I have this DataFrame and want only the records whose "Total" column is not NaN ,and records when A~E has more than two NaN:
A B C D E Total
1 1 3 5 5 8
1 4 3 5 5 NaN
3 6 NaN NaN NaN 6
2 2 5 9 NaN 8
..i.e. something like df.dropna(....) to get this resulting dataframe:
A B C D E Total
1 1 3 5 5 8
2 2 5 9 NaN 8
Here's my code
import pandas as pd
dfInputData = pd.read_csv(path)
dfInputData = dfInputData.dropna(axis=1,how = 'any')
RowCnt = dfInputData.shape[0]
But it looks like no modification has been made even error
Please help!! Thanks
Use boolean indexing with count all columns without Total for number of missing values and not misisng values in Total:
df = df[df.drop('Total', axis=1).isna().sum(axis=1).le(2) & df['Total'].notna()]
print (df)
A B C D E Total
0 1 1 3.0 5.0 5.0 8.0
3 2 2 5.0 9.0 NaN 8.0
Or filter columns between A:E:
df = df[df.loc[:, 'A':'E'].isna().sum(axis=1).le(2) & df['Total'].notna()]
print (df)
A B C D E Total
0 1 1 3.0 5.0 5.0 8.0
3 2 2 5.0 9.0 NaN 8.0
Given that, i have a dataset as below:
dict = {
"A": [math.nan,math.nan,1,math.nan,2,math.nan,3,5],
"B": np.random.randint(1,5,size=8)
}
dt = pd.DataFrame(dict)
My favorite output is, if the in column A we have an Nan then multiply the value of the column B in the same row and replace it with Nan. So, given that, the below is my dataset:
A B
NaN 1
NaN 1
1.0 3
NaN 2
2.0 3
NaN 1
3.0 1
5.0 3
My favorite output is:
A B
2 1
2 1
1 3
4 2
2 3
2 1
3 1
5 3
My current solution is as below which does not work:
dt[pd.isna(dt["A"])]["A"] = dt[pd.isna(dt["A"])]["B"].apply( lambda x:2*x )
print(dt)
In your case with fillna
df.A.fillna(df.B*2, inplace=True)
df
A B
0 2.0 1
1 2.0 1
2 1.0 3
3 4.0 2
4 2.0 3
5 2.0 1
6 3.0 1
7 5.0 3
I've got 3 Dataframes I would like to merge or join by "label" and then being able to compare all columns
Examples of df are below:
df1
Label,col1,col2,col3
NF1,1,1,6
NF2,3,2,8
NF3,4,5,4
NF4,5,7,2
NF5,6,2,2
df2
Label,col1,col1,col3
NF1,8,4,5
NF2,4,7,8
NF3,9,7,8
df3
Label,col1,col1,col3
NF1,2,8,8
NF2,6,2,0
NF3,2,2,5
NF4,2,4,9
NF5,2,5,8
and what ill like to see is similar to
Label,df1_col1,df2_col1,df_col1,df1_col2,df2_col2,df3_col2,df1_col3,df2_col3,df_col3
NF1,1,8,2,1,4,8,6,5,8
NF2,3,4,6,2,7,2,8,8,0
NF3,4,9,2,5,7,2,4,8,5
NF4,5,,2,7,,4,2,,9
NF5,6,,2,2,,5,2,,8
but I'm to suggestions on how to make the comparisons more readable.
Thanks!
Use concat with list of DataFrames, add parameter keys for prefixes and sorting by columns names:
dfs = [df1, df2, df3]
k = ('df1','df2','df3')
df = (pd.concat([x.set_index('Label') for x in dfs], axis=1, keys=k)
.sort_index(axis=1, level=1)
.rename_axis('Label')
.reset_index())
df.columns = df.columns.map('_'.join).str.strip('_')
print (df)
Label df1_col1 df2_col1 df3_col1 df2_col1.1 df3_col1.1 df1_col2 \
0 NF1 1 8.0 2 4.0 8 1
1 NF2 3 4.0 6 7.0 2 2
2 NF3 4 9.0 2 7.0 2 5
3 NF4 5 NaN 2 NaN 4 7
4 NF5 6 NaN 2 NaN 5 2
df1_col3 df2_col3 df3_col3
0 6 5.0 8
1 8 8.0 0
2 4 8.0 5
3 2 NaN 9
4 2 NaN 8
You can use df.merge:
In [1965]: res = df1.merge(df2, on='Label', how='left', suffixes=('_df1', '_df2')).merge(df3, on='Label', how='left').rename(columns={'col1': 'col1_df3','col2':'col2_df3','col3':'col3_df3'})
In [1975]: res = res.reindex(sorted(res.columns), axis=1)
In [1976]: res
Out[1965]:
Label col1_df1 col1_df2 col1_df3 col2_df1 col2_df2 col2_df3 col3_df1 col3_df2 col3_df3
0 NF1 1 8.00 2 1 4.00 8 6 5.00 8
1 NF2 3 4.00 6 2 7.00 2 8 8.00 0
2 NF3 4 9.00 2 5 7.00 2 4 8.00 5
3 NF4 5 nan 2 7 nan 4 2 nan 9
4 NF5 6 nan 2 2 nan 5 2 nan 8
We can use Pandas' join method, by setting the Label column as the index and joining the dataframes :
dfs = [df1,df2,df3]
keys = ['df1','df2','df3']
#set Label as index
df1, *others = [frame.set_index("Label").add_prefix(f"{prefix}_")
for frame,prefix in zip(dfs,keys)]
#join df1 with others
outcome = df1.join(others,how='outer').rename_axis(index='Label').reset_index()
outcome
Label df1_col1 df1_col2 df1_col3 df2_col1 df2_col2 df2_col3 df3_col1 df3_col2 df3_col3
0 NF1 1 1 6 8.0 4.0 5.0 2 8 8
1 NF2 3 2 8 4.0 7.0 8.0 6 2 0
2 NF3 4 5 4 9.0 7.0 8.0 2 2 5
3 NF4 5 7 2 NaN NaN NaN 2 4 9
4 NF5 6 2 2 NaN NaN NaN 2 5 8
I have a dataframe which looks like this:
0 1 2 3 4 5 6
0 a(A) b c c d a a
1 b h w k d c(A) k
2 g e(A) s g h s f
3 f d s h(A) c w n
4 e g s b c e w
I want to get the index of the cell which contains (A) in each column.
