diff options
Diffstat (limited to 'meta/recipes-devtools/python/python3-hypothesis')
3 files changed, 246 insertions, 0 deletions
diff --git a/meta/recipes-devtools/python/python3-hypothesis/run-ptest b/meta/recipes-devtools/python/python3-hypothesis/run-ptest new file mode 100644 index 0000000000..54f6e7930f --- /dev/null +++ b/meta/recipes-devtools/python/python3-hypothesis/run-ptest | |||
@@ -0,0 +1,10 @@ | |||
1 | #!/bin/sh | ||
2 | |||
3 | # Upstream "fast" tests take about 5 minutes and currently do not run cleanly | ||
4 | # (tests/cover and tests/pytest) | ||
5 | # https://github.com/HypothesisWorks/hypothesis/tree/master/hypothesis-python/tests | ||
6 | # https://github.com/HypothesisWorks/hypothesis/blob/master/hypothesis-python/scripts/basic-test.sh#L18 | ||
7 | # | ||
8 | # Instead we run two test suites imported from examples/ | ||
9 | |||
10 | pytest --automake | ||
diff --git a/meta/recipes-devtools/python/python3-hypothesis/test_binary_search.py b/meta/recipes-devtools/python/python3-hypothesis/test_binary_search.py new file mode 100644 index 0000000000..21267c4ac2 --- /dev/null +++ b/meta/recipes-devtools/python/python3-hypothesis/test_binary_search.py | |||
@@ -0,0 +1,135 @@ | |||
1 | # This file is part of Hypothesis, which may be found at | ||
2 | # https://github.com/HypothesisWorks/hypothesis/ | ||
3 | # | ||
4 | # Most of this work is copyright (C) 2013-2021 David R. MacIver | ||
5 | # (david@drmaciver.com), but it contains contributions by others. See | ||
6 | # CONTRIBUTING.rst for a full list of people who may hold copyright, and | ||
7 | # consult the git log if you need to determine who owns an individual | ||
8 | # contribution. | ||
9 | # | ||
10 | # This Source Code Form is subject to the terms of the Mozilla Public License, | ||
11 | # v. 2.0. If a copy of the MPL was not distributed with this file, You can | ||
12 | # obtain one at https://mozilla.org/MPL/2.0/. | ||
13 | # | ||
14 | # END HEADER | ||
15 | # | ||
16 | # SPDX-License-Identifier: MPL-2.0 | ||
17 | |||
18 | """This file demonstrates testing a binary search. | ||
19 | |||
20 | It's a useful example because the result of the binary search is so clearly | ||
21 | determined by the invariants it must satisfy, so we can simply test for those | ||
22 | invariants. | ||
23 | |||
24 | It also demonstrates the useful testing technique of testing how the answer | ||
25 | should change (or not) in response to movements in the underlying data. | ||
26 | """ | ||
27 | |||
28 | from hypothesis import given, strategies as st | ||
29 | |||
30 | |||
31 | def binary_search(ls, v): | ||
32 | """Take a list ls and a value v such that ls is sorted and v is comparable | ||
33 | with the elements of ls. | ||
34 | |||
35 | Return an index i such that 0 <= i <= len(v) with the properties: | ||
36 | |||
37 | 1. ls.insert(i, v) is sorted | ||
38 | 2. ls.insert(j, v) is not sorted for j < i | ||
39 | """ | ||
40 | # Without this check we will get an index error on the next line when the | ||
41 | # list is empty. | ||
42 | if not ls: | ||
43 | return 0 | ||
44 | |||
45 | # Without this check we will miss the case where the insertion point should | ||
46 | # be zero: The invariant we maintain in the next section is that lo is | ||
47 | # always strictly lower than the insertion point. | ||
48 | if v <= ls[0]: | ||
49 | return 0 | ||
50 | |||
51 | # Invariant: There is no insertion point i with i <= lo | ||
52 | lo = 0 | ||
53 | |||
54 | # Invariant: There is an insertion point i with i <= hi | ||
55 | hi = len(ls) | ||
56 | while lo + 1 < hi: | ||
57 | mid = (lo + hi) // 2 | ||
58 | if v > ls[mid]: | ||
59 | # Inserting v anywhere below mid would result in an unsorted list | ||
60 | # because it's > the value at mid. Therefore mid is a valid new lo | ||
61 | lo = mid | ||
62 | # Uncommenting the following lines will cause this to return a valid | ||
