df1 = pd.DataFrame({'Key':['OK340820.1','OK340821.1'],'Length':[50000,67000]})
df2 = pd.DataFrame({'Key':['OK340820','OK340821'],'Length':[np.nan,np.nan]})
If df2.Key is a substring of df1.Key, set Length of df2 as value of Length in df1
I tried doing this:
df2['Length']=np.where(df2.Key.isin(df1.Key.str.extract(r'(.+?(?=\.))')), df1.Length, '')
But it's not returning the matches.
Map df2.Key to a "prepared" Key values of df1:
df2['Length'] = df2.Key.map(dict(zip(df1.Key.str.replace(r'\..+', '', regex=True), df1.Length)))
In [45]: df2
Out[45]:
Key Length
0 OK340820 50000
1 OK340821 67000
You can use a regex to extract the string, then map the values:
import re
pattern = '|'.join(map(re.escape, df2['Key']))
s = pd.Series(df1['Length'].values, index=df1['Key'].str.extract(f'({pattern})', expand=False))
df2['Length'] = df2['Key'].map(s)
Updated df2:
Key Length
0 OK340820 50000
1 OK340821 67000
Or with a merge:
import re
pattern = '|'.join(map(re.escape, df2['Key']))
(df2.drop(columns='Length')
.merge(df1, how='left', left_on='Key', suffixes=(None, '_'),
right_on=df1['Key'].str.extract(f'({pattern})', expand=False))
.drop(columns='Key_')
)
Alternative if the Key in df1 is always in the form XXX.1 and removing the .1 is enough:
df2['Length'] = df2['Key'].map(df1.set_index(df1['Key'].str.extract('([^.]+)', expand=False))['Length'])
Another possible solution, which is based on pandas.DataFrame.update:
df2.update(df1.assign(Key = df1['Key'].str.extract('(.*)\.')))
Output:
Key Length
0 OK340820 50000.0
1 OK340821 67000.0
Related
I have a dataframe called datafe from which I want to combine the hyphenated words.
for example input dataframe looks like this:
,author_ex
0,Marios
1,Christodoulou
2,Intro-
3,duction
4,Simone
5,Speziale
6,Exper-
7,iment
And the output dataframe should be like:
,author_ex
0,Marios
1,Christodoulou
2,Introduction
3,Simone
4,Speziale
5,Experiment
I have written a sample code to achieve this but I am not able to get out of the recursion safely.
def rm_actual(datafe, index):
stem1 = datafe.iloc[index]['author_ex']
stem2 = datafe.iloc[index + 1]['author_ex']
fixed_token = stem1[:-1] + stem2
datafe.drop(index=index + 1, inplace=True, axis=0)
newdf=datafe.reset_index(drop=True)
newdf.iloc[index]['author_ex'] = fixed_token
return newdf
def remove_hyphens(datafe):
for index, row in datafe.iterrows():
flag = False
token=row['author_ex']
if token[-1:] == '-':
datafe=rm_actual(datafe, index)
flag=True
break
if flag==True:
datafe=remove_hyphens(datafe)
if flag==False:
return datafe
datafe=remove_hyphens(datafe)
print(datafe)
Is there any possibilities I can get out of this recursion with expected output?
Another option:
Given/Input:
author_ex
0 Marios
1 Christodoulou
2 Intro-
3 duction
4 Simone
5 Speziale
6 Exper-
7 iment
Code:
import pandas as pd
# read/open file or create dataframe
df = pd.DataFrame({'author_ex':['Marios', 'Christodoulou', 'Intro-', \
'duction', 'Simone', 'Speziale', 'Exper-', 'iment']})
# check input format
print(df)
# create new column 'Ending' for True/False if column 'author_ex' ends with '-'
df['Ending'] = df['author_ex'].shift(1).str.contains('-$', na=False, regex=True)
# remove the trailing '-' from the 'author_ex' column
df['author_ex'] = df['author_ex'].str.replace('-$', '', regex=True)
# create new column with values of 'author_ex' and shifted 'author_ex' concatenated together
df['author_ex_combined'] = df['author_ex'] + df.shift(-1)['author_ex']
# create a series true/false but shifted up
index = (df['Ending'] == True).shift(-1)
# set the last row to 'False' after it was shifted
index.iloc[-1] = False
# replace 'author_ex' with 'author_ex_combined' based on true/false of index series
df.loc[index,'author_ex'] = df['author_ex_combined']
# remove rows that have the 2nd part of the 'author_ex' string and are no longer required
df = df[~df.Ending]
# remove the extra columns
df.drop(['Ending', 'author_ex_combined'], axis = 1, inplace=True)
# output final dataframe
print('\n\n')
print(df)
# notice index 3 and 6 are missing
Outputs:
author_ex
0 Marios
1 Christodoulou
2 Introduction
4 Simone
5 Speziale
6 Experiment
I can do the following if I want to extract rows whose column "A" contains the substring "hello".
