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25
4
我有一个 32x32 的 float 矩阵。我想得到其中所有元素的总和。但是当我使用循环时:
s = 0
for i in range(len(Temp)):
for j in range(len(Temp)):
s += Temp[i][j]
我得到 2.795541576006144e-13
当我使用 NumPy .sum(..) 方法时:
np.sum(Temp)
我得到 3.197442310920451e-13
结果截然不同，到底哪个是对的？
附言无法显示我在上面的示例中使用的数组，因为它说我的帖子主要是代码
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13.65805725 -12.19869213 10.13557074 -7.67002338 4.8085357
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-5.51985755 1.90225724]
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-108.05962098 104.6236814 -91.70342372 76.46641941 -56.22157372
35.72600205 -12.31192028]
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131.67102913 -127.48432464 111.74094511 -93.17460164 68.50618579
-43.53225945 15.00211827]
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49.66559654 -17.11579326]
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16.68022253 16.87531285 -49.68854288 80.68981522 -109.85290024
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-169.51516835 164.12514504 -143.85689279 119.95431635 -88.19584455
56.04405417 -19.31394188]
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169.51516835 -164.12514504 143.85689279 -119.95431635 88.19584455
-56.04405417 19.31394188]
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53.8903126 -18.57171792]
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122.89004841 -109.75923138 91.19604314 -69.01197782 43.2653908
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-119.98524672 138.44575074 -152.75423157 165.88049418 -154.47072474
150.22239348 -145.44581679 127.48432464 -106.30213618 78.15814356
-49.66559654 17.11579326]
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107.03229395 -116.13680659 121.54753509 -120.77824689 116.13680659
-107.71402831 96.20477092 -79.93400035 60.48950447 -37.9224322
12.95637487 13.10791142 -38.59561147 62.6758721 -85.3283194
105.16794834 -121.34871544 133.89020378 -145.39546918 135.39472263
-131.67102913 127.48432464 -111.74094511 93.17460164 -68.50618579
43.53225945 -15.00211827]
[ 9.51375658 -28.35981606 46.0538423 -62.38050004 76.99164451
-87.83913358 95.31101401 -99.75148412 99.12014562 -95.31101401
88.39861851 -78.95321507 65.60013876 -49.64245339 31.12213581
-10.63302206 -10.75738489 31.67459976 -51.43675893 70.02714198
-86.30910468 99.58831705 -109.88085054 119.32297784 -111.11557725
108.05962098 -104.6236814 91.70342372 -76.46641941 56.22157372
-35.72600205 12.31192028]
[ -7.0693589 21.07324444 -34.22109206 46.35289322 -57.20995302
65.27036457 -70.8224726 74.12203957 -73.65291273 70.8224726
-65.68609937 58.66753144 -48.74530062 36.88767066 -23.12583318
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-80.29554219 77.74240875 -68.14179118 56.81967556 -41.7763981
26.54681442 -9.1485821 ]
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40.44930027 -36.12728748 30.01720791 -22.71531543 14.24081776
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5.51985755 -1.90225724]
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5.51985755 -1.90225724]
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49.44575074 -47.87353893 41.96150782 -34.98938345 25.72577893
-16.34744761 5.63366904]
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65.27036457 -70.8224726 74.12203957 -73.65291273 70.8224726
-65.68609937 58.66753144 -48.74530062 36.88767066 -23.12583318
