes up the outline, then this Delete. For example, in Fig. 15a, the pixel indicated by the symbol x ₀ is the pixel that is being considered for deletion, and the pixels indicated by the symbols x ₁ to _{x 8} are the 8 pixels around the pixel x ₀ , the pixels indicated by hatching. If pixel represents the nucleus, then pixel x ₀
Of the 8 pixels around the , pixels x ₁ , x ₆ , x ₇ , x ₈ are the pixels that constitute the nucleus, and even if pixel x ₀ is removed, x ₁ , x ₆ , x ₇ , x The connection of ₈ is not lost, so remove pixel x ₀ . Also, (b) the number of connections is 1
, this pixel is not removed. For example, as shown in figure b, if we consider pixel x ₀ to be considered as to whether or not to delete it, this pixel x ₀ forms a nucleus with only pixel x ₇ out of the surrounding eight pixels x ₁ to _{x 8} . Pixel x ₀ is not deleted because it is connected to the pixel that is.

(3) Perform scanning (1) and (2) once from the upper left to the lower right of the image data.

Step (2) is performed as preprocessing to calculate the opening angle of the thinned corner skeleton.The pixels that make up the wide nucleus are deleted from the nucleus outline, and the central part of the nucleus is In the case of (a) of (2), if the connection is not lost, that part has not been completely thinned yet, and the pixels that have not lost the connection are This is an extra pixel, so delete it. In the case of (b) of (2), the loss of the connection means that the part has been completely thinned, so if you delete it, the thin line will be severed and the final judgment will be affected. An error will occur. Therefore, the deletion is not performed.

(4) If all of the remaining pixels have 2 or less connections with the pixels that constitute the outline of the nucleus among the surrounding 8 pixels, for example, pixel x ₀ in Figure 2a, x ₂ and
When all pixels are connected only to x ₆ , that is, when they are all composed of straight lines, it is assumed that line thinning has been completed, and no further pixels are deleted. Repeat operations (1), (2), and (3) until this condition is met. However, among the remaining pixels, x ₃ , like pixel x ₀ in Figure 2b,
If there is a pixel connected to x ₆ and x ₈ and the number of connections is 3, or like pixel x ₀ in Figure 2 c
If there is a line that is connected to x ₁ , x ₈ , x _{5 ,} x ₇ and the number of connections is 4, originally x ₀ would be deleted to further thin the line, but these x ₀ is a branching point or intersection, so do not delete it. Note that the number of deleted pixels is stored as a negative value of the number of operations, since it is necessary to unify the nuclear skeleton, which will be described later.
The reason why this negative value is stored will be explained in detail later.

The nuclear skeleton thus obtained is shown in FIG. Among the pixels that make up the skeleton, those that have three or more connections with the surrounding pixels that make up the skeleton, that is, those that serve as branching points or intersections, are called nodes 7, 8, and 9, and are called nodes 7, 8, and 9. The skeletons are called trunks 10 and 11, and the others are called branches 12, 13, 14, 15, and 16.
Note that the number of pixels (negative value) deleted to obtain branch 16 is assumed to be 4. Although there are originally some pixels deleted to obtain other branches, trunks, etc., for convenience of explanation, only branch 16 will be considered as the number of deleted pixels.

According to Hildrich's 8-connected thinning method, the nuclear skeleton is unified because multiple branches are generated as shown in Figure 3 due to unevenness of the nucleus other than the original opening of the nucleus. This is because the opening angle θ of the nucleus cannot be calculated unless branches due to extra unevenness of the nucleus are removed. This unification is basically performed by deleting the shorter branch among multiple branches connected to the same node. In addition, in FIG. 3, the pixels forming the nuclear skeleton are represented by "1". In this state, still
It has not been determined which are branches, which are trunks, and which are nodes. Therefore, first, branches, nodes, and trunks are determined as follows.

