I have a df with multiple columns. One of my column is extra_type. Now i want to create a new column based on the values of extra_type column. For example
extra_type
NaN
legbyes
wides
byes
Now i want to create a new column with 1 and 0 if extra_type is not equal to wide then 1 else 0
I tried like this
df1['ball_faced'] = df1[df1['extra_type'].apply(lambda x: 1 if [df1['extra_type']!= 'wides'] else 0)]
It not working this way.Any help on how to make this work is appreciated
expected output is like below
extra_type ball_faced
NaN 1
legbyes 1
wides 0
byes 1
Note that there's no need to use apply() or a lambda as in the original question, since comparison of a pandas Series and a string value can be done in a vectorized manner as follows:
df1['ball_faced'] = df1.extra_type.ne('wides').astype(int)
Output:
extra_type ball_faced
0 NaN 1
1 legbyes 1
2 wides 0
3 byes 1
Here are links to docs for ne() and astype().
For some useful insights on when to use apply (and when not to), see this SO question and its answers. TL;DR from the accepted answer: "If you're not sure whether you should be using apply, you probably shouldn't."
df['ball_faced'] = df.extra_type.apply(lambda x: x != 'wides').astype(int)
extra_type
ball_faced
0
NaN
1
1
legbyes
1
2
wides
0
3
byes
1
Related
I have the following problem. I have a dataframe which look like this.
Dataframe1
start end
0 0 2
1 3 7
2 8 9
and another dataframe which looks like this.
Dataframe2
data
1 ...
4 ...
8 ...
11 ...
What I am trying to achieve is following:
For each row in Dataframe1 I want to check if there is any index value in Dataframe2 which is in range(start, end) of Dataframe1.
If the condition is True, I want to create a new column["condition"] where the outcome is stored.
Since there is the possiblity to deal with large amounts of data I tried using numpy.select.
Like this:
range_start = df1.start
range_end = df1.end
condition = [
df2.index.to_series().between(range_start, range_end)
]
choice = ["True"]
df1["condition"] = np.select(condition, choice, default=0)
This gives me an error:
ValueError: Can only compare identically-labeled Series objects
I also tried a list comprehension. That didn't work either. All the things I tried are failing because I am dealing with a series (--> range_start, range_end). There has to be a way to make this work I think..
I already searched stackoverflow for this paricular problem. But I wasn't able to find a solution to this problem. It could be, that I'm just to inexperienced for this type of problem, to search for the right solution.
So maybe you can help me out here.
Thank you!
expected output:
start end condition
0 0 2 True
1 3 7 True
2 8 9 True
Use DataFrame.drop_duplicates for remove duplicates by both columns and index, create all combinations by DataFrame.merge with cross join and last test at least one match by GroupBy.any:
df3 = (df1.drop_duplicates(['start','end'])
.merge(df2.index.drop_duplicates().to_frame(), how='cross'))
df3['condition'] = df3[0].between(df3.start, df3.end)
df3 = df1.join(df3.groupby(['start','end'])['condition'].any(), on=['start','end'])
print (df3)
start end condition
0 0 2 True
1 3 7 True
2 8 9 True
If all pairs in df1 are unique is possible use:
df3 = (df1.merge(df2.index.to_frame(), how='cross'))
df3['condition'] = df3[0].between(df3.start, df3.end)
df3 = df3.groupby(['start','end'], as_index=False)['condition'].any()
print (df3)
start end condition
0 0 2 True
1 3 7 True
2 8 9 True
I can think of 2 ways of doing this:
Apply df.query to match each row, then collect the index of each result
Set the column domain to be the index, and then reorder based on the index (but this would lose the index which I want, so may be trickier)
However I'm not sure these are good solutions (I may be missing something obvious)
Here's an example set up:
domain_vals = list("ABCDEF")
df_domain_vals = list("DECAFB")
df_num_vals = [0,5,10,15,20,25]
df = pd.DataFrame.from_dict({"domain": df_domain_vals, "num": df_num_vals})
This gives df:
domain num
0 D 0
1 E 5
2 C 10
3 A 15
4 F 20
5 B 25
1: Use df.query on each row
So I want to reorder the rows according using the values in order of domain_vals for the column domain.
A possible way to do this is to repeatedly use df.query but this seems like an un-Pythonic (un-panda-ese?) solution:
>>> pd.concat([df.query(f"domain == '{d}'") for d in domain_vals])
domain num
3 A 15
5 B 25
2 C 10
0 D 0
1 E 5
4 F 20
2: Setting the column domain as the index
reorder = df.domain.apply(lambda x: domain_vals.index(x))
df_reorder = df.set_index(reorder)
df_reorder.sort_index(inplace=True)
df_reorder.index.name = None
Again this gives
>>> df_reorder
domain num
0 A 15
1 B 25
2 C 10
3 D 0
4 E 5
5 F 20
Can anyone suggest something better (in the sense of "less of a hack"). I understand that my solution works, I just don't think that calling pandas.concat along with a list comprehension is the right approach here.
Having said that, it's shorter than the 2nd option, so I presume there must be some equally simple way I can do this with pandas methods I've overlooked?
Another way is merge:
(pd.DataFrame({'domain':df_domain_vals})
.merge(df, on='domain', how='left')
)
I wanted to create DataFrame with 2 columns, one called 'id' , one called 'SalePrice'
submission = pd.DataFrame({'SalePrice':pre})
It looks like this
SalePrice
0 183242.025920
1 188796.451732
2 187878.763989
3 179789.672031
I know that I can name the index, but I need instead name it as a normal column name, on the same level as SalePrice. Anyone knows how to do that?
