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I"m trying to find the best way to format an sql query string. When I"m debugging my application I"d like to log to file all the sql query strings, and it is important that the string is properly formated.
def myquery(): sql = "select field1, field2, field3, field4 from table where condition1=1 and condition2=2" con = mymodule.get_connection() ...
- This is good for printing the sql string.
- It is not a good solution if the string is long and not fits the standard width of 80 characters.
def query(): sql = """ select field1, field2, field3, field4 from table where condition1=1 and condition2=2""" con = mymodule.get_connection() ...
Here the code is clear but when you print the sql query string you get all these annoying white spaces.
u" select field1, field2, field3, field4 _____from table ____where condition1=1 _____and condition2=2"
Note: I have replaced white spaces with underscore
_, because they are trimmed by the editor
def query(): sql = """select field1, field2, field3, field4 from table where condition1=1 and condition2=2""" con = mymodule.get_connection() ...
- I don"t like this option because it breaks the clearness of the well tabulated code.
def query(): sql = "select field1, field2, field3, field4 " "from table " "where condition1=1 " "and condition2=2 " con = mymodule.get_connection() ...
- I don"t like this option because all the extra typing in each line and is difficult to edit the query also.
For me the best solution would be Option 2 but I don"t like the extra whitespaces when I print the sql string.
Do you know of any other options?
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Python SQL query string formatting find: Questions
Finding the index of an item in a list
Given a list
["foo", "bar", "baz"] and an item in the list
"bar", how do I get its index (
1) in Python?
>>> ["foo", "bar", "baz"].index("bar") 1
Reference: Data Structures > More on Lists
Note that while this is perhaps the cleanest way to answer the question as asked,
index is a rather weak component of the
list API, and I can"t remember the last time I used it in anger. It"s been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about
list.index follow. It is probably worth initially taking a look at the documentation for it:
list.index(x[, start[, end]])
Return zero-based index in the list of the first item whose value is equal to x. Raises a
ValueErrorif there is no such item.
The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.
Linear time-complexity in list length
index call checks every element of the list in order, until it finds a match. If your list is long, and you don"t know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give
index a hint. For instance, in this snippet,
l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight
l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:
>>> import timeit >>> timeit.timeit("l.index(999_999)", setup="l = list(range(0, 1_000_000))", number=1000) 9.356267921015387 >>> timeit.timeit("l.index(999_999, 999_990, 1_000_000)", setup="l = list(range(0, 1_000_000))", number=1000) 0.0004404920036904514
Only returns the index of the first match to its argument
A call to
index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.
>>> [1, 1].index(1) 0 >>> [i for i, e in enumerate([1, 2, 1]) if e == 1] [0, 2] >>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1) >>> next(g) 0 >>> next(g) 2
Most places where I once would have used
index, I now use a list comprehension or generator expression because they"re more generalizable. So if you"re considering reaching for
index, take a look at these excellent Python features.
Throws if element not present in list
A call to
index results in a
ValueError if the item"s not present.
>>> [1, 1].index(2) Traceback (most recent call last): File "<stdin>", line 1, in <module> ValueError: 2 is not in list
If the item might not be present in the list, you should either
- Check for it first with
item in my_list(clean, readable approach), or
- Wrap the
indexcall in a
try/exceptblock which catches
ValueError(probably faster, at least when the list to search is long, and the item is usually present.)
One thing that is really helpful in learning Python is to use the interactive help function:
>>> help(["foo", "bar", "baz"]) Help on list object: class list(object) ... | | index(...) | L.index(value, [start, [stop]]) -> integer -- return first index of value |
which will often lead you to the method you are looking for.
The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use
for i, j in enumerate(["foo", "bar", "baz"]): if j == "bar": print(i)
index() function only returns the first occurrence, while
enumerate() returns all occurrences.
As a list comprehension:
[i for i, j in enumerate(["foo", "bar", "baz"]) if j == "bar"]
Here"s also another small solution with
itertools.count() (which is pretty much the same approach as enumerate):
from itertools import izip as zip, count # izip for maximum efficiency [i for i, j in zip(count(), ["foo", "bar", "baz"]) if j == "bar"]
This is more efficient for larger lists than using
$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ["foo", "bar", "baz"]*500) if j == "bar"]" 10000 loops, best of 3: 174 usec per loop $ python -m timeit "[i for i, j in enumerate(["foo", "bar", "baz"]*500) if j == "bar"]" 10000 loops, best of 3: 196 usec per loop
Python"s equivalent of && (logical-and) in an if-statement
Here"s my code:
def front_back(a, b): # +++your code here+++ if len(a) % 2 == 0 && len(b) % 2 == 0: return a[:(len(a)/2)] + b[:(len(b)/2)] + a[(len(a)/2):] + b[(len(b)/2):] else: #todo! Not yet done. :P return
I"m getting an error in the IF conditional.
What am I doing wrong?
You would want
and instead of
if foo == "abc" and bar == "bac" or zoo == "123": # do something
How do you get the logical xor of two variables in Python?
How do you get the logical xor of two variables in Python?
For example, I have two variables that I expect to be strings. I want to test that only one of them contains a True value (is not None or the empty string):
str1 = raw_input("Enter string one:") str2 = raw_input("Enter string two:") if logical_xor(str1, str2): print "ok" else: print "bad"
^ operator seems to be bitwise, and not defined on all objects:
>>> 1 ^ 1 0 >>> 2 ^ 1 3 >>> "abc" ^ "" Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: unsupported operand type(s) for ^: "str" and "str"
If you"re already normalizing the inputs to booleans, then != is xor.
bool(a) != bool(b)