scatter plots with string arrays in matplotlib

You could use np.unique(..., return_inverse=True) to get representative ints for each string. For example,

In [117]: uniques, X = np.unique(['foo', 'baz', 'bar', 'foo', 'baz', 'bar'], return_inverse=True)

In [118]: X
Out[118]: array([2, 1, 0, 2, 1, 0])

Note that X has dtype int32, as np.unique can handle at most 2**31 unique strings.

---

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import mpl_toolkits.mplot3d.axes3d as axes3d

N = 12
arr = np.arange(N*2).reshape(N,2)
words = np.array(['foo', 'bar', 'baz', 'quux', 'corge'])
df = pd.DataFrame(words[arr % 5], columns=list('XY'))
df['Z'] = np.linspace(1, 1000, N)
Z = np.log10(df['Z'])
Xuniques, X = np.unique(df['X'], return_inverse=True)
Yuniques, Y = np.unique(df['Y'], return_inverse=True)

fig = plt.figure()
ax = fig.add_subplot(1, 1, 1, projection='3d')
ax.scatter(X, Y, Z, s=20, c='b')
ax.set(xticks=range(len(Xuniques)), xticklabels=Xuniques,
       yticks=range(len(Yuniques)), yticklabels=Yuniques) 
plt.show()

Try converting the characters to numbers for the plotting and then use the characters again for the axis labels.

Using hash

You could use the hash function for the conversion;

from mpl_toolkits.mplot3d import Axes3D
xlab = myDataFrame.columnX.values
ylab = myDataFrame.columnY.values

X =[hash(l) for l in xlab] 
Y =[hash(l) for l in xlab] 

Z= myDataFrame.columnY.values #float

fig = figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(X, Y, np.log10(Z), s=20, c='b')
ax.set_xticks(X)
ax.set_xticklabels(xlab)
ax.set_yticks(Y)
ax.set_yticklabels(ylab)
show()

As M4rtini has pointed out in the comments, it't not clear what the spacing/scaling of string coordinates should be; the hash function could give unexpected spacings.

Nondegenerate uniform spacing

If you wanted to have the points uniformly spaced then you would have to use a different conversion. For example you could use

X =[i for i in range(len(xlab))]

though that would cause each point to have a unique x-position even if the label is the same, and the x and y points would be correlated if you used the same approach for Y.

Degenerate uniform spacing

A third alternative is to first get the unique members of xlab (using e.g. set) and then map each xlab to a position using the unique set for the mapping; e.g.

xmap = dict((sn, i)for i,sn in enumerate(set(xlab)))
X = [xmap[l] for l in xlab]

Scatter does this automatically now (from at least matplotlib 2.1.0):

plt.scatter(['A', 'B', 'B', 'C'], [0, 1, 2, 1])