CODEWITHSUNDEEP
python
frank lowe
frank lowe

Reputation: 29

searching for numbers in a list

If I have output that looks like this

[[121, 120, 119, 118, 117, 116, 115, 114, 113, 112, 111],
 [82, 81, 80, 79, 78, 77, 76, 75, 74, 73, 110],
 [83, 50, 49, 48, 47, 46, 45, 44, 43, 72, 109],
 [84, 51, 26, 25, 24, 23, 22, 21, 42, 71, 108],
 [85, 52, 27, 10, 9, 8, 7, 20, 41, 70, 107],
 [86, 53, 28, 11, 2, 1, 6, 19, 40, 69, 106],
 [87, 54, 29, 12, 3, 4, 5, 18, 39, 68, 105],
 [88, 55, 30, 13, 14, 15, 16, 17, 38, 67, 104],
 [89, 56, 31, 32, 33, 34, 35, 36, 37, 66, 103],
 [90, 57, 58, 59, 60, 61, 62, 63, 64, 65, 102],
 [91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101]]

how do I get it to where I can find and print a certain number and the numbers surrounding it? This is my code.

dim = 11
dx, dy = [0, 1, 0, -1], [1, 0, -1, 0]
x, y, c = 0, -1, dim**2
m = [[0 for i in range(dim)] for j in range(dim)]
for i in range(dim + dim - 1):
        for j in range((dim + dim - i) // 2):
                x += dx[i % 4]
                y += dy[i % 4]
                m[x][y] = c
                c -= 1
print(m)
b = c.index(num)
print(b)
a =('\n'.join([' '.join([str(v) for v in r])for r in m]))
print(a)

Upvotes: 0

Views: 561

Answers (2)

Alex Martelli
Alex Martelli

Reputation: 881715

Assuming that list-of-lists is called lol:-):

def neighbors(lol, anum):
    for i, row in enumerate(lol[1:-1]):
        try: where = row.index(anum)
        except ValueError: continue
        if where==0 or where==len(row)-1: continue
        for j in range(i, i+3):
            print(lol[j][where-1], lol[j][where], lol[j][where+1])
        print()

This embodies several assumptions, such as: (A) you don't care for hits on the first or last row or column since they don't have all the neighbors you want to print, and also (B) you don't care about multiple "hits" in a single row but (C) do care about "hits" in multiple rows.

Of course all such assumptions can be changed but that requires you to be much more precise in your specs than you've been so far:-).

The print format assumes either Python 3 or a from __future__ import print_function if you're stuck with Python 2.

Upvotes: 1

Kasravnd
Kasravnd

Reputation: 107297

First in this case as c is a int value hasn't attribute index . so the following command will raise an AttributeError .

b = c.index(num)

and for get the desire result all that you need here is reversing the r list in following command , and for print you can use format :

a =('\n'.join([' '.join(["{:3}".format(v) for v in r[::-1]])for r in m]))

so the result will be :

111 112 113 114 115 116 117 118 119 120 121
110  73  74  75  76  77  78  79  80  81  82
109  72  43  44  45  46  47  48  49  50  83
108  71  42  21  22  23  24  25  26  51  84
107  70  41  20   7   8   9  10  27  52  85
106  69  40  19   6   1   2  11  28  53  86
105  68  39  18   5   4   3  12  29  54  87
104  67  38  17  16  15  14  13  30  55  88
103  66  37  36  35  34  33  32  31  56  89
102  65  64  63  62  61  60  59  58  57  90
101 100  99  98  97  96  95  94  93  92  91

Upvotes: 0

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