Resolving Dates on the X-Axis (Part 2)

Today I conclude with my solution for resolving dates as x-axis tick labels.

I think part of the confusion is that to this point, the x-coordinates of the points being plotted are equal to the x-axis tick labels. This need not be the case, though, and is really not even desired. I want to leave the tick labels as datetime so matplotlib can automatically scale it. This should also allow matplotlib to plot the x-values in the proper place.

Documentation on plt.xticks() reads:

The first segment suggests I can define the tick locations and tick labels with the first two arguments. For now, those are identical. Adding c as the first two arguments in L11 (see Part 1) gives this:

Ah ha! Can I now insert a subset as a different time range for the x-coordinates?

I think we’re onto something! I commented out the print lines in the interest of space.

Finally, let’s reformat the x-axis labels to something more readable and verify datatype:

Success! I am able to eliminate hours, minutes, and seconds. Interestingly, the axis labels now show up as string but matplotlib is still able to understand their values and plot the points correctly (I suspect the latter takes place before the former). Changing the date range on the axis helps because this graph should look different from the previous one.

To put in more object-oriented language:

I suspect the confusion between the plt and fig, ax approaches is widespread. For a better explanation, see here or here.

Resolving Dates on the X-Axis (Part 1)

Having previously discussed how to use np.linspace() to get evenly-spaced x-axis labels, my final challenge for this episode of “better understanding matplotlib’s plotting capability” is to do something similar with datetimes.

This will be a generalization of what I discussed in the last post and as mentioned in the fourth paragraph, articulation of exactly what I am trying to achieve is of the utmost importance.

I begin with the following code and a new method pd.date_range():

L5 generates a datetime index that I can convert to a list using the list() constructor (see output just above graph). Each element of the subsequent list is datatype pd.Timestamp, which is the pandas replacement for the Python datetime.datetime object. Observe that the first and second arguments are start date and end date, which are included in the Timestamp sequence. Also notice that the list has five elements, which is consistent with the third argument of pd.date_range().

Given a start date, end date, and n labels, this suggests I can generate (n – 1) evenly-spaced time intervals. Great start!

The enthusiasm fades when looking down at the graph, however. First, I get nine instead of five tick labels. Second, my desired format is yyyy-mm-dd as contained in L5. I do not know how/where the program makes either determination.

Another problem is that if I change the third argument (L5) to 15 to get more tick labels, a ValueError results: “x and y must have same first dimension, but have shapes (15,) and (5,).” That makes sense because I now have an unequal number of x- and y-coordinates. This date_range is really intended to be used only for tick labels and not as the source of x-coordinates. I may need to create a separate date_range (or make another list of x-coordinates) for plt.plot() and then create something customizable for evenly-spaced datetime tick labels.

I will continue next time.

Resolving the X-Axis (Part 2)

I left off last time with a promising solution for setting x-axis labels using the Matplotlib.Ticker.FixedLocator Class. Unfortunately, the example at the bottom shows this doesn’t work for all values, which calls the solution into question.

What’s going on? Take a look at the following code snippet:

This shows for equally-spaced tick labels having integer coordinates, only certain numbers of labels are possible: 2, 3, 4, 5, 7, 10, and 20. I did not get six because it’s not mathematically possible. The same holds true for 8-9 and 11-19. When multiple equally-spaced lists are possible, I was really aiming for the one with the last element closest to the final date in the list.

In order to code this stuff accurately, I need to articulate exactly what I’m trying to achieve. I failed to do that.

Aside from the FixedLocator Class, another way to approach this is with np.linspace(a, b, c). This automatically creates a linear space of c-point subdivisions between a and b inclusive (i.e. a and b always included as the first and last values):

Note how each list begins and ends with 0 (a) and 19 (b), respectively.

How do the plots look with different numbers of x-axis labels?

In the interest of space, I will describe rather than show the output. We get 20 subplots where the number of tick labels increases from zero to 19 by an increment of one for each subplot. The graphs are identical—the only thing that changes is the number of equally-spaced tick labels. Outstanding!

Some highlights of this code are as follows:

• The figure and axes are drawn in L8.
• L8 also includes the figsize argument to make the graphs larger (see second paragraph of Part 1).
• plt.sca(), as originally shown in L35 of this second code snippet, rotates x-axis labels for each subplot (a simple thing that took major work to figure out).
• L10 and L13 are basically applying the np.linspace() exercise shown above to the x-axis labels on the subplots.

I’m quite happy with the progress made here!

Resolving the X-Axis (Part 1)

As it turns out (see here and here), some of the matplotlib debugging came down to better understanding the zip() method. I still have some further considerations to resolve.

