我有一个由多个温度分布组成的数据集，我想创建一个平均温度分布。为了做到这一点，我想如果我能计算出每个深度的平均温度，我就可以简单地在一张图上绘制出每个深度的平均温度。
我知道为了计算某个深度的平均温度，我可以使用：
import numpy as np
import datetime as dt
import matplotlib.pyplot as plt
#read in data here
temper2out = []
temperature2m = np.mean(temperature[depth==2])
temper2out.extend(list(temperature2m)
T2out = np.array(temper2out)
t2 = np.mean(T2out)
tout.extend(list(t2)) #I could then do this for every depth, but since it goes to 150m, that would be lengthy.
Tout = np.array(tout)
我知道这能给我带来一些结果，但这是一个冗长的方法。我也不知道该怎么策划。所以我的问题是，有没有更好的方法来完成这个任务？如果是，怎么做？很抱歉问了这么多，我对Python很陌生！
数据采用此格式（抱歉，我不知道如何获取pastebin）：
TagID ProfileNo ProfileDirn DateTime Lat Lon Depth Temperature
pv42-156-12 1 A 25/01/2012 00:00 52.90618 0.36051 2 5.04
pv42-156-12 1 A 25/01/2012 00:00 52.90618 0.36051 3 5.0382
pv42-156-12 1 A 25/01/2012 00:00 52.90618 0.36051 4 5.0262
pv42-156-12 1 A 25/01/2012 00:00 52.90618 0.36051 5 5.0105
pv42-156-12 1 A 25/01/2012 00:00 52.90618 0.36051 6 4.993
pv42-156-12 1 A 25/01/2012 00:00 52.90618 0.36051 7 4.9866
pv42-156-12 1 A 25/01/2012 00:00 52.90618 0.36051 8 4.9709
pv42-156-12 1 A 25/01/2012 00:00 52.90618 0.36051 9 4.97
pv42-156-12 2 A 25/01/2012 00:00 52.90493 0.35768 2 4.98
pv42-156-12 2 A 25/01/2012 00:00 52.90493 0.35768 3 4.9783
pv42-156-12 2 A 25/01/2012 00:00 52.90493 0.35768 4 4.9739
pv42-156-12 2 A 25/01/2012 00:00 52.90493 0.35768 5 4.9733
pv42-156-12 2 A 25/01/2012 00:00 52.90493 0.35768 6 4.9733
pv42-156-12 2 A 25/01/2012 00:00 52.90493 0.35768 7 4.9717
pv42-156-12 2 A 25/01/2012 00:00 52.90493 0.35768 8 4.97
pv42-156-12 3 A 25/01/2012 00:00 52.89433 0.35407 2 5.05
pv42-156-12 3 A 25/01/2012 00:00 52.89433 0.35407 3 5.05
pv42-156-12 3 A 25/01/2012 00:00 52.89433 0.35407 6 5.05
pv42-156-12 3 A 25/01/2012 00:00 52.89433 0.35407 7 5.05
pv42-156-12 3 A 25/01/2012 00:00 52.89433 0.35407 8 5.05
pv42-156-12 4 A 25/01/2012 00:00 52.89037 0.35476 2 5.17
pv42-156-12 4 A 25/01/2012 00:00 52.89037 0.35476 3 5.1521
pv42-156-12 4 A 25/01/2012 00:00 52.89037 0.35476 4 5.1423
pv42-156-12 4 A 25/01/2012 00:00 52.89037 0.35476 5 5.1274
pv42-156-12 4 A 25/01/2012 00:00 52.89037 0.35476 6 5.1075
pv42-156-12 4 A 25/01/2012 00:00 52.89037 0.35476 7 5.1056
pv42-156-12 4 A 25/01/2012 00:00 52.89037 0.35476 8 5.0828
pv42-156-12 4 A 25/01/2012 00:00 52.89037 0.35476 9 5.05
pv42-156-12 5 A 25/01/2012 00:00 52.88979 0.35553 2 5.36
pv42-156-12 5 A 25/01/2012 00:00 52.88979 0.35553 3 5.3392
pv42-156-12 5 A 25/01/2012 00:00 52.88979 0.35553 4 5.3263
pv42-156-12 5 A 25/01/2012 00:00 52.88979 0.35553 5 5.2937
pv42-156-12 5 A 25/01/2012 00:00 52.88979 0.35553 6 5.2828
pv42-156-12 5 A 25/01/2012 00:00 52.88979 0.35553 7 5.2729
pv42-156-12 5 A 25/01/2012 00:00 52.88979 0.35553 9 5.26
