首先，我是Python的新手，所以如果我只见树木不见森林，请原谅我。我的问题是读取一个巨大的 float 文件并将它们存储在数组中以进行快速数学后处理。
让我们假设该文件看起来与此类似:
!!
-3.2297390 0.4474691 3.5690145 3.5976372 6.9002712 7.7787466 14.2159269 14.3291490
16.7660723 17.1258704 18.9469059 19.1716808 20.0700721 21.4088414
-3.2045361 0.4123081 3.5625981 3.5936954 6.8901539 7.7543415 14.2764611 14.3623976
16.7955934 17.1560337 18.9527369 19.1251184 20.0700709 21.3515145
-3.2317597 0.4494166 3.5799182 3.6005429 6.8838705 7.7661897 14.2576455 14.3295731
16.7550357 17.0986678 19.0187779 19.1687722 20.0288587 21.3818250
-3.1921346 0.3949598 3.5636878 3.5892085 6.8833690 7.7404542 14.3061281 14.3855389
16.8063645 17.1697110 18.9549920 19.1134580 20.0613223 21.3196066
这里有 4 (nb) 个 block ，每个 block 有 14 (nk) 个 float 。我希望它们排列在数组 elements[nb][nk] 中，以便我可以轻松地循环访问 block 的某些 float 。
我认为它应该是这样的，但它根本不起作用:
nb=4
nk=14
with open("datafile") as file:
elements = []
n = 0
while '!!' not in file:
while n <= (nb-1):
elements.append([])
current = map(float,file.read().split()) # here I would need something to assure only 14 (nk) floats are read in
elements[n].append(current)
n += 1
print(elements[0][1])
如果有一些想法和建议，那就太好了。谢谢!
编辑:
这里是一个数据文件，其中数字紧随其后， block nb 之后没有明确的分隔符。这里nb=2，nk=160。如何在每个第 160 个数字之后将读取拆分为 float ？
!!
-7.2578105433 -7.2578105433 -6.7774609392 -6.7774609392 -6.3343986693 -6.3343986693 -5.8537216826 -5.8537216826
-5.6031029888 -5.6031029888 -2.9103190893 -2.9103190893 -1.7962279174 -1.7962279174 -0.8136720023 -0.8136720023
-0.1418500769 -0.1418500769 2.9923464558 2.9923464558 3.5797768050 3.5797768050 3.8793240270 3.8793240270
4.0774192689 4.0774192689 4.2378755781 4.2378755781 4.2707165126 4.2707165126 4.3290523910 4.3290523910
4.4487102661 4.4487102661 4.5341883539 4.5341883539 4.7946098470 4.7946098470 4.9518205998 4.9518205998
4.9592549825 4.9592549825 5.1648268937 5.1648268937 5.2372127454 5.2372127454 5.9377062691 5.9377062691
6.2971992823 6.2971992823 6.6324702419 6.6324702419 6.7948808733 6.7948808733 7.0835270703 7.0835270703
7.6252686579 7.6252686579 7.7886279100 7.7886279100 7.8514022664 7.8514022664 7.9188180854 7.9188180854
7.9661386138 7.9661386138 8.2830991934 8.2830991934 8.4581462733 8.4581462733 8.5537201519 8.5537201519
10.2738010533 10.2738010533 11.4495306517 11.4495306517 11.4819579346 11.4819579346 11.5788238984 11.5788238984
11.9411469341 11.9411469341 12.5006172267 12.5006172267 12.5055546075 12.5055546075 12.6659410418 12.6659410418
12.8741094000 12.8741094000 12.9560279595 12.9560279595 12.9780521671 12.9780521671 13.2195973082 13.2195973082
13.2339969658 13.2339969658 13.3594047155 13.3594047155 13.4530024795 13.4530024795 13.4556342387 13.4556342387
13.5784994631 13.5784994631 14.6887369915 14.6887369915 14.9019726334 14.9019726334 15.1279383300 15.1279383300
15.1953349879 15.1953349879 15.3209538297 15.3209538297 15.4042612992 15.4042612992 15.4528348692 15.4528348692
15.4542742538 15.4542742538 15.5291462589 15.5291462589 15.5415591416 15.5415591416 16.0741610117 16.0741610117
16.1117432607 16.1117432607 16.3566675522 16.3566675522 17.7569123657 17.7569123657 18.4416346230 18.4416346230
18.9525843134 18.9525843134 19.0591624486 19.0591624486 19.1069867477 19.1069867477 19.1853525353 19.1853525353
19.4020021909 19.4020021909 19.4718240723 19.4718240723 19.6384650104 19.6384650104 19.6919638323 19.6919638323
