I have data frame same like below:
Name Rating Review Price
1 The park NaN NaN 5040
2 The Westin Good 7.6 NaN 6045
3 Courtyard NaN NaN 4850
4 Radisson Excellent 9.8 NaN 7050
5 Banjara Average 6.7 NaN 5820
6 Mindspace NaN NaN 8000
My required output is like this:
Name Review Rating Price
1 The park NaN NaN 5040
2 The Westin Good 7.6 6045
3 Courtyard NaN NaN 4850
4 Radisson Excellent 9.8 7050
5 Banjara Average 6.7 5820
6 Mindspace NaN NaN 8000
I use this split function:
df[["review","ratings"]] = df["rating"].str.split(expand=True)
But I got 'Columns must be same length as key' this type error.
How to split this type of data can anyone help me?
Problem is there is multiple space, not only one at least in one splitted value.
You can add n=1 for split after first space:
df[["review","ratings"]] = df["Rating"].str.split(expand=True, n=1)
Or use rsplit with n=1 for split by last space:
df[["review","ratings"]] = df["Rating"].str.rsplit(expand=True, n=1)
Another idea is use Series.str.extract with regex for get all values before space before float:
df[["review","ratings"]] = df["Rating"].str.extract('(.*)\s+(\d+\.\d+)')
I have a panda-series of the form
0 NaN
1 NaN
2 NaN
3 NaN
4 NaN
5 NaN
6 NaN
7 NaN
8 0.0
9 1.0
10 NaN
11 NaN
12 NaN
13. 0.0
...
The values can either be 0.0 or 1.0. From my knowledge of the data however, the 0's come in groups. Meaning, the entries 0-8 should be 0, then entries 9-12 should all be 1's, and then 13+ will be 0's. Therefore, the best way to impute the NaN's would be to do some kind of nearest neighbor I believe. However, it should return a 0 or 1 obviously and not an average value. Please let me know of anyway to do this!
I am new to data science and trying to solve a course exercise for movie recommender system and I want to drop the rows based on total count of values for columns for each row.
i.e.
if someone gave rating to too much movies he should be dropped to filter out the final results.
Though, I found a traditional way of doing it but I am not satisfied as it will be really helpful if someone would help me find a more pythonic way of solving the problem.
Here is the table named userRatings
title Zeus and Roxanne (1997) unknown Á köldum klaka (Cold Fever) (1994)
user_id
0 NaN NaN NaN
1 NaN 4.0 NaN
2 NaN NaN NaN
3 NaN NaN NaN
4 NaN NaN NaN
5 NaN 4.0 NaN
6 NaN NaN NaN
7 NaN NaN NaN
8 NaN NaN NaN
9 NaN NaN NaN
[10 rows x 1664 columns]
And here is the code i tried to solve the problem:
for index in userRatings.index:
if userRatings.loc[index].count() > 500:
userRatings = userRatings.drop(index)
I'm assuming you have a Pandas DataFrame... if so, one alternative would be something like this:
valid_rating_ixs = userRatings.sum(axis=1) <= 500
userRantings_cleaned = userRatings[valid_rating_ixs]
Note that in my code above and also your code, you may be including columns that are not ratings (e.g. user_id). Maybe you need to check that you are using only the relevant columns in your data frame
I created a dataframe df1:
df1 = pd.read_csv('FBK_var_conc_1.csv', names = ['Cycle', 'SQ'])
df1 = df1['SQ'].copy()
df1 = df1.to_frame()
df1.head(n=10)
SQ
0 2430.0
1 2870.0
2 2890.0
3 3270.0
4 3350.0
5 3520.0
6 26900.0
7 26300.0
8 28400.0
9 3230.0
And then created a second dataframe df2, that I want to fill with the row values of df 1:
df2 = pd.DataFrame()
for x in range(12):
y='Experiment %d' % (x+1)
df2[y]= df1.iloc[3*x:3*x+3]
df2
I get the column names from Experiment 1 - Experiment 12 in df2 and the first column i filled with the right values, but all following columns are filled with N/A.
> Experiment 1 Experiment 2 Experiment 3 Experiment 4 Experiment 5 Experiment 6 Experiment 7 Experiment 8 Experiment 9 Experiment
> 10 Experiment 11 Experiment 12
> 0 2430.0 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
> 1 2870.0 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
> 2 2890.0 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
I've been looking at this for the last 2 hours but can't figure out why the columns after column 1 aren't filled with values.
