CODEWITHSUNDEEP
pythondictionary
powerPixie
powerPixie

Reputation: 708

Extracting elements counting 0 in a python dictionary

This the dataframe code:

    import pandas as pd
    storage = pd.DataFrame({'red_wire':[2,2,8,1,5],'white_wire':[5,5,3,5,2],'black_wire':[0,10,4,2,1]})

And the dataframe structure with data:

            red_wire  white_wire  black_wire
    0         2           5           0
    1         2           5          10
    2         8           3           4
    3         1           5           2
    4         5           2           1

Consider we have two dictionaries. They have tags outside with the color of the wires that must be inside the box:

    box1 = {'red_wire':2,'white_wire':5,'black_wire':10}
    box2 = {'red_wire':2,'white_wire':5,'black_wire':0}

I made this code to print the boxes contents, my intention is to control and print the wire codes, some boxes may have distinct color tags. That's why I came out with this code:

    for key,value in box1.items():  #getting box tags and quantities

        col_name = [column for column in storage.columns if key == column] #columns in storage df

        for x,i in zip(storage.columns,storage.index): #accesing data in dataframe

            if x in col_name and value:

                print(x,col_name,value)

The output is :

red_wire ['red_wire'] 2
white_wire ['white_wire'] 5
black_wire ['black_wire'] 10

calling the box2:

for key,value in box2.items():

    col_name = [column for column in storage.columns if key == column]#columns in dict  

    for x,i in zip(storage.columns,storage.index):#accesing data in dataframe

        if x in col_name and value:

            print(x,col_name,value)

red_wire ['red_wire'] 2
white_wire ['white_wire'] 5

In box2 I was expecting to get black_wire ['black_wire'] 0, but it assumes it is a value to skip. I want to get the prints of the 0 articles in the dictionary evaluation and only skip values if the tag is not present on the box.

Some users provided me good solutions to get the 0 with the 'is not None' after the conditional sentence:

 if x in col_name and value is not None:

It prints:

    red_wire ['red_wire'] 2
    white_wire ['white_wire'] 5
    black_wire ['black_wire'] 0

But what if I ask specifically by 0:

for key,value in box2.items():

col_name = [column for column in storage.columns if key == column]#columns in dict  

for x,i in zip(storage.columns,storage.index):#accesing data in dataframe

    if x in col_name and value == storage[x][i]:

            print(x,col_name,value)

It prints:

red_wire ['red_wire'] 2
white_wire ['white_wire'] 5

"is not None", doesn't make any difference

#is not None 1st
if x in col_name and value is not None and value == storage[x][i]:
            print(x,col_name,value)


red_wire ['red_wire'] 2
white_wire ['white_wire'] 5

#is not None after
if x in col_name and value == storage[x][i] and value is not None:

            print(x,col_name,value)


red_wire ['red_wire'] 2
white_wire ['white_wire'] 5

Upvotes: 2

Views: 130

Answers (1)

Bsquare ℬℬ
Bsquare ℬℬ

Reputation: 4487

You have 2 issues in your source code.

0 is a valid value

In your first version, you used value in your conditional, and like said in comments, bool(0) == false, so you add to specify is not None to manage the fact there is a 0 value, and there is NO value at all.

Columns and values indexes mixing

In your final version, you are iterating on columns, using the Panda indexing system, called i in your loop:

for x,i in zip(storage.columns,storage.index):#accesing data in dataframe

But you use this index to seek the value in corresponding Dataframe:

value == storage[x][i]

You are lucky it works for red_wire, and white_wire, and even black_wire for the box1.

Maybe you want to ensure the value belongs to the corresponding Dataframe part; in which case you can change your source code this way:

value in storage[x].values

This way you don't need at all to check if value if not None; and you don't need either to iterate in index; your source code could be simplified to:

for x in storage.columns: if x in col_name and value in storage[x].values:

Let me know if it reaches you needs.

Upvotes: 2

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