pandas 在同一个图形 seaborn 上绘制多列
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plot multiple columns on same graph seaborn
提问by pinkvirus
I have a pandas dataFrame that I am plotting with seaborn:
我有一个我正在用 seaborn 绘制的 Pandas 数据框：
g = sns.FacetGrid(readCov, col='chr', col_wrap = 4, size=4)
g.map(plt.scatter, 'pos', 'bergC9', hue = edgecolor='white')
g.set(xlim= (0, 250000))
This works great and gives me a single graph for each 'chr' that is in the 'chr' column. However, I would like each graph to have multiple columns on it. Currently only one is displayed, the one called 'bergC9'. I want to put more columns on the same graph with different colors.
这很好用，并为“chr”列中的每个“chr”提供了一个图表。但是，我希望每个图表上都有多个列。目前只显示一个，一个叫做'bergC9'。我想在同一个图表上用不同的颜色放置更多的列。
Any ideas?
有任何想法吗？
Thanks!
谢谢！
edit: input data file
编辑：输入数据文件
chr description pos bergB7 bergC9 EvolB20
1 1 '"ID=PBANKA_010290;Name=PBANKA_010290;descript... 108389 0.785456 0.899275 0.803017
2 1 '"ID=PBANKA_010300;Name=PBANKA_010300;descript... 117894 1.070673 0.964203 0.989372
3 1 '"ID=PBANKA_010310;Name=PBANKA_010310;descript... 119281 1.031106 1.042189 0.883518
4 1 '"ID=PBANKA_010320;Name=PBANKA_010320;descript... 122082 0.880109 1.031673 1.026539
5 1 '"ID=PBANKA_010330;Name=PBANKA_010330;descript... 126075 0.948105 0.969198 0.849213
EDIT: I would like a scatterplot that has pos as the x-axis and bergB7, bergC9, EvolB20 etc, which are all 'strains' as the y-axis, thus several strains on the same graph. I was able to accomplish this by reformatting my data set so it now has a 'strain' parameter or column and concatenated all of the y data. Now I can use the hue syntax with 'strain'. I would like to not have to reformat all of my data sets. I thought that it may be possible to create a loop that would reference all the columns I want plotted, but I tried several syntaxes to no avail. There are other ways I've thought of to accomplish this, but these create new datasets and I know is not the way to go programmatically. I am a new user and would like to start out correctly.
编辑：我想要一个散点图，它以 pos 作为 x 轴和 bergB7、bergC9、EvolB20 等，它们都是“应变”作为 y 轴，因此在同一张图上有几个应变。我能够通过重新格式化我的数据集来实现这一点，因此它现在有一个“应变”参数或列，并连接了所有 y 数据。现在我可以使用带有“应变”的色调语法。我不想重新格式化我的所有数据集。我认为可以创建一个循环来引用我想要绘制的所有列，但我尝试了几种语法都无济于事。我还想过其他方法来实现这一点，但这些方法会创建新的数据集，我知道这不是以编程方式进行的方法。我是一个新用户，想正确开始。
This is what the output should look like (subset of 15 graph panel shown): (I cannot post the image because my 'reputation' is not high enough)
这就是输出的样子（显示了 15 个图形面板的子集）：（我无法发布图像，因为我的“声誉”不够高）
回答by cphlewis
Edited the data to have two chrcases. Should work for any number of "strains" columns. The data does need reformatting; from the seaborn documentation:
编辑数据有两种chr情况。应该适用于任意数量的“菌株”列。数据确实需要重新格式化；来自seaborn文档：
To use these features, your data has to be in a Pandas DataFrame and it must take the form of what Hadley Whickam calls “tidy” data. In brief, that means your dataframe should be structured such that each column is a variable and each row is an observation.
要使用这些功能，您的数据必须位于 Pandas DataFrame 中，并且必须采用 Hadley Whickam 所说的“整洁”数据的形式。简而言之，这意味着您的数据框的结构应该使每一列都是一个变量，每一行都是一个观察值。
but pandas does it easily:
但Pandas很容易做到：
import matplotlib.pyplot as plt
import pandas as pd
import seaborn as sns
readCov = pd.DataFrame([ (1, '\'"ID=PBANKA_010290;Name=PBANKA_010290;descript...', 108389, 0.785456, 0.899275, 0.803017),
(1, '\'"ID=PBANKA_010300;Name=PBANKA_010300;descript...', 117894, 1.070673, 0.964203, 0.9893719999999999),
(1, '\'"ID=PBANKA_010310;Name=PBANKA_010310;descript...', 119281, 1.0311059999999999, 1.042189, 0.883518),
(2, '\'"ID=PBANKA_010320;Name=PBANKA_010320;descript...', 122082, 0.880109, 1.031673, 1.0265389999999999),
(2, '\'"ID=PBANKA_010330;Name=PBANKA_010330;descript...', 126075, 0.948105, 0.969198, 0.8492129999999999)],
columns=[u'chr', u'description', u'pos', u'bergB7', u'bergC9', u'EvolB20'],
)
meltCov = pd.melt(readCov,id_vars=['chr','description','pos'], var_name='strain')
g = sns.FacetGrid(meltCov, col='chr', hue='strain')
g.map(plt.scatter, 'pos','value')
g.set_xticklabels(rotation=45)
g.add_legend()
#this plots a figure per script automatically
from os.path import realpath, basename
s = basename(realpath(__file__))
fig = plt.gcf()
fig.savefig(s.split('.')[0])
plt.show()