63 | # insertion point which is not always minimal. | ||
64 | # elif v == ls[mid]: | ||
65 | # return mid | ||
66 | else: | ||
67 | # Either v == ls[mid] in which case mid is a valid insertion point | ||
68 | # or v < ls[mid], in which case all valid insertion points must be | ||
69 | # < hi. Either way, mid is a valid new hi. | ||
70 | hi = mid | ||
71 | assert lo + 1 == hi | ||
72 | # We now know that there is a valid insertion point <= hi and there is no | ||
73 | # valid insertion point < hi because hi - 1 is lo. Therefore hi is the | ||
74 | # answer we were seeking | ||
75 | return hi | ||
76 | |||
77 | |||
78 | def is_sorted(ls): | ||
79 | """Is this list sorted?""" | ||
80 | for i in range(len(ls) - 1): | ||
81 | if ls[i] > ls[i + 1]: | ||
82 | return False | ||
83 | return True | ||
84 | |||
85 | |||
86 | Values = st.integers() | ||
87 | |||
88 | # We generate arbitrary lists and turn this into generating sorting lists | ||
89 | # by just sorting them. | ||
90 | SortedLists = st.lists(Values).map(sorted) | ||
91 | |||
92 | # We could also do it this way, but that would be a bad idea: | ||
93 | # SortedLists = st.lists(Values).filter(is_sorted) | ||
94 | # The problem is that Hypothesis will only generate long sorted lists with very | ||
95 | # low probability, so we are much better off post-processing values into the | ||
96 | # form we want than filtering them out. | ||
97 | |||
98 | |||
99 | @given(ls=SortedLists, v=Values) | ||
100 | def test_insert_is_sorted(ls, v): | ||
101 | """We test the first invariant: binary_search should return an index such | ||
102 | that inserting the value provided at that index would result in a sorted | ||
103 | set.""" | ||
104 | ls.insert(binary_search(ls, v), v) | ||
105 | assert is_sorted(ls) | ||
106 | |||
107 | |||
108 | @given(ls=SortedLists, v=Values) | ||
109 | def test_is_minimal(ls, v): | ||
110 | """We test the second invariant: binary_search should return an index such | ||
111 | that no smaller index is a valid insertion point for v.""" | ||
112 | for i in range(binary_search(ls, v)): | ||
113 | ls2 = list(ls) | ||
114 | ls2.insert(i, v) | ||
115 | assert not is_sorted(ls2) | ||
116 | |||
117 | |||
118 | @given(ls=SortedLists, v=Values) | ||
119 | def test_inserts_into_same_place_twice(ls, v): | ||
120 | """In this we test a *consequence* of the second invariant: When we insert | ||
121 | a value into a list twice, the insertion point should be the same both | ||
122 | times. This is because we know that v is > the previous element and == the | ||
123 | next element. | ||
124 | |||
125 | In theory if the former passes, this should always pass. In practice, | ||
126 | failures are detected by this test with much higher probability because it | ||
127 | deliberately puts the data into a shape that is likely to trigger a | ||
128 | failure. | ||
129 | |||
130 | This is an instance of a good general category of test: Testing how the | ||
131 | function moves in responses to changes in the underlying data. | ||
132 | """ | ||
133 | i = binary_search(ls, v) | ||
134 | ls.insert(i, v) | ||
135 | assert binary_search(ls, v) == i | ||
diff --git a/meta/recipes-devtools/python/python3-hypothesis/test_rle.py b/meta/recipes-devtools/python/python3-hypothesis/test_rle.py new file mode 100644 index 0000000000..4d618865ac --- /dev/null +++ b/meta/recipes-devtools/python/python3-hypothesis/test_rle.py | |||
@@ -0,0 +1,101 @@ | |||
1 | # This file is part of Hypothesis, which may be found at | ||
2 | # https://github.com/HypothesisWorks/hypothesis/ | ||
3 | # | ||
4 | # Most of this work is copyright (C) 2013-2021 David R. MacIver | ||
5 | # (david@drmaciver.com), but it contains contributions by others. See | ||
6 | # CONTRIBUTING.rst for a full list of people who may hold copyright, and | ||