df[df['A'].str.contains("hello")]
How can I select rows whose column is the substring for another word? e.g.
df["hello".contains(df['A'].str)]
Here's an example dataframe
df = pd.DataFrame.from_dict({"A":["hel"]})
df["hello".contains(df['A'].str)]
IIUC, you could apply str.find:
import pandas as pd
df = pd.DataFrame(['hell', 'world', 'hello'], columns=['A'])
res = df[df['A'].apply("hello".find).ne(-1)]
print(res)
Output
A
0 hell
2 hello
As an alternative use __contains__
res = df[df['A'].apply("hello".__contains__)]
print(res)
Output
A
0 hell
2 hello
Or simply:
res = df[df['A'].apply(lambda x: x in "hello")]
print(res)
I have a pandas dataframe with one column that contains an empty list in each cell.
I need to duplicate the dataframe, and append it at the bottom of the original dataframe, but with additional information in the list.
Here is a minimal code example:
df_main = pd.DataFrame([['a', []], ['b', []]], columns=['letter', 'mylist'])
> df_main
letter mylist
0 a []
1 b []
df_copy = df_main.copy()
for index, row in df_copy.iterrows():
row.mylist = row.mylist.append(1)
pd.concat([ df_copy,df_main], ignore_index=True)
> result:
letter mylist
0 a None
1 b None
2 a [1]
3 b [1]
As you can see there is a problem that the [] empty list was replaced by a None
Just to make sure, this is what I would like to have:
letter mylist
0 a []
1 b []
2 a [1]
3 b [1]
How can I achieve that?
append method on list return a None value, that's why None appears in the final dataframe. You may have use + operator for reassignment like this:
import pandas as pd
df_main = pd.DataFrame([['a', []], ['b', []]], columns=['letter', 'mylist'])
df_copy = df_main.copy()
for index, row in df_copy.iterrows():
row.mylist = row.mylist + list([1])
pd.concat([df_main, df_copy], ignore_index=True).head()
Output of this block of code:
letter mylist
0 a []
1 b []
2 a [1]
3 b [1]
A workaround to solve your problem would be to create a temporary column mylist2 with np.empty((len(df), 0)).tolist()) and use np.where() to change the None values of mylist to an empty list and then drop the empty column.
import pandas as pd, numpy as np
df_main = pd.DataFrame([['a', []], ['b', []]], columns=['letter', 'mylist'])
df_copy = df_main.copy()
for index, row in df_copy.iterrows():
row.mylist = row.mylist.append(1)
df = (pd.concat([df_copy,df_main], ignore_index=True)
.assign(mylist2=np.empty((len(df), 0)).tolist()))
df['mylist'] = np.where((df['mylist'].isnull()), df['mylist2'], df['mylist'])
df= df.drop('mylist2', axis=1)
df
Out[1]:
letter mylist
0 a []
1 b []
2 a [1]
3 b [1]
Not only does append method on list return a None value as indicated in the first answer, but both df_main and df_copy contain pointers to the same lists. So after:
for index, row in df_copy.iterrows():
row.mylist.append(1)
both dataframes have updated lists with one element. For your code to work as expected you can create a new list after you copy the dataframe:
df_copy = df_main.copy()
for index, row in df_copy.iterrows():
row.mylist = []
This question is another great example why we should not put objects in a dataframe.
df.groupby('columns').apply(''.join()), join all the cells to a string.
This is for a junior dataprocessor. In the past, I've tried many ways.
import pandas as pd
data = {'key':['a','b','c','a','b','c','a'], 'profit':
[12,3,4,5,6,7,9],'income':['j','d','d','g','d','t','d']}
df = pd.DataFrame(data)
df = df.set_index(‘key’)
#df2 is expected result
data2 = {'a':['12j5g9d'],'b':['3d6d'],'c':['4d7t']}
df2 = pd.DataFrame(data2)
df2 = df2.set_index(‘key’)
Here's a simple solution, where we first translate the integers to strings and then concatenate profit and income, then finally we concatenate all strings under the same key:
data = {'key':['a','b','c','a','b','c','a'], 'profit':
[12,3,4,5,6,7,9],'income':['j','d','d','g','d','t','d']}
df = pd.DataFrame(data)
df['profit_income'] = df['profit'].apply(str) + df['income']
res = df.groupby('key')['profit_income'].agg(''.join)
print(res)
output:
key
a 12j5g9d
b 3d6d
c 4d7t
Name: profit_income, dtype: object
This question can be solved couple different ways:
First add an extra column by concatenating the profit and income columns.
