如何使用 Python 从 Mathematica 的 .dat 文件中获取列表的导数？
How to take the derivative of list inside a .dat file from Mathematica using Python?
我有一个来自 Mathematica 的 .dat 文件。它有两列：第一列代表淬火前的相对化学势（$\theta_{0}$），第二列代表完成的平均功（$\braket{W}$）。
0.0001 1.968765727
0.03151592653589794 1.526403743
0.06293185307179587 1.315922249
0.0943477796076938 1.181680326
0.12576370614359172 1.092500991
0.15717963267948964 1.034055677
0.1885955592153876 0.9986171379
0.22001148575128554 0.9811207217
0.25142741228718346 0.9778559454
0.2828433388230814 0.9864639035
0.3142592653589793 1.005123779
0.3456751918948773 1.032472433
0.3770911184307752 1.06748456
0.4085070449666731 1.109396204
0.4399229715025711 1.157654853
0.471338898038469 1.211886853
0.5027548245743669 1.271876803
0.5341707511102649 1.337555877
0.5655866776461628 1.408997391
0.5970026041820607 1.486418909
0.6284185307179586 1.570190989
0.6598344572538566 1.660731115
0.6912503837897546 1.759140881
0.7226663103256524 1.865661911
0.7540822368615504 1.982382946
0.7854981633974484 2.110586591
0.8169140899333462 2.252795219
0.8483300164692442 2.410700136
0.8797459430051422 2.589771972
0.9111618695410401 2.794071187
0.942577796076938 3.033575331
0.9739937226128359 3.323391524
1.0054096491487339 3.699588638
1.0368255756846319 3.905513036
1.0682415022205298 3.365482708
1.0996574287564276 2.902485831
1.1310733552923256 2.502188583
1.1624892818282235 2.154786056
1.1939052083641215 1.851708306
1.2253211349000195 1.585798632
1.2567370614359172 1.351316619
1.2881529879718152 1.122552869
1.3195689145077132 0.9609250081
1.3509848410436112 0.7970216327
1.3824007675795091 0.649843702
1.4138166941154071 0.5169416572
1.4452326206513049 0.3961682839
1.4766485471872028 0.2855755599
1.5080644737231008 0.1836374567
1.5394804002589988 -0.0843703707
1.5708963267948968 0.0002668209559
1.6023122533307945 0.2516937775
1.6337281798666925 0.492273888
1.6651441064025905 0.7253692214
1.6965600329384884 0.9500449098
1.7279759594743864 1.166259793
1.7593918860102844 1.374218726
1.7908078125461822 1.573771245
1.8222237390820801 1.765135571
1.853639665617978 1.948242241
1.885055592153876 2.123273373
1.916471518689774 2.290564007
1.9478874452256718 2.450236891
1.9793033717615698 2.601330042
2.010719298297468 2.745413105
2.0421352248333657 2.881912668
2.073551151369264 2.986664867
2.1049670779051617 3.13340867
2.13638300444106 3.248402463
2.1677989309769576 3.355004915
2.1992148575128554 3.457560413
2.2306307840487536 3.55222602
2.2620467105846513 3.64009798
2.2934626371205495 3.722207193
2.3248785636564473 3.796860739
2.356294490192345 3.866975081
2.3877104167282432 3.931456933
2.419126343264141 3.989416888
2.450542269800039 4.038039032
2.481958196335937 4.089254811
2.5133741228718347 4.118176178
2.544790049407733 4.167857922
2.5762059759436307 4.200698856
2.607621902479529 4.22397227
2.6390378290154266 4.244919941
2.670453755551325 4.258992793
2.7018696820872226 4.267133062
2.7332856086231203 4.268776953
2.7647015351590185 4.263238875
2.7961174616949163 4.249339865
2.8275333882308145 4.225337816
2.858949314766712 4.18946278
2.89036524130261 4.139231773
2.921781167838508 4.070571301
2.953197094374406 3.978907532
2.984613020910304 3.856762428
3.016028947446202 3.693401012
3.0474448739820996 3.473508407
3.078860800517998 3.17101391
3.1102767270538956 2.735631484
当我在 Mathematica 中绘制此图时，我注意到 $\theta_{0} = 1.02678266592038$ 处有一个尖点。为了验证这是否真的是一个不连续点，我需要对 $\theta_{0}$ 求导数 $\braket{W}$。我已经在 Mathematica 中完成了，但我的结果图并不令人信服。因此，我想在 python 尝试一下。但是，我对如何开始它一无所知。任何建议都会有很大帮助。谢谢！
已编辑：
数学代码
a = Import['file.dat']
b = Interpolation[a, Method -> "Spline"]
c = b'
我不确定为什么您在检查导数时遇到问题。按照完全相同的方法，您指出，导数显示了不连续的位置。您可以在此处使用它来获取 Mathematica 格式的数据
Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["http://i.stack.imgur.com/lfXo1.png"]
然后
ip = Interpolation[data, InterpolationOrder -> 3, Method -> "Spline"];
Plot[{ip[x], .1 ip'[x]}, {x, 0.0001, 3.11}]
话虽这么说，你可以用
