PHP Imagick identifFormat () function

| |
Escape sequence list:Please follow the link to see the escape sequence list. Link: Below are some important inline formatting characters such as escape sequences for the corresponding image property:
  • % hcurrent image height in pixels
  • % i image file name (note: becomes the output filename for "info:")
  • % kCALCULATION: number of unique colors
  • % mimage file format (file magic)
  • % nnumber of images in the current image sequence
  • % w current width in pixels
  • % xx resolution (density)
  • % yy resolution (density)
  • % zimage depth (as shown, if not changed, image save depth)
  • % Up units Permissions
  • % @CALCULATED: trimming bounding box (no actual trimming), etc.
  • Syntax:
    string Imagick::identifyFormat ($embedText)
    Parameters:This function takes one parameter $embedText, which contains a string containing formatting sequences.Return Value:This function returns the image format or FALSE on error.The following program illustrates the Imagick function: : identifFormat() in PHP:Program:This program uses the Imagick::identifFormat() function to find the format of the image data.
    // Declaring a new Imagick object $imagick = new Imagick ( " " );
    // Save the string to a variable $embedText = "Output of’ Trim box:% @ number of unique colors:% k’ is: < br / > " ;
    // Use the Imagick::identifFormat() function to replace the built-in
    // format symbols with the appropriate image property $embedText . = $imagick -> identifyFormat ( "Trim box:% @ number of unique colors:% k" );
    // Show output echo $embedText ;
    Output of ’Trim box:% @ number of unique colors:% k’ is: Trim box: 656x144 + 5 + 15 number of unique colors: 2955
    Link: php

    PHP Imagick identifFormat () function: StackOverflow 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?

    Answer #1:

    >>> ["foo", "bar", "baz"].index("bar")

    Reference: Data Structures > More on Lists

    Caveats follow

    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 ValueError if 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

    An 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)
    >>> timeit.timeit("l.index(999_999, 999_990, 1_000_000)", setup="l = list(range(0, 1_000_000))", number=1000)

    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)
    >>> [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)
    >>> next(g)

    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

    1. Check for it first with item in my_list (clean, readable approach), or
    2. Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)

    Answer #2:

    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.

    Answer #3:

    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 enumerate():

    for i, j in enumerate(["foo", "bar", "baz"]):
        if j == "bar":

    The 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 enumerate():

    $ 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

    Answer #4:

    To get all indexes:

    indexes = [i for i,x in enumerate(xs) if x == "foo"]

    Answer #5:

    index() returns the first index of value!

    | index(...)
    | L.index(value, [start, [stop]]) -> integer -- return first index of value

    def all_indices(value, qlist):
        indices = []
        idx = -1
        while True:
                idx = qlist.index(value, idx+1)
            except ValueError:
        return indices
    all_indices("foo", ["foo";"bar";"baz";"foo"])

    PHP Imagick identifFormat () function: StackOverflow Questions

    InsecurePlatformWarning: A true SSLContext object is not available. This prevents urllib3 from configuring SSL appropriately

    Tried to perform REST GET through python requests with the following code and I got error.

    Code snip:

    import requests
    header = {"Authorization": "Bearer..."}
    url = az_base_url + az_subscription_id + "/resourcegroups/Default-Networking/resources?" + az_api_version
    r = requests.get(url, headers=header)


              InsecurePlatformWarning: A true SSLContext object is not available. 
              This prevents urllib3 from configuring SSL appropriately and may cause certain SSL connections to fail. 
              For more information, see

    My python version is 2.7.3. I tried to install urllib3 and requests[security] as some other thread suggests, I still got the same error.

    Wonder if anyone can provide some tips?

    Answer #1:

    The docs give a fair indicator of what"s required., however requests allow us to skip a few steps:

    You only need to install the security package extras (thanks @admdrew for pointing it out)

    $ pip install requests[security]

    or, install them directly:

    $ pip install pyopenssl ndg-httpsclient pyasn1

    Requests will then automatically inject pyopenssl into urllib3

    If you"re on ubuntu, you may run into trouble installing pyopenssl, you"ll need these dependencies:

    $ apt-get install libffi-dev libssl-dev

    Answer #2:

    If you are not able to upgrade your Python version to 2.7.9, and want to suppress warnings,

    you can downgrade your "requests" version to 2.5.3:

    pip install requests==2.5.3

    Bugfix disclosure / Warning introduced in 2.6.0

    Dynamic instantiation from string name of a class in dynamically imported module?

    In python, I have to instantiate certain class, knowing its name in a string, but this class "lives" in a dynamically imported module. An example follows:

    loader-class script:

    import sys
    class loader:
      def __init__(self, module_name, class_name): # both args are strings
          modul = sys.modules[module_name]
          instance = modul.class_name() # obviously this doesn"t works, here is my main problem!
        except ImportError:
           # manage import error

    some-dynamically-loaded-module script:

    class myName:
      # etc...

    I use this arrangement to make any dynamically-loaded-module to be used by the loader-class following certain predefined behaviours in the dyn-loaded-modules...

    Answer #1:

    You can use getattr

    getattr(module, class_name)

    to access the class. More complete code:

    module = __import__(module_name)
    class_ = getattr(module, class_name)
    instance = class_()

    As mentioned below, we may use importlib

    import importlib
    module = importlib.import_module(module_name)
    class_ = getattr(module, class_name)
    instance = class_()

    Answer #2:


    Import the root module with importlib.import_module and load the class by its name using getattr function:

    # Standard import
    import importlib
    # Load "module.submodule.MyClass"
    MyClass = getattr(importlib.import_module("module.submodule"), "MyClass")
    # Instantiate the class (pass arguments to the constructor, if needed)
    instance = MyClass()


    You probably don"t want to use __import__ to dynamically import a module by name, as it does not allow you to import submodules:

    >>> mod = __import__("os.path")
    >>> mod.join
    Traceback (most recent call last):
      File "<stdin>", line 1, in <module>
    AttributeError: "module" object has no attribute "join"

    Here is what the python doc says about __import__:

    Note: This is an advanced function that is not needed in everyday Python programming, unlike importlib.import_module().

    Instead, use the standard importlib module to dynamically import a module by name. With getattr you can then instantiate a class by its name:

    import importlib
    my_module = importlib.import_module("module.submodule")
    MyClass = getattr(my_module, "MyClass")
    instance = MyClass()

    You could also write:

    import importlib
    module_name, class_name = "module.submodule.MyClass".rsplit(".", 1)
    MyClass = getattr(importlib.import_module(module_name), class_name)
    instance = MyClass()

    This code is valid in python ‚â• 2.7 (including python 3).

    pandas loc vs. iloc vs. at vs. iat?

    Recently began branching out from my safe place (R) into Python and and am a bit confused by the cell localization/selection in Pandas. I"ve read the documentation but I"m struggling to understand the practical implications of the various localization/selection options.

    Is there a reason why I should ever use .loc or .iloc over at, and iat or vice versa? In what situations should I use which method?

    Note: future readers be aware that this question is old and was written before pandas v0.20 when there used to exist a function called .ix. This method was later split into two - loc and iloc - to make the explicit distinction between positional and label based indexing. Please beware that ix was discontinued due to inconsistent behavior and being hard to grok, and no longer exists in current versions of pandas (>= 1.0).

    Answer #1:

    loc: only work on index
    iloc: work on position
    at: get scalar values. It"s a very fast loc
    iat: Get scalar values. It"s a very fast iloc


    at and iat are meant to access a scalar, that is, a single element in the dataframe, while loc and iloc are ments to access several elements at the same time, potentially to perform vectorized operations.


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