First, in FIG. 3, a pixel with a value of "1" and a pixel with a value of "1" among the eight pixels around it are found to have one connection. Since this is the end point of the branch, the branch is traced from this pixel as shown in FIG. 4, and a number 2 is assigned until a pixel (node) with a number of connections of 3 or more is found. Assign a number to each branch in the same way.
Also, since pixels with a connection number of 3 or more are considered to be pixels forming a node, their value is set to "100".
After that, scan the image again as shown in Figure 5,
The nodes are numbered starting from 101. At this time, as shown in FIG. 4, if nodes with a value of "100" are connected to each other, they are regarded as the same node and given the same number, such as 102 and 103 in FIG. 5. Next, the pixels that make up the trunk are left as "1", so as shown in Figure 5, the pixels that form the end points of the trunk (the values connected to the pixels with a value of "100" or more are 1). 1000 or more pixels, for example 100, as shown in Figure 6 until the other end point.
Add 1,1002 to each.

After that, focus on the node, and create a table as shown in FIG. 7 with the number and length of the branch connected to this node, and the number and length of the trunk connected to this node. In this case, the lengths of branches and trunks are calculated by setting the distance between pixels to 1 in the vertical and horizontal directions and √2 in the diagonal direction. Therefore, as is clear from FIG. 16, the length of branch 2 is calculated from node 101 as 1+√2+√2
It is calculated as ≒4. As is clear from FIG. 17, the length of branch 3 is calculated from node 101 and is √
2+√2+√2+1+√2+1+1+√2≒10. In the case of branch 4, when calculated from Figure 18, √2 + √2 ≒ 3, but as mentioned above, when extracting the skeleton, 4 pixels are deleted to obtain branch 4, so it is set to 3. , the true length of the branch cannot be determined. Therefore, in order to obtain the true length of branch 7, 4 is added to make it 7. This is because Hildrich's 8-connected thinning method expresses the features of the shape well, but it does not express the exact length. Branch 5 is calculated from node 103 as shown in FIG.
2+√2+1+1+1+1+√2+√2+1≒12
becomes. As is clear from FIG. 20, branch 6 is calculated from node 103 as √2+√2+1+
√2+1+1+√2+1+1+√2+1+1≒14
becomes. As is clear from FIG. 21, the length of trunk 1002 is calculated from node 102 as √2+
1+1+√2+1+√2+1+√2≒10.
As is clear from FIG. 22, the length of trunk 1001 is calculated from node 101 as √2+√2+
√2+1+√2≒7.

Next, since nodes with the same number are considered to constitute the same node, as shown in Figure 8, one column of these pixels contains the branch written in the column of another pixel that is considered to be the same node. Rewrite the number and length of the trunk and the number and length of the trunk. At this time, the number of pixels constituting the same node is also written in the same column. This allows each trunk,
Determination of branches and nodes and their lengths are completed.

After that, attention is paid to a node having only one trunk among each node, that is, a node on the tip side of the skeleton, for example, node 101. As shown in FIG. 9, 1 is written in the check bit of this node, and of the branches 2 and 3 of this node, the longer branch 3 is left and the shorter branch 2 is deleted. As a result, the node 101 is no longer connected to 3 or more surrounding 8 pixels, so it is no longer a node, and the trunk 1002 is no longer a trunk.
Branch 3, node 101 and trunk 1 so far
This means that a new branch 3 is configured where 001 is connected to node 102. Therefore, the trunk 1001 of the node 102 is deleted, and the number 3 of the trunk newly connected to the node 102 is added to the branch number column of the node 102, and the length of the original branch 3 is 10, and the trunk 1001 is added to the length column of the trunk 1001. Write the length 7 of the node 101 plus the number 1 of the node 101, that is, the length 18 of the new branch 3. As a result, branch 3, node 101, and trunk 1001 are treated as one branch from now on. This state is shown in FIG. In the figure, the part labeled 3,1001 is the original trunk 1001 part, and 3,101 is the original node 101.

As a result of processing in this way, we focused on node 102, which became the tip node, and
The lengths of branches 3 and 4 connected to 02 are compared. At this time, since branch 3 is longer, branch 4 is deleted as shown in FIG. In this way, since the branch to be deleted is determined according to the length of the branch, it is necessary to accurately determine the length of branch 4, and the number of deleted pixels is added to that length as in branch 4. are doing. As a result of deletion in this way, branch 3, node 102, and trunk 1002 can be treated as one branch connected to node 103, so trunk 10 of node 103 is deleted as shown in FIG.
02 is deleted, the number 3 of the new branch connected to node 103 is entered in the branch number column of node 103, and the length of branch 3 is 18 in the length column.
, the length 10 of the trunk 1002, and the node 103
The value obtained by adding the number 3 of , that is, the new branch 3
Write the length of 31. This state is shown in FIG. In the figure, the original trunk 1002 is designated by the reference numeral 3,1002.