0 0
1 2
2 NaN
3 3
4 NaN
5 1
6 NaN
I tried this code but the result doesn't reach my expectation.
df.apply(lambda x: (x.str.contains(r'(A)')==True).idxmax(), axis=0)
Result looks like this:
0 0
1 2
2 0
3 3
4 0
5 1
6 0
I think it returns the first index if there is no (A) in that column.
How should I fix it?
Use Series.where for set default missing value for overwrite default 0 value of DataFrame.idxmax:
mask = df.apply(lambda x: x.str.contains('A'))
s1 = mask.idxmax().where(mask.any())
print (s1)
0 0.0
1 2.0
2 NaN
3 3.0
4 NaN
5 1.0
6 NaN
dtype: float64
You could do what you're doing but explicitly check if the rows contain any matches:
In [51]: pred = df.applymap(lambda x: '(A)' in x)
In [52]: pred.idxmax() * np.where(pred.any(), 1, np.nan)
Out[52]:
0 0.0
1 2.0
2 NaN
3 3.0
4 NaN
5 1.0
6 NaN
dtype: float64
Or alternatively, using DataFrame.where directly:
In [211]: pred.where(pred).idxmax()
Out[211]:
0 0.0
1 2.0
2 NaN
3 3.0
4 NaN
5 1.0
6 NaN
dtype: float64
A slightly cheatier one-liner is to use DataFrame.where on the identity:
In [78]: df.apply(lambda x: x.str.contains('A')).where(lambda x: x).idxmax()
Out[78]:
0 0.0
1 2.0
2 NaN
3 3.0
4 NaN
5 1.0
6 NaN
Add an if condition at the end of the apply:
>>> df.apply(lambda x: x.str.contains('A').idxmax() if 'A' in x[x.str.contains('A').idxmax()] else np.nan)
0 0.0
1 2.0
2 NaN
3 3.0
4 NaN
5 1.0
6 NaN
dtype: float64
>>>
I have a pandas.DataFrame that contain string, float and int types.
Is there a way to set all strings that cannot be converted to float to NaN ?
For example:
A B C D
0 1 2 5 7
1 0 4 NaN 15
2 4 8 9 10
3 11 5 8 0
4 11 5 8 "wajdi"
to:
A B C D
0 1 2 5 7
1 0 4 NaN 15
2 4 8 9 10
3 11 5 8 0
4 11 5 8 NaN
You can use pd.to_numeric and set errors='coerce'
pandas.to_numeric
df['D'] = pd.to_numeric(df.D, errors='coerce')
Which will give you:
A B C D
0 1 2 5.0 7.0
1 0 4 NaN 15.0
2 4 8 9.0 10.0
3 11 5 8.0 0.0
4 11 5 8.0 NaN
Deprecated solution (pandas <= 0.20 only):
df.convert_objects(convert_numeric=True)
pandas.DataFrame.convert_objects
Here's the dev note in the convert_objects source code: # TODO: Remove in 0.18 or 2017, which ever is sooner. So don't make this a long term solution if you use it.
Here is a way:
df['E'] = pd.to_numeric(df.D, errors='coerce')
And then you have:
A B C D E
0 1 2 5.0 7 7.0
1 0 4 NaN 15 15.0
2 4 8 9.0 10 10.0
3 11 5 8.0 0 0.0
4 11 5 8.0 wajdi NaN
You can use pd.to_numeric with errors='coerce'.
In [30]: df = pd.DataFrame({'a': [1, 2, 'NaN', 'bob', 3.2]})
In [31]: pd.to_numeric(df.a, errors='coerce')
Out[31]:
0 1.0
1 2.0
2 NaN
3 NaN
4 3.2
Name: a, dtype: float64
Here is one way to apply it to all columns:
for c in df.columns:
df[c] = pd.to_numeric(df[c], errors='coerce')
(See comment by NinjaPuppy for a better way.)