Desired output:
Experiment 1 Experiment 2 Experiment 3 Experiment 4 Experiment 5 Experiment 6 Experiment 7 Experiment 8 Experiment 9 Experiment 10 Experiment 11 Experiment 12
2430 3270 26900 3230 2940 243000 256000 249000 2880 26100 3890 33400
2870 3350 26300 3290 3180 242000 254000 250000 3390 27900 3730 30700
2890 3520 28400 3090 3140 253000 260000 237000 3510 27400 3760 29600
I found the issue.
I had to use .values
So the final line of the loop has to be:
df2[y] = df1.iloc[3*x:3*x+3].values
and I get the right output
(Editing to clarify my application, sorry for any confusion)
I run an experiment broken up into trials. Each trial can produce invalid data or valid data. When there is valid data the data take the form of a list of numbers which can be of zero length.
So an invalid trial produces None and a valid trial can produce [] or [1,2] etc etc.
Ideally, I'd like to be able to save this data as a frame_table (call it data). I have another table (call it trials) that is easily converted into a frame_table and which I use as a selector to extract rows (trials). I would then like to pull up by data using select_as_multiple.
Right now, I'm saving the data structure as a regular table as I'm using an object array. I realize folks are saying this is inefficient, but I can't think of an efficient way to handle the variable length nature of data.
I understand that I can use NaNs and make a (potentially very wide) table whose max width is the maximum length of my data array, but then I need a different mechanism to flag invalid trials. A row with all NaNs is confusing - does it mean that I had a zero length data trial or did I have an invalid trial?
I think there is no good solution to this using Pandas. The NaN solution leads me to potentially extremely wide tables and an additional column marking valid/invalid trials
If I used a database I would make the data a binary blob column. With Pandas my current working solution is to save data as an object array in a regular frame and load it all in and then pull out the relevant indexes based on my trials table.
This is slightly inefficient, since I'm reading my whole data table in one go, but it's the most workable/extendable scheme I have come up with.
But I welcome most enthusiastically a more canonical solution.
Thanks so much for all your time!
EDIT: Adding code (Jeff's suggestion)
import pandas as pd, numpy
mydata = [numpy.empty(n) for n in range(1,11)]
df = pd.DataFrame(mydata)
In [4]: df
Out[4]:
0
0 [1.28822975392e-231]
1 [1.28822975392e-231, -2.31584192385e+77]
2 [1.28822975392e-231, -1.49166823584e-154, 2.12...
3 [1.28822975392e-231, 1.2882298313e-231, 2.1259...
4 [1.28822975392e-231, 1.72723381477e-77, 2.1259...
5 [1.28822975392e-231, 1.49166823584e-154, 1.531...
6 [1.28822975392e-231, -2.68156174706e+154, 2.20...
7 [1.28822975392e-231, -2.68156174706e+154, 2.13...
8 [1.28822975392e-231, -1.3365130604e-315, 2.222...
9 [1.28822975392e-231, -1.33651054067e-315, 2.22...
In [5]: df.info()
<class 'pandas.core.frame.DataFrame'>
Int64Index: 10 entries, 0 to 9
Data columns (total 1 columns):
0 10 non-null values
dtypes: object(1)
df.to_hdf('test.h5','data')
--> OK
df.to_hdf('test.h5','data1',table=True)
--> ...
