matplotlib; quyền hạn phân số mười; ký hiệu khoa học

Question

Tôi xử lý dữ liệu mô phỏng và gần đây đã sử dụng matplotlib và đã gặp phải điều gì đó (một lỗi?) gây phiền toái.

Tôi đã cho phép matplotlib tự động đặt nhãn đánh dấu và loại của chúng (khoa học, v.v.) và với một số dữ liệu tôi nhận được nhãn ticker khoa học lạ.

Khi tìm kiếm giải pháp cho điều này, tôi thấy rằng bạn có thể gọi set_powerlimits ((n, m)) để đặt giới hạn dữ liệu sẽ được hiển thị bằng ký pháp khoa học. Nhưng tôi đã gặp phải vấn đề này (nếu tôi nhớ chính xác) với dữ liệu trải rộng một số đơn đặt hàng, thì dữ liệu của tôi ở khắp nơi, vì vậy tôi cần một giải pháp có lập trình, không phải là giải pháp khó. xem: http://matplotlib.org/api/ticker_api.html

Dưới đây tôi đã bao gồm dữ liệu mẫu, mã và ảnh chụp màn hình.

#! /usr/bin/env python 
from matplotlib import pyplot as plt 

data = [ 
[1.83186088e-08,0.03275], 
[1.07139009e-07,0.03275], 
[2.06376627e-07,0.03275], 
[3.03918517e-07,0.03275], 
[4.06032883e-07,0.03275], 
[5.01194017e-07,0.03275], 
[6.02195723e-07,0.03275], 
[7.03536925e-07,0.03275], 
[8.04625154e-07,0.03275], 
[9.06401951e-07,0.03275], 
[1.00041895e-06,0.03275], 
[1.10230745e-06,0.03275], 
[1.2042525e-06,0.03275], 
[1.30647822e-06,0.03275], 
[1.40109887e-06,0.03275], 
[1.50380097e-06,0.03275], 
[1.60683242e-06,0.03275], 
[1.70208505e-06,0.03275], 
[1.80545692e-06,0.03275], 
[1.90090648e-06,0.03275], 
[2.00453092e-06,0.03275], 
[2.10018627e-06,0.03275], 
[2.20401747e-06,0.03275], 
[2.30009359e-06,0.03275], 
[2.4043033e-06,0.03275], 
[2.50066449e-06,0.03275], 
[2.60513728e-06,0.03275], 
[2.70165405e-06,0.03275], 
[2.80635938e-06,0.03275], 
[2.90331342e-06,0.03275], 
[3.00021199e-06,0.03275], 
[3.10546819e-06,0.03275], 
[3.20257899e-06,0.03275], 
[3.30032923e-06,0.0327499999], 
[3.40612833e-06,0.0327499999], 
[3.50401732e-06,0.0327499997], 
[3.60153069e-06,0.0327499996], 
[3.70700708e-06,0.0327499993], 
[3.80456907e-06,0.0327499988], 
[3.90259984e-06,0.0327499982], 
[4.00084149e-06,0.0327499973], 
[4.10700266e-06,0.0327499959], 
[4.2047462e-06,0.0327499942], 
[4.30209468e-06,0.0327499918], 
[4.40018204e-06,0.0327499886], 
[4.50712875e-06,0.032749984], 
[4.60630591e-06,0.0327499785], 
[4.70519881e-06,0.0327499715], 
[4.80398305e-06,0.0327499628], 
[4.90251297e-06,0.0327499521], 
[5.00182752e-06,0.032749939], 
[5.10157551e-06,0.0327499232], 
[5.20157575e-06,0.0327499043], 
[5.30145192e-06,0.0327498822], 
[5.40127044e-06,0.0327498565], 
[5.500537e-06,0.0327498272], 
[5.60773155e-06,0.0327497911], 
[5.70660709e-06,0.0327497534], 
[5.80610521e-06,0.0327497112], 
[5.90651786e-06,0.0327496642], 
[6.00749437e-06,0.0327496124], 
[6.10822094e-06,0.0327495566], 
[6.20042255e-06,0.0327495018], 
[6.30049028e-06,0.0327494386], 
[6.40035803e-06,0.0327493715], 
[6.50035477e-06,0.0327493004], 
[6.60056805e-06,0.0327492251], 
[6.70029936e-06,0.0327491461], 
[6.80054193e-06,0.0327490625], 
[6.90130872e-06,0.0327489743], 
[7.00202598e-06,0.0327488818], 
[7.10217348e-06,0.0327487855], 
[7.20243015e-06,0.0327486847], 
[7.30199609e-06,0.0327485801], 
[7.40193254e-06,0.0327484707], 
[7.50188319e-06,0.0327483567], 
[7.60306205e-06,0.0327482367], 
[7.70357184e-06,0.0327481129], 
[7.80343389e-06,0.0327479853], 
[7.90330165e-06,0.0327478532], 
[8.00348513e-06,0.0327477162], 
[8.10167039e-06,0.0327475777], 
[8.206328e-06,0.0327474253], 
[8.3020567e-06,0.0327472819], 
[8.40527826e-06,0.0327471228], 
[8.50095898e-06,0.0327469714], 
[8.60536828e-06,0.0327468019], 
[8.70106059e-06,0.0327466426], 
[8.80396558e-06,0.032746467], 
[8.90727378e-06,0.0327462865], 
[9.00225164e-06,0.0327461166], 
[9.10359892e-06,0.0327459311], 
[9.20470894e-06,0.0327457418], 
[9.30582982e-06,0.0327455481], 
[9.40750123e-06,0.0327453488], 
[9.50134495e-06,0.0327451608], 
[9.60358199e-06,0.0327449513], 
[9.70705637e-06,0.0327447344], 
[9.80377546e-06,0.0327445269], 
[9.90091941e-06,0.032744314], 
] 

