blocks|key|1194291|text|#+n1,+n2+=+cell+numbers
cellsize+=+100.0
x1,x2+=+n1%25gs,+n2%25gs
y1,y2+=+n1//gs,+n2//gs
dist+=+sqrt(+float(x1-x2)**2+%2B+float(y1-y2)**2)++#+pythagoras+theorem
dist+*=+cellsize|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1194292|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code># n1, n2 = cell numbers
cellsize = 100.0
x1,x2 = n1%gs, n2%gs
y1,y2 = n1//gs, n2//gs
dist = sqrt( float(x1-x2)**2 + float(y1-y2)**2) # pythagoras theorem
dist *= cellsize
</code></pre>

blocks|key|3510796|text|您应该考虑坐标，而不是单元格编号|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3510797|gs+=+5+##+Cells+per+side
S+=+gs*gs+##+Grid+Size
r0+=+100+##+distance+between+two+cells

for+i+in+range(0,+S):
++++for+j+in+range(0,+S):
++++++++if+i+==+j:+continue
+++++++++++xi+=+int(i/gs)
+++++++++++yi+=+i+%25+gs
+++++++++++xj+=+int(j/gs)
+++++++++++yj+=+j+%25+gs
+++++++++++dist+=+r0+*+(abs(xi-xj)+%2B+abs(yi-yj))|code-block|syntax|javascript|3510798|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

you should consider co-ordinate and not the cell number

<pre><code>gs = 5 ## Cells per side
S = gs*gs ## Grid Size
r0 = 100 ## distance between two cells

for i in range(0, S):
 for j in range(0, S):
 if i == j: continue
 xi = int(i/gs)
 yi = i % gs
 xj = int(j/gs)
 yj = j % gs
 dist = r0 * (abs(xi-xj) + abs(yi-yj))
</code></pre>

blocks|key|3510804|text|这是实现这一点的一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3510805|r+=+100

grid+=+((0,+++1,++2,++3,++4),
++++++++(5,+++6,++7,++8,++9),
++++++++(10,+11,+12,+13,+14),
++++++++(15,+16,+17,+18,+19),
++++++++(20,+21,+22,+23,+24))

def+coord(n):
++++for+x,+line+in+enumerate(grid):
++++++++if+n+not+in+line:
++++++++++++continue
++++++++y+=+line.index(n)
++++++++return+x,+y

def+dist(n,+m):
++++++++xn,+yn+=+coord(n)
++++++++xm,+ym+=+coord(m)
++++++++return+r+*+(abs(xn+-+xm)+%2B+abs(yn+-+ym))

print(dist(0,+3))++#+300
print(dist(2,+7))++#+100
print(dist(11,+19))+#+400|code-block|syntax|javascript|3510806|这个想法是先得到你的数字的坐标，然后再计算“距离”。|3510807|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

this is a way to do that:

<pre><code>r = 100

grid = ((0, 1, 2, 3, 4),
 (5, 6, 7, 8, 9),
 (10, 11, 12, 13, 14),
 (15, 16, 17, 18, 19),
 (20, 21, 22, 23, 24))

def coord(n):
 for x, line in enumerate(grid):
 if n not in line:
 continue
 y = line.index(n)
 return x, y

def dist(n, m):
 xn, yn = coord(n)
 xm, ym = coord(m)
 return r * (abs(xn - xm) + abs(yn - ym))

print(dist(0, 3)) # 300
print(dist(2, 7)) # 100
print(dist(11, 19)) # 400
</code></pre>

the idea is to get the coordinates of your numbers first and then calculating the 'distance'.

