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I"m looking for a way to update dict dictionary1 with the contents of dict update wihout overwriting levelA
dictionary1={"level1":{"level2":{"levelA":0,"levelB":1}}}
update={"level1":{"level2":{"levelB":10}}}
dictionary1.update(update)
print dictionary1
{"level1": {"level2": {"levelB": 10}}}
I know that update deletes the values in level2 because it"s updating the lowest key level1.
How could I tackle this, given that dictionary1 and update can have any length?
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Update value of a nested dictionary of varying depth __del__: Questions
How can I make a time delay in Python?
5 answers
I would like to know how to put a time delay in a Python script.
Answer #1
import time
time.sleep(5) # Delays for 5 seconds. You can also use a float value.
Here is another example where something is run approximately once a minute:
import time
while True:
print("This prints once a minute.")
time.sleep(60) # Delay for 1 minute (60 seconds).
Answer #2
You can use the sleep()
function in the time
module. It can take a float argument for sub-second resolution.
from time import sleep
sleep(0.1) # Time in seconds
Update value of a nested dictionary of varying depth __del__: Questions
How to delete a file or folder in Python?
5 answers
How do I delete a file or folder in Python?
Answer #1
os.remove()
removes a file.os.rmdir()
removes an empty directory.shutil.rmtree()
deletes a directory and all its contents.
Path
objects from the Python 3.4+ pathlib
module also expose these instance methods:
pathlib.Path.unlink()
removes a file or symbolic link.pathlib.Path.rmdir()
removes an empty directory.
Answer #2
os.remove()
removes a file.os.rmdir()
removes an empty directory.shutil.rmtree()
deletes a directory and all its contents.
Path
objects from the Python 3.4+ pathlib
module also expose these instance methods:
pathlib.Path.unlink()
removes a file or symbolic link.pathlib.Path.rmdir()
removes an empty directory.
Answer #3
Python syntax to delete a file
import os
os.remove("/tmp/<file_name>.txt")
Or
import os
os.unlink("/tmp/<file_name>.txt")
Or
pathlib Library for Python version >= 3.4
file_to_rem = pathlib.Path("/tmp/<file_name>.txt")
file_to_rem.unlink()
Path.unlink(missing_ok=False)
Unlink method used to remove the file or the symbolik link.
If missing_ok is false (the default), FileNotFoundError is raised if the path does not exist.
If missing_ok is true, FileNotFoundError exceptions will be ignored (same behavior as the POSIX rm -f command).
Changed in version 3.8: The missing_ok parameter was added.
Best practice
- First, check whether the file or folder exists or not then only delete that file. This can be achieved in two ways :
a.os.path.isfile("/path/to/file")
b. Useexception handling.
EXAMPLE for os.path.isfile
#!/usr/bin/python
import os
myfile="/tmp/foo.txt"
## If file exists, delete it ##
if os.path.isfile(myfile):
os.remove(myfile)
else: ## Show an error ##
print("Error: %s file not found" % myfile)
Exception Handling
#!/usr/bin/python
import os
## Get input ##
myfile= raw_input("Enter file name to delete: ")
## Try to delete the file ##
try:
os.remove(myfile)
except OSError as e: ## if failed, report it back to the user ##
print ("Error: %s - %s." % (e.filename, e.strerror))
RESPECTIVE OUTPUT
Enter file name to delete : demo.txt Error: demo.txt - No such file or directory. Enter file name to delete : rrr.txt Error: rrr.txt - Operation not permitted. Enter file name to delete : foo.txt
Python syntax to delete a folder
shutil.rmtree()
Example for shutil.rmtree()
#!/usr/bin/python
import os
import sys
import shutil
# Get directory name
mydir= raw_input("Enter directory name: ")
## Try to remove tree; if failed show an error using try...except on screen
try:
shutil.rmtree(mydir)
except OSError as e:
print ("Error: %s - %s." % (e.filename, e.strerror))
Is there a simple way to delete a list element by value?
5 answers
I want to remove a value from a list if it exists in the list (which it may not).
a = [1, 2, 3, 4]
b = a.index(6)
del a[b]
print(a)
The above case (in which it does not exist) shows the following error:
Traceback (most recent call last):
File "D:zjm_codea.py", line 6, in <module>
b = a.index(6)
ValueError: list.index(x): x not in list
So I have to do this:
a = [1, 2, 3, 4]
try:
b = a.index(6)
del a[b]
except:
pass
print(a)
But is there not a simpler way to do this?
Answer #1
To remove an element"s first occurrence in a list, simply use list.remove
:
>>> a = ["a", "b", "c", "d"]
>>> a.remove("b")
>>> print(a)
["a", "c", "d"]
Mind that it does not remove all occurrences of your element. Use a list comprehension for that.
>>> a = [10, 20, 30, 40, 20, 30, 40, 20, 70, 20]
>>> a = [x for x in a if x != 20]
>>> print(a)
[10, 30, 40, 30, 40, 70]
We hope this article has helped you to resolve the problem. Apart from Update value of a nested dictionary of varying depth, check other __del__-related topics.
Want to excel in Python? See our review of the best Python online courses 2023. If you are interested in Data Science, check also how to learn programming in R.
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Milan | 2023-04-01
Maybe there are another answers? What Update value of a nested dictionary of varying depth exactly means?. I am just not quite sure it is the best method
Prague | 2023-04-01
I was preparing for my coding interview, thanks for clarifying this - Update value of a nested dictionary of varying depth in Python is not the simplest one. I am just not quite sure it is the best method
Warsaw | 2023-04-01
Simply put and clear. Thank you for sharing. Update value of a nested dictionary of varying depth and other issues with __delete__ was always my weak point 😁. I just hope that will not emerge anymore