import pandas as pd
data = {'key':['a','b','c','a','b','c','a'], 'profit':
[12,3,4,5,6,7,9],'income':['j','d','d','g','d','t','d']}
df = pd.DataFrame(data)
df = df.set_index('key')
df['profinc']=df['profit'].astype(str)+df['income']
1) Using sum
df2=df.groupby('key').profinc.sum()
2) Using apply and join
df2=df.groupby('key').profinc.apply(''.join)
Results from both of the above would be the same:
key
a 12j5g9d
b 3d6d
c 4d7t
I have a dataframe with column names, and I want to find the one that contains a certain string, but does not exactly match it. I'm searching for 'spike' in column names like 'spike-2', 'hey spike', 'spiked-in' (the 'spike' part is always continuous).
I want the column name to be returned as a string or a variable, so I access the column later with df['name'] or df[name] as normal. I've tried to find ways to do this, to no avail. Any tips?
Just iterate over DataFrame.columns, now this is an example in which you will end up with a list of column names that match:
import pandas as pd
data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)
spike_cols = [col for col in df.columns if 'spike' in col]
print(list(df.columns))
print(spike_cols)
Output:
['hey spke', 'no', 'spike-2', 'spiked-in']
['spike-2', 'spiked-in']
Explanation:
df.columns returns a list of column names
[col for col in df.columns if 'spike' in col] iterates over the list df.columns with the variable col and adds it to the resulting list if col contains 'spike'. This syntax is list comprehension.
If you only want the resulting data set with the columns that match you can do this:
df2 = df.filter(regex='spike')
print(df2)
Output:
spike-2 spiked-in
0 1 7
1 2 8
2 3 9
This answer uses the DataFrame.filter method to do this without list comprehension:
import pandas as pd
data = {'spike-2': [1,2,3], 'hey spke': [4,5,6]}
df = pd.DataFrame(data)
print(df.filter(like='spike').columns)
Will output just 'spike-2'. You can also use regex, as some people suggested in comments above:
print(df.filter(regex='spike|spke').columns)
Will output both columns: ['spike-2', 'hey spke']
You can also use df.columns[df.columns.str.contains(pat = 'spike')]
data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)
colNames = df.columns[df.columns.str.contains(pat = 'spike')]
print(colNames)
This will output the column names: 'spike-2', 'spiked-in'
More about pandas.Series.str.contains.
# select columns containing 'spike'
df.filter(like='spike', axis=1)
You can also select by name, regular expression. Refer to: pandas.DataFrame.filter
df.loc[:,df.columns.str.contains("spike")]
Another solution that returns a subset of the df with the desired columns:
df[df.columns[df.columns.str.contains("spike|spke")]]
You also can use this code:
spike_cols =[x for x in df.columns[df.columns.str.contains('spike')]]
Getting name and subsetting based on Start, Contains, and Ends:
# from: https://stackoverflow.com/questions/21285380/find-column-whose-name-contains-a-specific-string
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.str.contains.html
# from: https://cmdlinetips.com/2019/04/how-to-select-columns-using-prefix-suffix-of-column-names-in-pandas/
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.filter.html
import pandas as pd
data = {'spike_starts': [1,2,3], 'ends_spike_starts': [4,5,6], 'ends_spike': [7,8,9], 'not': [10,11,12]}
df = pd.DataFrame(data)
print("\n")
print("----------------------------------------")
colNames_contains = df.columns[df.columns.str.contains(pat = 'spike')].tolist()
print("Contains")
print(colNames_contains)
print("\n")
print("----------------------------------------")
colNames_starts = df.columns[df.columns.str.contains(pat = '^spike')].tolist()
print("Starts")
print(colNames_starts)
print("\n")
print("----------------------------------------")
colNames_ends = df.columns[df.columns.str.contains(pat = 'spike$')].tolist()
print("Ends")
print(colNames_ends)
print("\n")
print("----------------------------------------")
df_subset_start = df.filter(regex='^spike',axis=1)
print("Starts")
print(df_subset_start)
print("\n")
print("----------------------------------------")
df_subset_contains = df.filter(regex='spike',axis=1)
print("Contains")
print(df_subset_contains)
print("\n")
print("----------------------------------------")
df_subset_ends = df.filter(regex='spike$',axis=1)
print("Ends")
print(df_subset_ends)