I would like to enlarge the graph so the axis isn’t so crowded when every label is included.

First though, I want the x-axis tick labels and locations to be handled automatically. I want z labels spaced evenly throughout the time interval from first Friday to last Friday. Alternatively, I may want to try plotting labels only where new trades begin.

When left to plot the x-axis tick labels automatically, others were seeing consistent tick labels on the 1st and 15th of each month as discussed in the third paragraph of Part 7. That would be acceptable, but for some unknown reason, I got asymmetric labels on the 1st and 22nd of each month as shown near the bottom here.

I stumbled upon the Matplotlib.ticker.FixedLocator Class, which is seen in L10 below:

The highlighted number is the number of tick labels that I expect to see. I determined this by trial and error (it requires the minus one). I want constant spacing across these labels and eventually, I’d like the program to calculate the optimal number.

Let’s break this down to see how it works (or not):

     > [x for x in range(len(a)) if x%((len(a)-1)//(5-1))== 0]

This is a pretty complicated piece of code for a beginner (me). First, we have to recognize it as a list comprehension: it will generate a list. A list will direct the program to place tick labels at specified locations as shown just above the first graph here.

The list will be generated as follows:

• len(a) is 20 (items in list a).
• range(20) creates a range object starting from 0 and going to 19.
• x for x in range(len(a)) will iterate over the range object and place each entry in the list…
• …if the subsequent condition is met.
• Subsequent condition involves the modulo operator (%), which calls for division remainder.
• Numerator is x.
• Denominator is (len(a)-1) // (5-1)
• // is floor division, which truncates any remainder.
• len(a)-1 // (5 – 1) = (20 – 1) // 4 = 19 / 4 = 4.75, which gets truncated to 4.
• Subsequent condition therefore calls for multiples of 4 (x / 4 will have remainder == 0).
• From 0 to 19, multiples of 4 are: 0, 4, 8, 12, and 16.
• These are exactly the dates shown as x-axis labels (count L3 – L5 dates with first being #0).

If I populate the highlighted number as 1, then I’ll get division by zero (not good). I’d never want just one tick label anyway. Two works along with 3, 4, and 5.

What about 6?

I count seven tick labels.

Houston, we have a problem.

Understanding the Python Zip() Method (Part 2)

Zip() returns an iterator. Last time, I discussed how elements may be unpacked by looping over the iterator. Today I will discuss element unpacking through assignment.

As shown in case [40] below, without the for loop each tuple may be assigned to a variable:

[37] shows that when assigned to one variable, the zip method transfers a zip object. Trying to assign to two or three variables does not work because zip(a, b, c) contains four tuples. As just mentioned, [40] works and if I print yp, m, and n, the other three tuples can be seen:

I got one response that reads:

     > But since you hand your zip iterables that all have 4 elements, your
     > zip iterator will also have 4 elements.

Regardless of the number of variables on the left, on the right I am handing zip three iterables with four elements each.

     > This means if you try to assign it to (xp, yp, m), it will complain
     > that 4 elements can’t fit into 3 variables.

This holds true for three and two variables as shown in [39] and [38], respectively, but not for one variable ([37]). Why?

Maybe it would help to press forward with [37]:

If assigned to one variable, the zip() object still needs to be unpacked (which may also be accomplished with a for loop). If assigned to four variables, each variable receives one 3-element tuple at once.

In figuring this out, I was missing the intermediate step in the [un]packing. zip(a, b, c) produces this series:

     (‘1-6-2017’, 265, ‘d’), (‘1-13-2017’, -10, ”), (‘1-20-2017’, 130, ‘d’), (‘1-27-2017’, 330, ”)
     or
     (a0, b0, c0), (a1, b1, c1), (a2, b2, c2), (a3, b3, c3)

xp, yp, m = zip(a, b, c) tries to unpack that series of four tuples into three variables. This does not fit and a ValueError results.

for xp, yp, m in zip(a, b, c) unpacks one tuple (ax, bx, cx) at a time into xp, yp and m.

Despite my confusion (I’m not alone as a Python beginner), zip() is always working the same. The difference is what gets unpacked: an entire sequence or one iteration of a sequence. zip(a, b, c) always generates a sequence of tuples (ax, bx, cx).

When unpacking in a for loop, one iteration of the sequence—a tuple—gets unpacked:

     xp, yp, m = (ax, bx, cx)

When unpacking outside a for loop, the entire sequence gets unpacked:

     xp, yp, m, n = ((a0, b0, c0), (a1, b1, c1), (a2, b2, c2), (a3, b3, c3))