This leaves only the node 103, to which three branches 3, 5, and 6 are connected. Then, out of these three branches, leave the two longest ones and delete the others.
FIG. 12 shows the skeleton unified in this way.

When integrating the skeleton into one in this way,
The basic principle is to compare the lengths of branches and delete the shorter branch. Therefore, since it is necessary to know the true length of the branch, the number of pixels deleted when obtaining this branch is added to the calculated length of the branch.

The process of calculating the opening angle of the nucleus is performed as follows. Now, it is assumed that a nuclear skeleton as shown in FIG. 13 is obtained by extracting the nuclear skeleton and unifying it as described above. Then, the midpoint A (o, o) of this skeleton, the end points B (l, m), and C
Find (p, q). Next, find the inner product of vectors AB and AC using the components. That is, |AB|・|AC|・cosθ=pl+qm. Here |AB| is √ ² + ² and |AC
| is √ ² + ² , so cosθ is More demanded. Calculate angle θ from cosθ. This angle θ is defined as the opening angle of this nucleus. When determining this opening angle, it is necessary to determine the midpoint A (0, 0) of the unified nuclear skeleton as described above. In this case, the midpoint is calculated based on the length of the nuclear skeleton, but if the number of deleted pixels is ignored when calculating the nuclear skeleton, the exact length of the nuclear skeleton cannot be determined, and this makes it difficult to accurately calculate the midpoint. It is not possible to find the midpoint A. In the end, it is not possible to obtain an accurate opening angle. From this point of view, when determining the length of each branch, etc., it is necessary to add the number of deleted pixels to determine the true length of the branch or trunk.

The process of comparing the angles is to compare the opening angle of the nucleus obtained by calculation with the angle φ (approximately 100°) at the boundary between the neutron annular nucleus cell and the metamyelocyte, and calculate the angle as shown in Fig. 14. If it is larger than φ, it is determined to be a metamyelocyte, and if it is smaller than φ, it is determined to be a neutrophil ring nucleus cell.

This method distinguishes between neutrophil ring-shaped karyocytes and metamyelocytes based on the opening angle of the nucleus, so it is more effective than the conventional method of distinguishing based on the ratio of the long axis to short axis of the nucleus. , can be determined with high precision and reliability.

[Brief explanation of the drawing]

Fig. 1 is a block diagram of the apparatus used to carry out the discrimination method according to the present invention, Figs.
The figure shows the extracted nuclear skeleton, Figures 4 to 12 show the process of unifying the nuclear skeletons, and Figure 13 shows the process of calculating the opening angle of the unified nuclear skeleton. Figure 14 is a diagram showing the distribution of neutrophil ring cells and metamyelocytes, Figure 1.
Figures 5a and b are diagrams showing the basic principle of wire thinning, No. 16
Figures 22 to 22 are diagrams showing how to calculate the branch and trunk lengths of data obtained by thinning;
This figure and FIG. 24 are diagrams showing changes in data during the process of unification. 3...Analog-digital converter, A...1
Midpoint of the unified skeleton, B, C... end points of the unified skeleton, φ... boundary angle between the neutrophil ring nucleus cell and metamyelocyte.

Claims (1)

[Claims] 1. Even if each pixel data obtained by digitizing a video signal obtained by imaging a white blood cell is sequentially scanned to find pixel data corresponding to the outline of the nucleus of the white blood cell, and that pixel data is deleted, this pixel data will not be lost. If the connectivity of the pixel data constituting the above nucleus around it is not lost, delete it, and if the number of connections between the pixel data searched above and the pixel data constituting the above nucleus around it is 1, the above The step of obtaining the thinning data of the nucleus by sequentially not deleting the searched pixels, and the step of deleting the branched ones of the thinning data with short lengths. the step of calculating the nucleus opening angle between the midpoint of the unified thinning data and its both ends;
When the angle of nuclear opening is smaller than 100°, it is determined to be a neutrophil ring-karyocyte, and when the above-mentioned nuclear opening angle is larger than 100° and smaller than 180°, it is judged to be a metamyelocyte. How to identify myelocytes.