7 | # consult the git log if you need to determine who owns an individual | ||
8 | # contribution. | ||
9 | # | ||
10 | # This Source Code Form is subject to the terms of the Mozilla Public License, | ||
11 | # v. 2.0. If a copy of the MPL was not distributed with this file, You can | ||
12 | # obtain one at https://mozilla.org/MPL/2.0/. | ||
13 | # | ||
14 | # END HEADER | ||
15 | # | ||
16 | # SPDX-License-Identifier: MPL-2.0 | ||
17 | |||
18 | """This example demonstrates testing a run length encoding scheme. That is, we | ||
19 | take a sequence and represent it by a shorter sequence where each 'run' of | ||
20 | consecutive equal elements is represented as a single element plus a count. So | ||
21 | e.g. | ||
22 | |||
23 | [1, 1, 1, 1, 2, 1] is represented as [[1, 4], [2, 1], [1, 1]] | ||
24 | |||
25 | This demonstrates the useful decode(encode(x)) == x invariant that is often | ||
26 | a fruitful source of testing with Hypothesis. | ||
27 | |||
28 | It also has an example of testing invariants in response to changes in the | ||
29 | underlying data. | ||
30 | """ | ||
31 | |||
32 | from hypothesis import assume, given, strategies as st | ||
33 | |||
34 | |||
35 | def run_length_encode(seq): | ||
36 | """Encode a sequence as a new run-length encoded sequence.""" | ||
37 | if not seq: | ||
38 | return [] | ||
39 | # By starting off the count at zero we simplify the iteration logic | ||
40 | # slightly. | ||
41 | result = [[seq[0], 0]] | ||
42 | for s in seq: | ||
43 | if ( | ||
44 | # If you uncomment this line this branch will be skipped and we'll | ||
45 | # always append a new run of length 1. Note which tests fail. | ||
46 | # False and | ||
47 | s | ||
48 | == result[-1][0] | ||
49 | # Try uncommenting this line and see what problems occur: | ||
50 | # and result[-1][-1] < 2 | ||
51 | ): | ||
52 | result[-1][1] += 1 | ||
53 | else: | ||
54 | result.append([s, 1]) | ||
55 | return result | ||
56 | |||
57 | |||
58 | def run_length_decode(seq): | ||
59 | """Take a previously encoded sequence and reconstruct the original from | ||
60 | it.""" | ||
61 | result = [] | ||
62 | for s, i in seq: | ||
63 | for _ in range(i): | ||
64 | result.append(s) | ||
65 | return result | ||
66 | |||
67 | |||
68 | # We use lists of a type that should have a relatively high duplication rate, | ||
69 | # otherwise we'd almost never get any runs. | ||
70 | Lists = st.lists(st.integers(0, 10)) | ||
71 | |||
72 | |||
73 | @given(Lists) | ||
74 | def test_decodes_to_starting_sequence(ls): | ||
75 | """If we encode a sequence and then decode the result, we should get the | ||
76 | original sequence back. | ||
77 | |||
78 | Otherwise we've done something very wrong. | ||
79 | """ | ||
80 | assert run_length_decode(run_length_encode(ls)) == ls | ||
81 | |||
82 | |||
83 | @given(Lists, st.data()) | ||
84 | def test_duplicating_an_element_does_not_increase_length(ls, data): | ||
85 | """The previous test could be passed by simply returning the input sequence | ||
86 | so we need something that tests the compression property of our encoding. | ||
87 | |||
88 | In this test we deliberately introduce or extend a run and assert | ||
89 | that this does not increase the length of our encoding, because they | ||
90 | should be part of the same run in the final result. | ||
91 | """ | ||
92 | # We use assume to get a valid index into the list. We could also have used | ||
93 | # e.g. flatmap, but this is relatively straightforward and will tend to | ||
94 | # perform better. | ||
95 | assume(ls) | ||
96 | i = data.draw(st.integers(0, len(ls) - 1)) | ||
97 | ls2 = list(ls) | ||
98 | # duplicating the value at i right next to it guarantees they are part of | ||
99 | # the same run in the resulting compression. | ||
100 | ls2.insert(i, ls2[i]) | ||
101 | assert len(run_length_encode(ls2)) == len(run_length_encode(ls)) | ||