Skip to main content

Return to Answer

code highlighting
Source Link
edit approved Nov 5, 2016 at 19:37
DomTomCat
edit approved Nov 5, 2016 at 19:37
DomTomCat
  • 8.6k
  • 1
  • 54
  • 66

If the size of the circles corresponds to the square of the parameter in s=parameters=parameter, then assign a square root to each element you append to your size array, like this: s=[1, 1.414, 1.73, 2.0, 2.24]s=[1, 1.414, 1.73, 2.0, 2.24] such that when it takes these values and returns them, their relative size increase will be the square root of the squared progression, which returns a linear progression.

If I were to square each one as it gets output to the plot: output=[1, 2, 3, 4, 5]output=[1, 2, 3, 4, 5]. Try list interpretation: s=[numpy.sqrt(i) for i in s]s=[numpy.sqrt(i) for i in s]

If the size of the circles corresponds to the square of the parameter in s=parameter, then assign a square root to each element you append to your size array, like this: s=[1, 1.414, 1.73, 2.0, 2.24] such that when it takes these values and returns them, their relative size increase will be the square root of the squared progression, which returns a linear progression.

If I were to square each one as it gets output to the plot: output=[1, 2, 3, 4, 5]. Try list interpretation: s=[numpy.sqrt(i) for i in s]

If the size of the circles corresponds to the square of the parameter in s=parameter, then assign a square root to each element you append to your size array, like this: s=[1, 1.414, 1.73, 2.0, 2.24] such that when it takes these values and returns them, their relative size increase will be the square root of the squared progression, which returns a linear progression.

If I were to square each one as it gets output to the plot: output=[1, 2, 3, 4, 5]. Try list interpretation: s=[numpy.sqrt(i) for i in s]

deleted 21 characters in body
Source Link
Eric Leschinski
Eric Leschinski
  • 154.5k
  • 96
  • 422
  • 337

If the size of the circles corresponds to the square of the parameter in s=parameter, then assign a square root to each element you append to your size array, like this: s=[1, 1.414, 1.73, 2.0, 2.24] such that when it takes these values and returns them, their relative size increase will be the square root of the squared progression, which returns a linear progression.

If I were to square each one as it gets output to the plot: output=[1, 2, 3, 4, 5]. Try list interpretation: s=[numpy.sqrt(i) for i in s]

Hope this helps.

If the size of the circles corresponds to the square of the parameter in s=parameter, then assign a square root to each element you append to your size array, like this: s=[1, 1.414, 1.73, 2.0, 2.24] such that when it takes these values and returns them, their relative size increase will be the square root of the squared progression, which returns a linear progression.

If I were to square each one as it gets output to the plot: output=[1, 2, 3, 4, 5]. Try list interpretation: s=[numpy.sqrt(i) for i in s]

Hope this helps.

If the size of the circles corresponds to the square of the parameter in s=parameter, then assign a square root to each element you append to your size array, like this: s=[1, 1.414, 1.73, 2.0, 2.24] such that when it takes these values and returns them, their relative size increase will be the square root of the squared progression, which returns a linear progression.

If I were to square each one as it gets output to the plot: output=[1, 2, 3, 4, 5]. Try list interpretation: s=[numpy.sqrt(i) for i in s]

Source Link
user34028
user34028
  • 143
  • 7

If the size of the circles corresponds to the square of the parameter in s=parameter, then assign a square root to each element you append to your size array, like this: s=[1, 1.414, 1.73, 2.0, 2.24] such that when it takes these values and returns them, their relative size increase will be the square root of the squared progression, which returns a linear progression.

If I were to square each one as it gets output to the plot: output=[1, 2, 3, 4, 5]. Try list interpretation: s=[numpy.sqrt(i) for i in s]

Hope this helps.