7.90104818 7.9934581 -23.53635093 38.22095996 -52.03486078
64.13345052 -74.00079549 81.64883784 -88.66497138 82.56632255
-80.29554219 77.74240875 -68.14179118 56.81967556 -41.7763981
26.54681442 -9.1485821 ]
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-87.83913358 95.31101401 -99.75148412 99.12014562 -95.31101401
88.39861851 -78.95321507 65.60013876 -49.64245339 31.12213581
-10.63302206 -10.75738489 31.67459976 -51.43675893 70.02714198
-86.30910468 99.58831705 -109.88085054 119.32297784 -111.11557725
108.05962098 -104.6236814 91.70342372 -76.46641941 56.22157372
-35.72600205 12.31192028]
[ -11.59254593 34.55653586 -56.11676921 76.01085923 -93.81459028
107.03229395 -116.13680659 121.54753509 -120.77824689 116.13680659
-107.71402831 96.20477092 -79.93400035 60.48950447 -37.9224322
12.95637487 13.10791142 -38.59561147 62.6758721 -85.3283194
105.16794834 -121.34871544 133.89020378 -145.39546918 135.39472263
-131.67102913 127.48432464 -111.74094511 93.17460164 -68.50618579
43.53225945 -15.00211827]
[ 13.22584025 -39.42526736 64.02316015 -86.72016373 107.03229395
-122.11226329 132.49952683 -138.67258244 137.79490787 -132.49952683
122.89004841 -109.75923138 91.19604314 -69.01197782 43.2653908
-14.78182146 -14.95470826 44.03341549 -71.50638666 97.35037734
-119.98524672 138.44575074 -152.75423157 165.88049418 -154.47072474
150.22239348 -145.44581679 127.48432464 -106.30213618 78.15814356
-49.66559654 17.11579326]
[ -14.35087294 42.77890794 -69.46917695 94.09686096 -116.13680659
132.49952683 -143.77036457 150.46852022 -149.51618781 143.77036457
-133.34347287 119.09570613 -98.95347314 74.88235957 -46.94568469
16.03921093 16.22680405 -47.77904004 77.58895088 -105.63131489
130.19157937 -150.22239348 165.74799991 -179.99082482 167.61050354
-163.00079549 157.81790776 -138.32855308 115.34453924 -84.8065277
53.8903126 -18.57171792]
[ 14.92440963 -44.48857909 72.24553223 -97.8574686 120.77824689
-137.79490787 149.51618781 -156.48203715 155.49164451 -149.51618781
138.67258244 -123.85539968 102.90817666 -77.87505423 48.82188225
-16.68022253 -16.87531285 49.68854288 -80.68981522 109.85290024
-135.39472263 156.22607389 -172.37216557 187.184209 -174.30910468
169.51516835 -164.12514504 143.85689279 -119.95431635 88.19584455
-56.04405417 19.31394188]
[ -14.92440963 44.48857909 -72.24553223 97.8574686 -120.77824689
137.79490787 -149.51618781 156.48203715 -155.49164451 149.51618781
-138.67258244 123.85539968 -102.90817666 77.87505423 -48.82188225
16.68022253 16.87531285 -49.68854288 80.68981522 -109.85290024
135.39472263 -156.22607389 172.37216557 -187.184209 174.30910468
-169.51516835 164.12514504 -143.85689279 119.95431635 -88.19584455
56.04405417 -19.31394188]
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133.34347287 -119.09570613 98.95347314 -74.88235957 46.94568469
-16.03921093 -16.22680405 47.77904004 -77.58895088 105.63131489
-130.19157937 150.22239348 -165.74799991 179.99082482 -167.61050354
163.00079549 -157.81790776 138.32855308 -115.34453924 84.8065277
-53.8903126 18.57171792]
[ -13.22584025 39.42526736 -64.02316015 86.72016373 -107.03229395
122.11226329 -132.49952683 138.67258244 -137.79490787 132.49952683
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14.78182146 14.95470826 -44.03341549 71.50638666 -97.35037734
119.98524672 -138.44575074 152.75423157 -165.88049418 154.47072474