19.7044699790 19.7044699790 19.8851141335 19.8851141335 20.6132283388 20.6132283388 21.4074471478 21.4074471478
-7.2568288331 -7.2568280628 -6.7765483088 -6.7765429702 -6.3336003082 -6.3334841531 -5.8529872639 -5.8528369047
-5.6024822566 -5.6024743589 -2.9101060346 -2.9100930470 -1.7964872791 -1.7959333994 -0.8153333579 -0.8144924713
-0.1440078470 -0.1421444935 2.9869228390 2.9935342026 3.5661875018 3.5733148387 3.8777649741 3.8828300867
4.0569348321 4.0745074351 4.2152251981 4.2276050415 4.2620483420 4.2649182323 4.3401804124 4.3402590222
4.4446178512 4.4509411587 4.5139270348 4.5526439516 4.7788285567 4.7810706248 4.9282976775 4.9397807768
4.9737752749 4.9900180286 5.1456209436 5.1507667583 5.2528363215 5.2835144984 5.9252188817 5.9670441193
6.2699491148 6.3270140700 6.5912060019 6.6576016532 6.7976670773 6.7982056614 7.0789050974 7.1023337244
7.6182108739 7.6309688587 7.7678148773 7.7874194913 7.8544608005 7.8594983757 7.9019395451 7.9100447766
7.9872550937 7.9902791771 8.2617740182 8.3147140843 8.4533756827 8.4672364683 8.5556163680 8.5558640539
10.2756173692 10.2760227976 11.4344757209 11.4355375519 11.4737803653 11.4760186102 11.5914333288 11.5953932241
11.9369518613 11.9380900159 12.4973099542 12.5002401499 12.5030167542 12.5031963862 12.6629548222 12.6634150863
12.8719844312 12.8728126622 12.9541436501 12.9568445777 12.9762780998 12.9764840239 13.2074024551 13.2108294169
13.2279146175 13.2308902307 13.3780648962 13.3839050348 13.4634576072 13.4650575047 13.4701414823 13.4718238883
13.5901622459 13.5971076111 14.6735704782 14.6840793519 14.8963924604 14.8968395615 15.1163287408 15.1219631271
15.1791724308 15.1817299995 15.2628531102 15.3027136606 15.3755066968 15.3802521520 15.3969012144 15.4139294088
15.5131322524 15.5315039463 15.5465532500 15.5629105034 15.5927166831 15.5966393750 16.0841067052 16.0883417123
16.1224821534 16.1226510159 16.3646268213 16.3665839987 17.7654543366 17.7657216551 18.4305335335 18.4342292730
18.9110142692 18.9215889808 18.9821593138 18.9838270736 19.1633959849 19.1637558341 19.2040877093 19.2056062802
19.3760597529 19.3846323861 19.4323552578 19.4329488797 19.6494790293 19.6813374885 19.6943820824 19.7202356536
19.7381237231 19.7414645409 19.9056461663 19.9197428869 20.6239183178 20.6285756411 21.4127637743 21.4128909767
最佳答案
这应该有效:
elements = []
with open("datafile") as file:
next(file)
for line in file:
elements.append([float(x) for x in line.split()])
next(line) 读取第一行。然后 for line in file: 迭代所有其他行。列表理解 [float(x) for x in line.split()] 遍历行中由空格分隔的所有条目。最后，elements.append() 将此列表附加到 elements，该列表成为可称为二维数组的列表列表。
访问第一行中的第一个条目:
>>> elements[0][0]
-3.229739
或者最后一行的最后一个条目:
>>> elements[3][13]
21.319606
或者:
>>> elements[-1][-1]
21.319606
更新
这会将文件读取到列表列表中，而不将换行符视为特殊:
nb = 2
nk = 160
with open("datafile") as fobj:
all_values = iter(x for x in fobj.read().split())
next(all_values)
elements = []
for x in range(nb):
elements.append([float(next(all_values)) for counter in range(nk)])
如果您喜欢嵌套列表推导式:
with open("datafile") as fobj:
all_values = iter(x for x in fobj.read().split())
next(all_values)
elements = [[float(next(all_values)) for counter in range(nk)] for x in range(nb)]
关于python - 在二维数组中排列不同数量的 float ，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/33981578/