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I'm trying to plot a DataFrame and I want to modify the markersize, but it seems like I cannot do it in the same plot() call.

ax0 = df_can_t.plot(kind='scatter', x='Year', y='China', 
figsize=(30,10), color = 'red', marker= '+', markersize = 14.0)

I am getting the error: AttributeError: Unknown property markersize.

However, it seems like 'markersize' is allowed (https://matplotlib.org/api/_as_gen/matplotlib.pyplot.plot.html) so I'm not sure what the problem is.

python matplotlib plot

share|improve this question

asked Mar 27 at 0:22

paseipasei

1466 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You are using scatter, not plot.
  
  – ImportanceOfBeingErnest
  Mar 27 at 0:34
  

  
  
  
  

add a comment | 

1

0

I'm trying to plot a DataFrame and I want to modify the markersize, but it seems like I cannot do it in the same plot() call.

ax0 = df_can_t.plot(kind='scatter', x='Year', y='China', 
figsize=(30,10), color = 'red', marker= '+', markersize = 14.0)

I am getting the error: AttributeError: Unknown property markersize.

However, it seems like 'markersize' is allowed (https://matplotlib.org/api/_as_gen/matplotlib.pyplot.plot.html) so I'm not sure what the problem is.

python matplotlib plot

share|improve this question

asked Mar 27 at 0:22

paseipasei

1466 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You are using scatter, not plot.
  
  – ImportanceOfBeingErnest
  Mar 27 at 0:34
  

  
  
  
  

add a comment | 

I'm trying to plot a DataFrame and I want to modify the markersize, but it seems like I cannot do it in the same plot() call.

ax0 = df_can_t.plot(kind='scatter', x='Year', y='China', 
figsize=(30,10), color = 'red', marker= '+', markersize = 14.0)

I am getting the error: AttributeError: Unknown property markersize.

However, it seems like 'markersize' is allowed (https://matplotlib.org/api/_as_gen/matplotlib.pyplot.plot.html) so I'm not sure what the problem is.

python matplotlib plot

share|improve this question

asked Mar 27 at 0:22

paseipasei

1466 bronze badges

ax0 = df_can_t.plot(kind='scatter', x='Year', y='China', 
figsize=(30,10), color = 'red', marker= '+', markersize = 14.0)

share|improve this question

asked Mar 27 at 0:22

paseipasei

1466 bronze badges

share|improve this question

asked Mar 27 at 0:22

paseipasei

1466 bronze badges

asked Mar 27 at 0:22

paseipasei

1466 bronze badges

asked Mar 27 at 0:22

paseipasei

1466 bronze badges

1 Answer
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0

With scatter plot, you the size parameter is just s, try:

ax0 = df_can_t.plot(kind='scatter', x='Year', y='China', 
figsize=(30,10), color = 'red', marker= '+', s = 14.0)

Here is a MVCE:

import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns

df = sns.load_dataset('iris')

ax = df.plot(kind='scatter', x='petal_width', y='sepal_width', s=10.0)

Output:

share|improve this answer

answered Mar 27 at 0:36

Scott BostonScott Boston

65.4k77 gold badges3939 silver badges6363 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  This works, thanks. I suppose pandas.pydata.org/pandas-docs/stable/reference/api/… says: "New in version 0.17.0: Each plot kind has a corresponding method on the DataFrame.plot accessor: df.plot(kind='line') is equivalent to df.plot.line()."
  
  – pasei
  Mar 27 at 1:24
  

  
  
  
  

add a comment | 

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0

With scatter plot, you the size parameter is just s, try:

ax0 = df_can_t.plot(kind='scatter', x='Year', y='China', 
figsize=(30,10), color = 'red', marker= '+', s = 14.0)

Here is a MVCE:

import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns

df = sns.load_dataset('iris')

ax = df.plot(kind='scatter', x='petal_width', y='sepal_width', s=10.0)

Output:

share|improve this answer

answered Mar 27 at 0:36

Scott BostonScott Boston

65.4k77 gold badges3939 silver badges6363 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  This works, thanks. I suppose pandas.pydata.org/pandas-docs/stable/reference/api/… says: "New in version 0.17.0: Each plot kind has a corresponding method on the DataFrame.plot accessor: df.plot(kind='line') is equivalent to df.plot.line()."
  
  – pasei
  Mar 27 at 1:24
  

  
  
  
  

add a comment | 

0

With scatter plot, you the size parameter is just s, try:

ax0 = df_can_t.plot(kind='scatter', x='Year', y='China', 
figsize=(30,10), color = 'red', marker= '+', s = 14.0)

Here is a MVCE:

import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns

df = sns.load_dataset('iris')

ax = df.plot(kind='scatter', x='petal_width', y='sepal_width', s=10.0)

Output:

share|improve this answer

answered Mar 27 at 0:36

Scott BostonScott Boston

65.4k77 gold badges3939 silver badges6363 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  This works, thanks. I suppose pandas.pydata.org/pandas-docs/stable/reference/api/… says: "New in version 0.17.0: Each plot kind has a corresponding method on the DataFrame.plot accessor: df.plot(kind='line') is equivalent to df.plot.line()."
  
  – pasei
  Mar 27 at 1:24
  

  
  
  
  

add a comment | 

With scatter plot, you the size parameter is just s, try:

ax0 = df_can_t.plot(kind='scatter', x='Year', y='China', 
figsize=(30,10), color = 'red', marker= '+', s = 14.0)

Here is a MVCE:

import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns

df = sns.load_dataset('iris')

ax = df.plot(kind='scatter', x='petal_width', y='sepal_width', s=10.0)

Output:

share|improve this answer

answered Mar 27 at 0:36

Scott BostonScott Boston

65.4k77 gold badges3939 silver badges6363 bronze badges

ax0 = df_can_t.plot(kind='scatter', x='Year', y='China', 
figsize=(30,10), color = 'red', marker= '+', s = 14.0)

import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns

df = sns.load_dataset('iris')

ax = df.plot(kind='scatter', x='petal_width', y='sepal_width', s=10.0)

share|improve this answer

answered Mar 27 at 0:36

Scott BostonScott Boston

65.4k77 gold badges3939 silver badges6363 bronze badges

answered Mar 27 at 0:36

Scott BostonScott Boston

65.4k77 gold badges3939 silver badges6363 bronze badges

answered Mar 27 at 0:36

Scott BostonScott Boston

65.4k77 gold badges3939 silver badges6363 bronze badges