times=[] 
vals=[] 
for elem in data: 
    times.append(elem[0]) 
    vals.append(elem[1]) 

plt.plot(times,vals) 
plt.show() 

screen_shot

Answer 1

Đây là một known problem. Bạn nên phân tích dữ liệu theo cách thủ công cho các giới hạn của nó, như bạn đã thực hiện trong ảnh chụp màn hình và tự sử dụng ax.set_ylim(min, max) sau khi vẽ đồ thị. Bạn cũng có thể tắt bù đắp với:

import matplotlib.ticker as mticker 

# plot some stuff 
# ... 

y_formatter = mticker.ScalarFormatter(useOffset=False) 
ax.yaxis.set_major_formatter(y_formatter) 

Answer 2

Bạn có thể thử bằng cách sử dụng Formatter Kỹ thuật:

times=[] 
vals=[] 
for elem in data: 
    times.append(elem[0]) 
    vals.append(elem[1]) 

plt.plot(times,vals) 
plt.show() 

formatter = matplotlib.ticker.EngFormatter(unit='S', places=3) 
formatter.ENG_PREFIXES[-6] = 'u' 
plt.axes().yaxis.set_major_formatter(formatter) 

Mà sẽ trông như thế này:

Answer 3

Tôi nghĩ rằng bạn lựa chọn tốt nhất là để sử dụng trục logaritmic, nhưng nếu bạn cần tạo đồ họa bằng trục tuyến tính, bạn phải tự đặt giới hạn nguồn. Bạn có thể tính giới hạn công suất bằng cách sử dụng math.log10:

import math 
from matplotlib import ticker 

# Compute the span of the data 
pow_min = math.floor(math.log10(min(vals))) 
pow_max = math.ceil(math.log10(max(vals))) 

# Create a scalar formatter without offset, in order to have 
# the right exponent over the yaxis 
fmt = ticker.ScalarFormatter(useOffset=False) 
fmt.set_powerlimits((pow_min, pow_max)) 

fig = plt.figure() 
ax1 = fig.add_subplot(1, 1, 1) 
ax1.plot(times, vals) 
ax1.yaxis.set_major_formatter(fmt) # Set the formatter