有人能帮我找到比循环更好的解决方案吗?假设下面的数据框由4列组成。我正在寻找一种方法,通过循环以外的其他方法获得与“Result”列中相同的值。你知道吗
逻辑如下:
如果priority1为负,则情况正好相反。你知道吗
这是我在一个非常丑陋和低效的循环之后的最终结果:
+-----+-----------+-----------+-----------+-----------+--------+
| | priority1 | priority2 | priority3 | priority4 | Result |
+-----+-----------+-----------+-----------+-----------+--------+
| 0 | | | | | |
| 1 | | 1 | -1 | -1 | |
| 2 | | | | | |
| 3 | | | | 1 | |
| 4 | | | | 1 | |
| 5 | | | | | |
| 6 | | | | -1 | |
| 7 | | | | | |
| 8 | | | | | |
| 9 | 1 | 1 | 1 | 1 | 1 |
| 10 | | | | | |
| 11 | | | | 1 | |
| 12 | | | 1 | | |
| 13 | | | | | |
| 14 | | | -1 | -1 | |
| 15 | | | | | |
| 16 | | | | | |
| 17 | | | | -1 | |
| 18 | | | | | |
| 19 | | | | | |
| 20 | | 1 | 1 | 1 | 2 |
| 21 | | | | | |
| 22 | | | -1 | -1 | |
| 23 | | | | | |
| 24 | | | | | |
| 25 | | | | -1 | |
| 26 | | | | | |
| 27 | | | 1 | 1 | 3 |
| 28 | | | | | |
| 29 | | | | | |
| 30 | | | | -1 | |
| 31 | | | | | |
| 32 | | | | | |
| 33 | | | -1 | -1 | |
| 34 | | | | | |
| 35 | | | 1 | 1 | 4 |
| 36 | | | | | |
| 37 | | | | | |
| 38 | | | | | |
| 39 | | | -1 | -1 | |
| 40 | | | | | |
| 41 | | | | | |
| 42 | | 1 | 1 | 1 | 2 |
| 43 | | | | | |
| 44 | | | | | |
| 45 | | | | -1 | |
| 46 | | | | | |
| 47 | | | | | |
| 48 | | | | | |
| 49 | | | | | |
| 50 | | -1 | -1 | -1 | |
| 51 | | | | | |
| 52 | | | | | |
| 53 | | 1 | 1 | 1 | 2 |
| 54 | | | | | |
| 55 | | | | | |
| 56 | | | | -1 | |
| 57 | | | | | |
| 58 | | | | | |
| 59 | | | -1 | -1 | |
| 60 | | | | | |
| 61 | | | | | |
| 62 | | | 1 | 1 | 3 |
| 63 | | | | | |
| 64 | -1 | -1 | -1 | -1 | -1 |
| 65 | | | | | |
| 66 | | | 1 | 1 | |
| 67 | | | | | |
| 68 | | | | | |
| 69 | | | | | |
| 70 | | | | -1 | |
| 71 | | | | | |
| 72 | | | | | |
| 73 | | | 1 | 1 | |
| 74 | | | | -1 | |
| 75 | | | | | |
| 76 | | | 1 | 1 | |
| 77 | | | | | |
| 78 | | -1 | -1 | -1 | -2 |
| 79 | | | 1 | | |
| 80 | | | 1 | 1 | |
| 81 | | | | | |
| 82 | | | -1 | -1 | -3 |
| 83 | | | 1 | 1 | |
| 84 | | | | | |
| 85 | | | | | |
| 86 | | | | | |
| 87 | | | -1 | -1 | -4 |
| 88 | | | | | |
| 89 | | -1 | -1 | -1 | -2 |
| 90 | | | | | |
| 91 | | | | | |
| 92 | | | | -1 | |
| 93 | | | | | |
| 94 | | | | | |
| 95 | | | 1 | 1 | |
| 96 | | | | | |
| 97 | | | | | |
| 98 | | | | -1 | |
| 99 | | | | 1 | |
| 100 | | | | | |
| 101 | | | -1 | -1 | -3 |
| 102 | | | | | |
| 103 | | | | | |
| 104 | | 1 | 1 | 1 | |
| 105 | | | | | |
| 106 | | | | 1 | |
| 107 | | | | | |
| 108 | | | -1 | -1 | |
| 109 | | | | | |
| 110 | | | | | |
| 111 | | | 1 | 1 | |
| 112 | | | | | |
| 113 | | | | | |
| 114 | | | -1 | -1 | |
| 115 | | | | | |
| 116 | | | 1 | 1 | |
| 117 | | | | | |
| 118 | | | | | |
| 119 | | -1 | -1 | -1 | -2 |
| 120 | | | | | |
| 121 | | | | | |
| 122 | | | | 1 | |
| 123 | | | | | |
| 124 | | | | 1 | |
| 125 | | | | | |
| 126 | | | | | |
| 127 | | | 1 | 1 | |
| 128 | | | | | |
| 129 | | | | | |
| 130 | | | -1 | -1 | -3 |
| 131 | | | | | |
| 132 | | | | | |
| 133 | | | | | |
| 134 | 1 | 1 | 1 | 1 | 1 |
| 135 | | | | -1 | |
| 136 | | | | | |
| 137 | | -1 | -1 | -1 | |
| 138 | | | 1 | | |
| 139 | | | | 1 | |
| 140 | | 1 | 1 | 1 | 2 |
| 141 | | | 1 | 1 | 3 |
| 142 | | | | | |
| 143 | | | | -1 | |
| 144 | | | | | |
| 145 | | | | 1 | 4 |
+-----+-----------+-----------+-----------+-----------+--------+
使用piRSquared的示例框架(帽尖!),我可能会这样做
这让我
设置
解决方案
它的工作原理
抓取
numpy
数组使用
True
表示非零每个连续的列继续是非零的
乘以
v
过滤出要加的数,然后求和原始时间测试 小数据
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