TypeError: Cannot serialize the column [0] because
its data contents are [mixed] object dtype
Here's a simple example along the lines of what you have described
In [17]: df = DataFrame(randn(10,10))
In [18]: df.iloc[5:10,7:9] = np.nan
In [19]: df.iloc[7:10,4:9] = np.nan
In [22]: df.iloc[7:10,-1] = np.nan
In [23]: df
Out[23]:
0 1 2 3 4 5 6 7 8 9
0 -1.671523 0.277972 -1.217315 -1.390472 0.944464 -0.699266 0.348579 0.635009 -0.330561 -0.121996
1 0.239482 -0.050869 0.488322 -0.668864 0.125534 -0.159154 1.092619 -0.638932 -0.091755 0.291824
2 0.432216 -1.101879 2.082755 -0.500450 0.750278 -1.960032 -0.688064 -0.674892 3.225115 1.035806
3 0.775353 -1.320165 -0.180931 0.342537 2.009530 0.913223 0.581071 -1.111551 1.118720 -0.081520
4 -0.255524 0.143255 -0.230755 -0.306252 0.748510 0.367886 -1.032118 0.232410 1.415674 -0.420789
5 -0.850601 0.273439 -0.272923 -1.248670 0.041129 0.506832 0.878972 NaN NaN 0.433333
6 -0.353375 -2.400167 -1.890439 -0.325065 -1.197721 -0.775417 0.504146 NaN NaN -0.635012
7 -0.241512 0.159100 0.223019 -0.750034 NaN NaN NaN NaN NaN NaN
8 -1.511968 -0.391903 0.257445 -1.642250 NaN NaN NaN NaN NaN NaN
9 -0.376762 0.977394 0.760578 0.964489 NaN NaN NaN NaN NaN NaN
In [24]: df['stop'] = df.apply(lambda x: x.last_valid_index(), 1)
In [25]: df
Out[25]:
0 1 2 3 4 5 6 7 8 9 stop
0 -1.671523 0.277972 -1.217315 -1.390472 0.944464 -0.699266 0.348579 0.635009 -0.330561 -0.121996 9
1 0.239482 -0.050869 0.488322 -0.668864 0.125534 -0.159154 1.092619 -0.638932 -0.091755 0.291824 9
2 0.432216 -1.101879 2.082755 -0.500450 0.750278 -1.960032 -0.688064 -0.674892 3.225115 1.035806 9
3 0.775353 -1.320165 -0.180931 0.342537 2.009530 0.913223 0.581071 -1.111551 1.118720 -0.081520 9
4 -0.255524 0.143255 -0.230755 -0.306252 0.748510 0.367886 -1.032118 0.232410 1.415674 -0.420789 9
5 -0.850601 0.273439 -0.272923 -1.248670 0.041129 0.506832 0.878972 NaN NaN 0.433333 9
6 -0.353375 -2.400167 -1.890439 -0.325065 -1.197721 -0.775417 0.504146 NaN NaN -0.635012 9
7 -0.241512 0.159100 0.223019 -0.750034 NaN NaN NaN NaN NaN NaN 3
8 -1.511968 -0.391903 0.257445 -1.642250 NaN NaN NaN NaN NaN NaN 3
9 -0.376762 0.977394 0.760578 0.964489 NaN NaN NaN NaN NaN NaN 3
Note that in 0.12 you should use table=True, rather than fmt (this is in the process of changing)
In [26]: df.to_hdf('test.h5','df',mode='w',fmt='t')
In [27]: pd.read_hdf('test.h5','df')
Out[27]:
0 1 2 3 4 5 6 7 8 9 stop
0 -1.671523 0.277972 -1.217315 -1.390472 0.944464 -0.699266 0.348579 0.635009 -0.330561 -0.121996 9
1 0.239482 -0.050869 0.488322 -0.668864 0.125534 -0.159154 1.092619 -0.638932 -0.091755 0.291824 9
2 0.432216 -1.101879 2.082755 -0.500450 0.750278 -1.960032 -0.688064 -0.674892 3.225115 1.035806 9
3 0.775353 -1.320165 -0.180931 0.342537 2.009530 0.913223 0.581071 -1.111551 1.118720 -0.081520 9
4 -0.255524 0.143255 -0.230755 -0.306252 0.748510 0.367886 -1.032118 0.232410 1.415674 -0.420789 9
5 -0.850601 0.273439 -0.272923 -1.248670 0.041129 0.506832 0.878972 NaN NaN 0.433333 9
6 -0.353375 -2.400167 -1.890439 -0.325065 -1.197721 -0.775417 0.504146 NaN NaN -0.635012 9
7 -0.241512 0.159100 0.223019 -0.750034 NaN NaN NaN NaN NaN NaN 3
8 -1.511968 -0.391903 0.257445 -1.642250 NaN NaN NaN NaN NaN NaN 3
9 -0.376762 0.977394 0.760578 0.964489 NaN NaN NaN NaN NaN NaN 3