How to plot points on graph with text at each point (python)

Question

I'm trying to replicate this:

I have a list of words, each of which has an x and y coordinate. I need to graph them just like the one above. What is the best way to do this? I know I could do something like...

y = [2.56422, 3.77284, 3.52623, 3.51468, 3.02199]
z = [0.15, 0.3, 0.45, 0.6, 0.75]
n = [hello, xyz, bbb, fasjd, boy]

fig, ax = plt.subplots()

for i, txt in enumerate(n):
    ax.annotate(txt, (z[i], y[i]))

But this doesn't seem too efficient for what I want to do. For each word, I have a function to pass it through to find the x coordinate and then another function to find its y coordinate. Maybe just a list of the words and then a function that loops through and plots each one as it goes through the list? Is this possible?

def plotWords(words):

    fig, ax = pyplot.subplots()

    for w in words:

        ax.annotate(w, code for x coordinate, code for y coordinate)

    return ax

Have you tried it this way, and if so are you running into efficiency issues? How many words are you trying to plot? This seems like a decent way to me. Since you tagged the question with pandas, a dataframe would also work nicely. — MattDMo
– MattDMo, Commented Aug 4, 2020 at 18:47
I haven't used the first method because I'll likely keep adding words to my list so I don't want to have to manually insert the x and y coordinates each time. I currently have this code: def plotWords(words, genderPC): fig, ax = pyplot.subplots() for w in words: ax.annotate(w, code for x coordinate, code for y coordinate) return ax but I'm getting an error. — curlypie99
– curlypie99, Commented Aug 4, 2020 at 20:26
The code formatting in comments is horrible so I added my current code to the original post. @MattDMo — curlypie99
– curlypie99, Commented Aug 4, 2020 at 20:28

r-beginners · Accepted Answer · 2020-08-05 01:57:41Z

1

I think you can use zip() to get each of them in a loop.

import matplotlib.pyplot as plt

y = [2.56422, 3.77284, 3.52623, 3.51468, 3.02199]
z = [0.15, 0.3, 0.45, 0.6, 0.75]
n = ['hello', 'xyz', 'bbb', 'fasjd', 'boy']

fig = plt.figure(figsize=(4,3),dpi=144)
ax = fig.add_subplot(111)

def plotWords(words,z,y):
    for w,xx,yy in zip(words,z,y):
        ax.annotate(w, xy=(xx,yy))
    ax.set_ylim(0,5)
    plt.show()
    return 

plotWords(n,z,y)

answered Aug 5, 2020 at 1:57

r-beginners

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Comments

gboffi · Accepted Answer · 2020-08-05 07:30:01Z

This figure

was produced running this script

import matplotlib.pyplot as plt 
import numpy as np 

# determine the sequence of random numbers 
np.random.seed(sum(ord(c) for c in 'gboffi')) 

# initialize the plotting
fig, ax = plt.subplots(figsize=(8,6), constrained_layout=1) 

# a list of long words 
N = 40 
data = '/usr/share/dict/italian' 
long_words = [w.strip() for w in open(data) if len(w)>12 and "'" not in w] 
words = np.random.choice(long_words, N) 

# let's compute the word characteristics
xs = np.random.random(N) + [len(w) for w in words] 
ys = np.random.random(N) + [sum(ord(c) for c in w)/len(w) for w in words] 
# determine the axes limits
xmn = min(xs) ; xmx = max(xs) ; dx = xmx-xmn 
ymn = min(ys) ; ymx = max(ys) ; dy = ymx-ymn 
ax.set_xlim(xmn-0.05*dx, xmx+0.05*dx) ; ax.set_ylim(ymn-0.05*dy, ymx+0.05*dx) 
# label the axes
plt.xlabel('Word Length', size='x-large')
plt.ylabel('Mean Value', size='x-large')

# for every word, plot a red "o" and place the text around it 
for w, x, y in zip(words, xs, ys): 
    ax.plot((x,),(y,), 'o', c='r') 
    ax.annotate(w, (x, y), ha='center', va='center')

# the dashed red lines, the position is a little arbitrary
plt.hlines(ymn+0.5*dy, xmn, xmx, ls='--', color='r') 
plt.vlines(xmn+0.4*dx, ymn, ymx, ls='--', color='r')                                                                

# remove everything except points, red lines and axes' labels
plt.box(False)
plt.tick_params(left=0, top=0, right=0, bottom=0,
                labelleft=0, labeltop=0, labelright=0, labelbottom=0) 
# we are done
plt.show()

@MattDMo First of all, thank you. Your comment made me curious and had me check: the OP has accepted an answer on every other question they asked w/o solicitation… who knows…

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