How to plot single legend that reflects multiple dataframe columns-CodePudding

I have two dataframes (data_frame_EHGP and data_frame_P) and I am plotting particular columns from each dataframe, everything in the same plot.

Considering the way this script is written since I am plotting multiple columns, for each data frame column I am getting multiple legends that are named the same.

Hence, if I am plotting 8 columns [0:7] from data_frame_EHGP and 200 columns from data_frame_P, I am getting legend printed out 8 200 times.

However, I only need two legends, one for each dataframe ("ehgp" and "P"), for each dataframe, does anyone knows how to do this in python?

figure, ax1 = plt.subplots()
ax1.plot(data_frame_EHGP[data_frame_EHGP.columns[0:7]],linewidth=1.5,zorder=1, label = "Ehgp", color="blue")
ax1.plot(data_frame_P[data_frame_P.columns[0:199]],linewidth=1,zorder=1, label = "P", color="yellow")

Below is an example of the plot of what the current script does.

The image can be found on this link

Data_frame_EHGP looks like:

    ehgp1   ehgp2   ehgp3   ehgp4   ehgp5   ehgp6   ehgp7   ehgp8   XXX
1   30.965454   31.433349   31.482033   31.689665   33.065981   31.987559   33.408349   32.527560   1
2   33.319992   28.847623   29.991308   28.278939   30.812096   32.167360   30.207887   29.110781   2
3   27.356627   28.016364   30.542680   29.993470   25.756890   27.512679   30.443207   32.031628   3
4   23.705713   25.333608   32.280713   31.634398   24.919924   29.810451   32.001503   26.082818   4
5   28.828271   30.459324   26.619850   22.957481   28.356956   34.134850   27.756692   26.343534   5
... ... ... ... ... ... ... ... ... ...
96  9.840532    9.367901    13.405007   9.212901    -8.923153   11.692112   10.067901   14.210006   96
97  12.915509   9.823141    11.148668   12.318141   -11.677649  10.089457   10.269456   11.762878   97
98  11.065410   14.490674   13.061462   17.482779   -12.301433  11.971989   14.136989   14.175410   98
99  13.826624   17.742150   9.436623    14.984782   -14.872851  12.721097   16.495571   12.690045   99
100 13.942704   15.877967   14.007177   16.033756   -14.451508  11.845861   14.470861   13.294019   100
100 rows × 9 columns

Data_frame_P looks like:

1   2   3   4   5   6   7   8   9   10  ... 192 193 194 195 196 197 198 199 200 XX
1   20.752032   17.214033   20.302032   17.978034   19.380034   18.332033   19.402032   16.400032   20.128034   23.466032   ... -16.551969  -19.117968  -19.605969  -22.445969  -20.027969  -17.903970  -17.321969  -19.077969  -20.423969  1
2   21.503306   17.041307   23.497308   21.215307   21.333306   17.163307   18.467308   18.175307   21.843307   24.061306   ... -15.122694  -23.030695  -19.664695  -22.506695  -23.582695  -18.934695  -15.450695  -19.122695  -20.678695  2
3   20.386994   17.658996   17.042996   21.590996   20.998994   14.058995   21.136996   15.312994   18.652994   22.478997   ... -14.725006  -23.575007  -19.483007  -27.095007  -22.103007  -15.343006  -17.277006  -22.949007  -22.683006  3
4   22.158765   20.308766   21.002767   23.750766   22.440766   18.982765   22.838764   18.824765   22.724765   22.424765   ... -13.441236  -20.723237  -19.165237  -23.101237  -25.441237  -16.459236  -14.917236  -19.773237  -24.341236  4
5   26.705375   18.711377   19.263375   22.919375   25.139375   18.381375   24.049377   17.477378   21.927377   23.665376   ... -14.344625  -21.994626  -14.216626  -23.820626  -25.922626  -16.102626  -15.236626  -21.174625  -21.314627  5
... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
96  24.688743   14.496743   20.840744   20.830742   21.262743   22.444745   23.212745   20.756743   19.688743   23.150746   ... -22.777258  -24.005259  -23.477259  -21.527258  -22.073258  -25.427259  -20.115259  -18.627258  -17.843258  96
97  26.707827   21.461824   21.787825   23.241826   24.829827   20.645826   22.425824   22.789827   18.131826   19.291826   ... -22.628177  -24.978176  -19.260176  -22.386177  -23.166176  -20.462176  -20.732176  -18.514176  -16.210176  97
98  23.157565   18.359568   17.527569   23.669570   22.339569   20.177569   26.799568   20.569567   17.489569   22.203569   ... -23.004434  -24.984434  -20.528433  -23.332434  -24.332434  -21.524434  -23.466434  -18.642433  -14.750434  98
99  23.090204   19.956203   17.970203   20.172202   19.236204   17.598203   22.784205   18.728203   15.854203   20.784202   ... -24.207799  -24.209799  -21.853799  -19.377799  -23.559799  -19.995799  -23.305799  -16.113799  -13.083799  99
100 24.470229   20.590229   21.172231   20.026229   19.946229   18.876229   22.534232   24.462230   15.550228   20.956229   ... -19.001773  -24.141772  -21.431771  -17.037772  -25.311773  -13.935772  -23.385772  -20.547771  -18.421771  100
100 rows × 201 columns

CodePudding user response：

Try this:

figure, ax1 = plt.subplots()
l1 = ax1.plot(data_frame_EHGP.iloc[:, 0:7], linewidth=1.5, zorder=1, label="Ehgp", color="blue")
l2 = ax1.plot(data_frame_P.iloc[:, 0:199], linewidth=1, zorder=1, label="P", color="yellow")
ax1.legend(handles=[l1[0], l2[0]])

More context

You are plotting multiple lines and assigning single label / color to all of them in single ax1.plot() statement. If you take a closer look ax1.plot() statement returns list with all the line plots which you selected from given dataframe. It looks like this:

[<matplotlib.lines.Line2D at 0x136da4e50>,
 <matplotlib.lines.Line2D at 0x136da4fa0>,
 <matplotlib.lines.Line2D at 0x136daf100>,
 <matplotlib.lines.Line2D at 0x136daf220>,
 <matplotlib.lines.Line2D at 0x136daf340>,
 <matplotlib.lines.Line2D at 0x136daf460>,
 <matplotlib.lines.Line2D at 0x136daf580>]

If you want to have all these lines to have the same color, label and only one legend item, you can plot legend to only one of them (e.g. first) which is what I did in: ax1.legend(handles=[l1[0], l2[0]])