Let suppose there is an array of cells each containing a specific set of valid characters.
Valid characters are either "Null/No value", "*", "X", "0", "1".
For simplicity sake, lets assume the array is 3 rows and 3 columns
Col1
Col2
Col 3
X
NV
0
*
1
X
1
NV
1
Glist = [[X,,0],[*,1,X],[1,,1]] /*assuming this is the right method to represent the 3X3 from above
How do I use count() function of array to determine how many times a certain value exists in the array elements?
I tried Glist.count(1) and it returned a value of 0
Thanks in advance.
Related
I have a dataframe with columns: A(continuous variable) and B(discrete 1 or 0). The df is initially sorted by A variable.
I need to order the dataframe so for each set of X rows, there are Y rows with value 1 in B column, and (X-Y) rows with 0 (B column) (when possible!). But these sets should have variable A in desceding order. X and Y are input by the user
Example:
X=4, Y=3
Rows 0-11 are ok, since the sets (0-3),(4-7) and (8-11) has 3 rows with 1 in column B and only one row with 0 AND variable A is descending. However, rows 12-15 are not ok, since there are 2 rows with 1(variable B) and two with 0. Row 17 would replace row 15 to make this set valid. There is no problem if the last rows has 0 in variable B, since there isn't any with value 1.
The code should be general enough to run on dataframes with different number of rows.
Any ideas?
Having a dataset like this:
word
0 TBH46T
1 BBBB
2 5AAH
3 CAAH
4 AAB1
5 5556
Which would be the most efficient way to select the rows where column word is a combination of numbers and letters?
The output would be like this:
word
0 TBH46T
2 5AAH
4 AAB1
A possible solution would be to create a new column using apply and regex in which store if column word has the desired structure. But I'm curious about if this could be achieved in a more straightforward way.
Use Series.str.contains for chain mask for match numeric and for match non numeric with & for bitwise AND:
df = df[df['word'].str.contains('\d') & df['word'].str.contains('\D')]
print (df)
word
0 TBH46T
2 5AAH
4 AAB1
I have a pandas data frame,
Currently the list column is a string, I want to delimit this by spaces and replicate rows for each primary key would be associated with each item in the list. Can you please advise me on how I can achieve this?
Edit:
I need to copy down the value column after splitting and stacking the list column
If your data frame is df you can do:
df.List.str.split(' ').apply(pd.Series).stack()
and you will get
Primary Key
0 0 a
1 b
2 c
1 0 d
1 e
2 f
dtype: object
You are splitting the variable List on spaces, turning the resulting list into a series, and then stacking it to turn it into long format, indexed on the primary key, along with a sequence for each item obtained from the split.
My version:
df['List'].str.split().explode()
produces
0 a
0 b
0 c
1 d
1 e
1 f
With regards to the Edit of the question, the following tweak will give you want you need I think:
df['List'] = df['List'].str.split()
df.explode('List')
Here is a solution.
df = df.assign(**{'list':df['list'].str.split()}).explode('list')
df['cc'] = df.groupby(level=0)['list'].cumcount()
df.set_index(['cc'],append=True)
I am trying to iterate over rows in a Pandas Dataframe using the itertuples()-method, which works quite fine for my case. Now i want to check if a specific value ('x') is in a specific tuple. I used the count() method for that, as i need to use the number of occurences of x later.
The weird part is, for some Tuples that works just fine (i.e. in my case (namedtuple[7].count('x')) + (namedtuple[8].count('x')) ), but for some (i.e. namedtuple[9].count('x')) i get an AttributeError: 'int' object has no attribute 'count'
Would appreciate your help very much!
Apparently, some columns of your DataFrame are of object type (actually a string)
and some of them are of int type (more generally - numbers).
To count occurrences of x in each row, you should:
Apply a function to each row which:
checks whether the type of the current element is str,
if it is, return count('x'),
if not, return 0 (don't attempt to look for x in a number).
So far this function returns a Series, with a number of x in each column
(separately), so to compute the total for the whole row, this Series should
be summed.
Example of working code:
Test DataFrame:
C1 C2 C3
0 axxv bxy 10
1 vx cy 20
2 vv vx 30
Code:
for ind, row in df.iterrows():
print(ind, row.apply(lambda it:
it.count('x') if type(it).__name__ == 'str' else 0).sum())
(in my opinion, iterrows is more convenient here).
The result is:
0 3
1 1
2 1
So as you can see, it is possible to count occurrences of x,
even when some columns are not strings.
I have the following:
C1 C2 C3
0 0 0 1
1 0 0 1
2 0 0 1
And i would like to get the corresponding column index value that has 1's, so the result
should be "C3".
I know how to do this by transposing the dataframe and then getting the index values, but this is not ideal for data in the dataframes i have, and i wonder there might be a more efficient solution?
I will save the result in a list because otherwise there could be more than one column with values equal to 1. You can use DataFrame.loc
if all column values must be 1 then you can use:
df.loc[:,df.eq(1).all()].columns.tolist()
Output:
['C3']
if this isn't necessary then use:
df.loc[:,df.eq(1).any()].columns.tolist()
or as suggested #piRSquared, you can select directly from df.columns:
[*df.columns[df.eq(1).all()]]