pv42-156-12 6 A 25/01/2012 00:00 52.88875 0.35713 2 5.15
pv42-156-12 6 A 25/01/2012 00:00 52.88875 0.35713 3 5.147
pv42-156-12 6 A 25/01/2012 00:00 52.88875 0.35713 4 5.142
pv42-156-12 6 A 25/01/2012 00:00 52.88875 0.35713 5 5.103
pv42-156-12 6 A 25/01/2012 00:00 52.88875 0.35713 6 5.065
pv42-156-12 6 A 25/01/2012 00:00 52.88875 0.35713 7 5.053
pv42-156-12 6 A 25/01/2012 00:00 52.88875 0.35713 8 5.043
pv42-156-12 6 A 25/01/2012 00:00 52.88875 0.35713 9 5.04
pv42-156-12 7 A 25/01/2012 00:00 52.88661 0.35636 2 5.31
pv42-156-12 7 A 25/01/2012 00:00 52.88661 0.35636 3 5.32
pv42-156-12 7 A 25/01/2012 00:00 52.88661 0.35636 4 5.3229
pv42-156-12 7 A 25/01/2012 00:00 52.88661 0.35636 5 5.3243
pv42-156-12 7 A 25/01/2012 00:00 52.88661 0.35636 6 5.33
pv42-156-12 7 A 25/01/2012 00:00 52.88661 0.35636 8 5.33
pv42-156-12 8 A 25/01/2012 00:00 52.88602 0.35538 2 5.282
pv42-156-12 8 A 25/01/2012 00:00 52.88602 0.35538 3 5.28
pv42-156-12 8 A 25/01/2012 00:00 52.88602 0.35538 5 5.28
pv42-156-12 8 A 25/01/2012 00:00 52.88602 0.35538 6 5.284
pv42-156-12 8 A 25/01/2012 00:00 52.88602 0.35538 7 5.284
pv42-156-12 8 A 25/01/2012 00:00 52.88602 0.35538 8 5.29
pv42-156-12 9 A 25/01/2012 00:00 52.88558 0.35458 2 5.26
pv42-156-12 9 A 25/01/2012 00:00 52.88558 0.35458 3 5.2627
pv42-156-12 9 A 25/01/2012 00:00 52.88558 0.35458 4 5.2709
pv42-156-12 9 A 25/01/2012 00:00 52.88558 0.35458 5 5.2755
pv42-156-12 9 A 25/01/2012 00:00 52.88558 0.35458 6 5.2791
pv42-156-12 9 A 25/01/2012 00:00 52.88558 0.35458 7 5.28
pv42-156-12 9 A 25/01/2012 00:00 52.88558 0.35458 8 5.28
pv42-156-12 10 A 25/01/2012 00:00 52.88433 0.35418 2 5.26
pv42-156-12 10 A 25/01/2012 00:00 52.88433 0.35418 3 5.2648
pv42-156-12 10 A 25/01/2012 00:00 52.88433 0.35418 4 5.2648
pv42-156-12 10 A 25/01/2012 00:00 52.88433 0.35418 5 5.2705
pv42-156-12 10 A 25/01/2012 00:00 52.88433 0.35418 6 5.2743
pv42-156-12 10 A 25/01/2012 00:00 52.88433 0.35418 7 5.2752
pv42-156-12 10 A 25/01/2012 00:00 52.88433 0.35418 8 5.28
最佳答案
import numpy as np
import matplotlib.pyplot as plt
data = np.genfromtxt('data', delimiter='\t', dtype=None, names=True)
min_depth, max_depth = data['Depth'].min(), data['Depth'].max()
temps = data['Temperature']
depths = data['Depth']
avg_temp = []
ds = np.arange(min_depth, max_depth+1)
for d in ds:
mask = (depths==d)
t = temps[mask].mean()
avg_temp.append(t)
plt.plot(ds, avg_temp)
plt.show()
使用pandas，任务更加简单：
import pandas as pd
import matplotlib.pyplot as plt
data = pd.read_table('data', delim_whitespace=True)
avg_temp = data.groupby('Depth')['Temperature'].agg('mean')
avg_temp.plot()
plt.show()
关于python - 在matplotlib中的绘图上绘制均值，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/25742368/