-150.22239348 145.44581679 -127.48432464 106.30213618 -78.15814356
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-107.03229395 116.13680659 -121.54753509 120.77824689 -116.13680659
107.71402831 -96.20477092 79.93400035 -60.48950447 37.9224322
-12.95637487 -13.10791142 38.59561147 -62.6758721 85.3283194
-105.16794834 121.34871544 -133.89020378 145.39546918 -135.39472263
131.67102913 -127.48432464 111.74094511 -93.17460164 68.50618579
-43.53225945 15.00211827]
[ -9.51375658 28.35981606 -46.0538423 62.38050004 -76.99164451
87.83913358 -95.31101401 99.75148412 -99.12014562 95.31101401
-88.39861851 78.95321507 -65.60013876 49.64245339 -31.12213581
10.63302206 10.75738489 -31.67459976 51.43675893 -70.02714198
86.30910468 -99.58831705 109.88085054 -119.32297784 111.11557725
-108.05962098 104.6236814 -91.70342372 76.46641941 -56.22157372
35.72600205 -12.31192028]
[ 7.0693589 -21.07324444 34.22109206 -46.35289322 57.20995302
-65.27036457 70.8224726 -74.12203957 73.65291273 -70.8224726
65.68609937 -58.66753144 48.74530062 -36.88767066 23.12583318
-7.90104818 -7.9934581 23.53635093 -38.22095996 52.03486078
-64.13345052 74.00079549 -81.64883784 88.66497138 -82.56632255
80.29554219 -77.74240875 68.14179118 -56.81967556 41.7763981
-26.54681442 9.1485821 ]
[ -4.35328971 12.97683985 -21.07324444 28.54397071 -35.22971513
40.19329204 -43.61226329 45.6441266 -45.35523972 43.61226329
-40.44930027 36.12728748 -30.01720791 22.71531543 -14.24081776
4.86544144 4.92234719 -14.49361334 23.53635093 -32.04290906
39.49318383 -45.56946487 50.27910609 -54.59961979 50.84409039
-49.44575074 47.87353893 -41.96150782 34.98938345 -25.72577893
16.34744761 -5.63366904]
[ 1.46992605 -4.38174259 7.11556386 -9.63811941 11.89561903
-13.57161385 14.72605916 -15.41213544 15.31459028 -14.72605916
13.65805725 -12.19869213 10.13557074 -7.67002338 4.8085357
-1.64285853 -1.66207325 4.89389432 -7.94725314 10.81956631
-13.33521624 15.38692526 -16.97717649 18.43603543 -17.16794834
16.69578682 -16.16491586 14.16866725 -11.81446896 8.6865325
-5.51985755 1.90225724]
最佳答案
他们都是对的。请注意，它是如此之小，以至于在这两种情况下它几乎为零(因为 e-13 部分)。
请参阅这篇关于浮点运算问题和限制的帖子:https://docs.python.org/3/tutorial/floatingpoint.html
关于python - 二维数组中所有元素的总和，不同的方法 - 不同的结果，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/51820019/
25
4
0
我基本上有三个表: hunt_c_usershunt_c_collected_eggshunt_c_achievements 我目前只使用 hunt_c_users 和 hunt_c_collecte
我已经计算了不同表中计数的总和。这会执行两次，每个 performanceID 一次。现在我想得到两个总和的总和。 下面是我目前做的两个总和的代码: SELECT SUM((COUNT (Bo
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所以，我需要聚合日期相同的行。 到目前为止，我的代码返回以下内容: date value source 0 2018-04-08 15:52:26.1
我有数字输入 数量约为 30 我需要将它们全部汇总到一个字段 我拥有的在下面 查看:
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实际上我的原始数据库为 SELECT sum(data1,data2) as database_value,sum(data3,data4) as database_not_value from t
我试图获取三个分数中每一个的值并将它们相加并显示在“总计:”中。我的问题是，我不知道如何做到这一点，以便每次其中一个分数值发生变化时，相应的总分值也会随之变化。 我可以在某处调用“onchange”来
如何获得按第一个值分组的元组列表中第二个和第三个值的总和？ 即: list_of_tuples = [(1, 3, 1), (1, 2, 4), (2, 1, 0), (2, 2, 0)] expec
我正在尝试将我的列表中的整数转换为列表的总和和平均值，并说明任何低于冰点 F<32 的温度。每当我尝试获取总和或平均值时，我都会收到错误提示“+: 'int' 和 'str' 不支持的操作数类型”。我
在我的 ios 项目中，我使用了两个实体 (CoreData):具有一对多关系的 Person 和 Gifts 我知道如何计算给一个人的礼物总和: NSDecimalNumber *orderSum=
我有两个表(输入和类别): CREATE TABLE categories ( iId INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT, sNam
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