Try create it with DataFrame constructor
submission = pd.DataFrame({'SalePrice':pre,'id':np.arange(len(per))})
Just use reset_index, same as #Andy L. suggested. here's the full code:
submission = pd.DataFrame({'SalePrice':[1,2,3,4]}).reset_index()
submission.rename(columns = {'index':'id'}, inplace=True)
print(submission)
The output:
id SalePrice
0 0 1
1 1 2
2 2 3
3 3 4
I am fairly new to python pandas and cannot find the answer to my problem in any older posts.
I have a simple dataframe that looks something like that:
dfA ={'stop':[1,2,3,4,5,1610,1611,1612,1613,1614,2915,...]
'seq':[B, B, D, A, C, C, A, B, A, C, A,...] }
Now I want to merge the 'seq' values from each group, where the difference between the next and previous value in 'stop' is equal to 1. When the difference is high like 5 and 1610, that is where the next cluster begins and so on.
What I need is to write all values from each cluster into separate rows:
0 BBDAC #join'stop' cluster 1-5
1 CABAC #join'stop' cluster 1610-1614
2 A.... #join'stop' cluster 2015 - ...
etc...
What I am getting with my current code is like:
True BDACABAC...
False BCA...
for the entire huge dataframe.
I understand the logic behid the whay it merges it, which is meeting the condition (not perfect, loosing cluster edges) I specified, but I am running out of ideas if I can get it joined and split properly into clusters somehow, not all rows of the dataframe.
Please see my code below:
dfB = dfA.groupby((dfA.stop - dfA.stop.shift(1) == 1))['seq'].apply(lambda x: ''.join(x)).reset_index()
Please help.
P.S. I have also tried various combinations with diff() but that didn't help either. I am not sure if groupby is any good for this solution as well. Please advise!
dfC = dfA.groupby((dfA['stop'].diff(periods=1)))['seq'].apply(lambda x: ''.join(x)).reset_index()
This somehow splitted the dataframe into smaller chunks, cluster-like, but I am not understanding the legic behind the way it did it, and I know the result makes no sense and is not what I intended to get.
I think you need create helper Series for grouping:
g = dfA['stop'].diff().ne(1).cumsum()
dfC = dfA.groupby(g)['seq'].apply(''.join).reset_index()
print (dfC)
stop seq
0 1 BBDAC
1 2 CABAC
2 3 A
Details:
First get differences by diff:
print (dfA['stop'].diff())
0 NaN
1 1.0
2 1.0
3 1.0
4 1.0
5 1605.0
6 1.0
7 1.0
8 1.0
9 1.0
10 1301.0
Name: stop, dtype: float64
Compare by ne (!=) for first values of groups:
print (dfA['stop'].diff().ne(1))
0 True
1 False
2 False
3 False
4 False
5 True
6 False
7 False
8 False
9 False
10 True
Name: stop, dtype: bool
Asn last create groups by cumsum:
print (dfA['stop'].diff().ne(1).cumsum())
0 1
1 1
2 1
3 1
4 1
5 2
6 2
7 2
8 2
9 2
10 3
Name: stop, dtype: int32
I just figured it out.
I managed to round the values of 'stop' to a nearest 100 and assigned it as a new column.
Then my previous code is working....
Thank you so much for quick answer though.
dfA['new_val'] = (dfA['stop'] / 100).astype(int) *100
I have a dataframe that contains information that is linked by an ID column. The rows are sequential with the odd rows containing a "start-point" and the even rows containing an "end" point. My goal is to collapse the data from these into a single row with columns for "start" and "end" following each other. The rows do have a "packet ID" that would link them if the sequential nature of the dataframe is not consistent.
example:
df:
0 1 2 3 4 5
0 hs6 106956570 106956648 ID_A1 60 -
1 hs1 153649721 153649769 ID_A1 60 -
2 hs1 865130744 865130819 ID_A2 0 -
3 hs7 21882206 21882237 ID_A2 0 -
4 hs1 74230744 74230819 ID_A3 0 +
5 hs8 92041314 92041508 ID_A3 0 +
The resulting dataframe that I am trying to achieve is:
new_df
0 1 2 3 4 5
0 hs6 106956570 106956648 hs1 153649721 153649769
1 hs1 865130744 865130819 hs7 21882206 21882237
2 hs1 74230744 74230819 hs8 92041314 92041508
with each row containing the information on both the start and the end-point.
I have tried to pass the IDs in to an array and use a for loop to pull the information out of the original dataframe into a new dataframe but this has not worked. I was looking at the melt documentation which would suggest that pd.melt(df, id_vars=[3], value_vars=[0,1,2]) may work but I cannot see how to get the corresponding row in to positions new_df[3,4,5].
I think that it may be something really simple that I am missing but any suggestions would be appreciated.
You can try this:
df_out = df.set_index([df.index%2, df.index//2])[df.columns[:3]]\
.unstack(0).sort_index(level=1, axis=1)
df_out.columns = np.arange(len(df_out.columns))
df_out
Output:
0 1 2 3 4 5
0 hs6 106956570 106956648 hs1 153649721 153649769
1 hs1 865130744 865130819 hs7 21882206 21882237
2 hs1 74230744 74230819 hs8 92041314 92041508