在这个区间内搜索一阶导数的根
FindRoot[ip'[x], {x, .8, 1.2}]
(* {x -> 1.03119} *)
我有一个来自 Mathematica 的 .dat 文件。它有两列：第一列代表淬火前的相对化学势（$\theta_{0}$），第二列代表完成的平均功（$\braket{W}$）。
0.0001 1.968765727
0.03151592653589794 1.526403743
0.06293185307179587 1.315922249
0.0943477796076938 1.181680326
0.12576370614359172 1.092500991
0.15717963267948964 1.034055677
0.1885955592153876 0.9986171379
0.22001148575128554 0.9811207217
0.25142741228718346 0.9778559454
0.2828433388230814 0.9864639035
0.3142592653589793 1.005123779
0.3456751918948773 1.032472433
0.3770911184307752 1.06748456
0.4085070449666731 1.109396204
0.4399229715025711 1.157654853
0.471338898038469 1.211886853
0.5027548245743669 1.271876803
0.5341707511102649 1.337555877
0.5655866776461628 1.408997391
0.5970026041820607 1.486418909
0.6284185307179586 1.570190989
0.6598344572538566 1.660731115
0.6912503837897546 1.759140881
0.7226663103256524 1.865661911
0.7540822368615504 1.982382946
0.7854981633974484 2.110586591
0.8169140899333462 2.252795219
0.8483300164692442 2.410700136
0.8797459430051422 2.589771972
0.9111618695410401 2.794071187
0.942577796076938 3.033575331
0.9739937226128359 3.323391524
1.0054096491487339 3.699588638
1.0368255756846319 3.905513036
1.0682415022205298 3.365482708
1.0996574287564276 2.902485831
1.1310733552923256 2.502188583
1.1624892818282235 2.154786056
1.1939052083641215 1.851708306
1.2253211349000195 1.585798632
1.2567370614359172 1.351316619
1.2881529879718152 1.122552869
1.3195689145077132 0.9609250081
1.3509848410436112 0.7970216327
1.3824007675795091 0.649843702
1.4138166941154071 0.5169416572
1.4452326206513049 0.3961682839
1.4766485471872028 0.2855755599
1.5080644737231008 0.1836374567
1.5394804002589988 -0.0843703707
1.5708963267948968 0.0002668209559
1.6023122533307945 0.2516937775
1.6337281798666925 0.492273888
1.6651441064025905 0.7253692214
1.6965600329384884 0.9500449098
1.7279759594743864 1.166259793
1.7593918860102844 1.374218726
1.7908078125461822 1.573771245
1.8222237390820801 1.765135571
1.853639665617978 1.948242241
1.885055592153876 2.123273373
1.916471518689774 2.290564007
1.9478874452256718 2.450236891
1.9793033717615698 2.601330042
2.010719298297468 2.745413105
2.0421352248333657 2.881912668
2.073551151369264 2.986664867
2.1049670779051617 3.13340867
2.13638300444106 3.248402463
2.1677989309769576 3.355004915
2.1992148575128554 3.457560413
2.2306307840487536 3.55222602
2.2620467105846513 3.64009798
2.2934626371205495 3.722207193
2.3248785636564473 3.796860739
2.356294490192345 3.866975081
2.3877104167282432 3.931456933
2.419126343264141 3.989416888
2.450542269800039 4.038039032
2.481958196335937 4.089254811
2.5133741228718347 4.118176178
2.544790049407733 4.167857922
2.5762059759436307 4.200698856
2.607621902479529 4.22397227
2.6390378290154266 4.244919941
2.670453755551325 4.258992793
2.7018696820872226 4.267133062
2.7332856086231203 4.268776953
2.7647015351590185 4.263238875
2.7961174616949163 4.249339865
2.8275333882308145 4.225337816
2.858949314766712 4.18946278
2.89036524130261 4.139231773
2.921781167838508 4.070571301
2.953197094374406 3.978907532
2.984613020910304 3.856762428
3.016028947446202 3.693401012
3.0474448739820996 3.473508407
3.078860800517998 3.17101391
3.1102767270538956 2.735631484
当我在 Mathematica 中绘制此图时，我注意到 $\theta_{0} = 1.02678266592038$ 处有一个尖点。为了验证这是否真的是一个不连续点，我需要对 $\theta_{0}$ 求导数 $\braket{W}$。我已经在 Mathematica 中完成了，但我的结果图并不令人信服。因此，我想在 python 尝试一下。但是，我对如何开始它一无所知。任何建议都会有很大帮助。谢谢！
已编辑： 数学代码
a = Import['file.dat']
b = Interpolation[a, Method -> "Spline"]
c = b'
我不确定为什么您在检查导数时遇到问题。按照完全相同的方法，您指出，导数显示了不连续的位置。您可以在此处使用它来获取 Mathematica 格式的数据
Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["http://i.stack.imgur.com/lfXo1.png"]
然后
ip = Interpolation[data, InterpolationOrder -> 3, Method -> "Spline"];
Plot[{ip[x], .1 ip'[x]}, {x, 0.0001, 3.11}]
话虽这么说，你可以用
在这个区间内搜索一阶导数的根FindRoot[ip'[x], {x, .8, 1.2}]
(* {x -> 1.03119} *)