blocks|key|4979690|text|这对你来说应该是可行的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4979691|n+=+5+#+row+length+in+array

def+distance(a,+b):
++++distance+=+(abs(a+//+n+-+b+//+n)+%2B+abs(a+%25+n+-+b+%25+n))+*+100
++++return+"distance+between+cell+%25s+and+%25s+is+%25s"+%25+(a,+b,+distance)

print(distance(0,+3))
print(distance(2,+7))
print(distance(11,+19))|code-block|syntax|javascript|4979692|输出：|4979693|distance+between+cell+0+and+3+is+300
distance+between+cell+2+and+7+is+100
distance+between+cell+11+and+19+is+400|4979694|其中a和b是您的单元格，n是数组中行的长度，在您的示例中是5。|offset|length|style|CODE|4979695|entityMap^0|0|0|0|0|2|1|4|1|C|1|T|1|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]|$M|11|N|12|O|P]|$M|13|N|14|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|R|$]]

This should work for you

<pre class="lang-py prettyprint-override"><code>n = 5 # row length in array

def distance(a, b):
 distance = (abs(a // n - b // n) + abs(a % n - b % n)) * 100
 return "distance between cell %s and %s is %s" % (a, b, distance)

print(distance(0, 3))
print(distance(2, 7))
print(distance(11, 19))

</code></pre>

Output:

<pre class="lang-sh prettyprint-override"><code>distance between cell 0 and 3 is 300
distance between cell 2 and 7 is 100
distance between cell 11 and 19 is 400
</code></pre>

Where <code>a</code> and <code>b</code> are your cells, and <code>n</code> is a length of the row in array, in your example is <code>5</code>.

blocks|key|4979706|text|我们只需要得到每个数字的行和列编号。然后用b/w之差乘以100就会得到答案。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|4979707|def+get_row_col(num):
++++for+i,g+in+enumerate(grid):
++++++++if+num+in+g:
++++++++++++col+=+g.index(num)
++++++++++++row+=+i
++++return+row,+col

num1+=+get_row_col(11)
num2+=+get_row_col(19)
print+(abs(num1[0]+-+num2[0])*100)+%2B+(abs(num1[1]-num2[1])*100)|code-block|syntax|javascript|4979708|可以增强此代码以检查数字是否存在。|4979709|entityMap^0|C|1|E|1|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

We just need to get the row and column number of every number. Then a difference b/w the two and multiplied by 100 will give you answer

<pre><code>def get_row_col(num):
 for i,g in enumerate(grid):
 if num in g:
 col = g.index(num)
 row = i
 return row, col

num1 = get_row_col(11)
num2 = get_row_col(19)
print (abs(num1[0] - num2[0])*100) + (abs(num1[1]-num2[1])*100)
</code></pre>

One can enhance this code to check if number is present or not.

Let's suppose I want to compute the distance between cells in a square grid <code>5x5</code>. The distance between two cells is <code>100m</code>.

Each cell of the grid is number between <code>0</code> and <code>24</code>

<pre><code>0 1 2 3 4
5 6 7 8 9
10 11 12 13 14
15 16 17 18 19
20 21 22 23 24 
</code></pre>

For instance:

<pre><code>distance between cell 0 and 3 is 300
distance between cell 2 and 7 is 100
distance between cell 11 and 19 is 400
</code></pre>

I have to count the distance as different between <code>x</code> and <code>y</code> location of the cells.

<pre><code>gs = 5 ## Cells per side
S = gs*gs ## Grid Size
r0 = 100 ## distance between two cells

for i in range(0, S):
 for j in range(0, S):
 if i == j: continue
 x = int(floor(i/gs))
 y = int(floor(j/gs))
 dist = x*r0 + abs(j-i)*r0
</code></pre>

but it is not the right solution

Python: how to compute the distance between cells?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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假设我想要计算一个正方形网格5x5中的单元格之间的距离。两个单元格之间的距离是100m。网格的每个像元的编号都在0和24之间0   1  2  3  45   6  7  8  910 11 12 13 1415 16 17 18 1920 21 22 23 24 例如：distance between cell 0 ...

问Python:如何计算单元格之间的距离？
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问Python:如何计算单元格之间的距离？EN