I have a DataFrame with two columns and want to combine their contents into another column. In the 3rd column, I will like to replace all entries that are 'hello' by the corresponding non 'hello' term. But my just returns df['C'] as the string addition of the df['A'] and df['B'].
df = pd.DataFrame({'A' : ['here', 'there', 'hello', 'hello', 'hello'],
'B' : ['hello', 'hello', 'go', 'click', 'clack']})
df['C'] = df['A'] + df['B']
I will be glad if someone can help me.
Related
Hi, given this dataframe is it possible to fetch the Number value associated with certain conditions using df.loc? This is what i came up with so far.
if df.loc[(df["Tags"]=="Brunei") & (df["Type"]=="Host"),"Number"]:
I want the output to be 1. Is this the correct way to do it?
You're in the right way, but you have to pass ".values[0]" in the end of the .loc statement to extract the only value that you got in the pandas Series.
df = pd.DataFrame({
'Tags': ['Brunei', 'China'],
'Type': ['Host', 'Address'],
'Number': [1, 1192]
}
)
display(df)
series = df.loc[(df["Tags"]=="Brunei") & (df["Type"]=="Host"),"Number"]
print(type(series))
value = df.loc[(df["Tags"]=="Brunei") & (df["Type"]=="Host"),"Number"].values[0]
print(type(value))
I have a series of web addresses, which I want to split them by the first '.'. For example, return 'google', if the web address is 'google.co.uk'
d1 = {'id':['1', '2', '3'], 'website':['google.co.uk', 'google.com.au', 'google.com']}
df1 = pd.DataFrame(data=d1)
d2 = {'id':['4', '5', '6'], 'website':['google.co.jp', 'google.com.tw', 'google.kr']}
df2 = pd.DataFrame(data=d2)
df_list = [df1, df2]
I use enumerate to iterate the dataframe list
for i, df in enumerate(df_list):
df_list[i]['website_segments'] = df['website'].str.split('.', n=1, expand=True)
Received error: ValueError: Wrong number of items passed 2, placement implies 1
You are splitting the website which gives you a list-like data structure. Think [google, co.uk]. You just want the first element of that list so:
for i, df in enumerate(df_list):
df_list[i]['website_segments'] = df['website'].str.split('.', n=1, expand=True)[0]
Another alternative is to use extract. It is also ~40% faster for your data:
for i, df in enumerate(df_list):
df_list[i]['website_segments'] = df['website'].str.extract('(.*?)\.')
I am trying to groupby for the following specializations but I am not getting the expected result (or any for that matter). The data stays ungrouped even after this step. Any idea what's wrong in my code?
cols_specials = ['Enterprise ID','Specialization','Specialization Branches','Specialization Type']
specials = pd.read_csv(agg_specials, engine='python')
specials = specials.merge(roster, left_on='Enterprise ID', right_on='Enterprise ID', how='left')
specials = specials[cols_specials]
specials = specials.groupby(['Enterprise ID'])['Specialization'].transform(lambda x: '; '.join(str(x)))
specials.to_csv(end_report_specials, index=False, encoding='utf-8-sig')
Please try using agg:
import pandas as pd
df = pd.DataFrame(
[
['john', 'eng', 'build'],
['john', 'math', 'build'],
['kevin', 'math', 'asp'],
['nick', 'sci', 'spi']
],
columns = ['id', 'spec', 'type']
)
df.groupby(['id'])[['spec']].agg(lambda x: ';'.join(x))
resiults in:
if you need to preserve starting number of lines, use transform. transform returns one column:
df['spec_grouped'] = df.groupby(['id'])[['spec']].transform(lambda x: ';'.join(x))
df
results in:
I needed your advice regarding how to map columns between data-frames:
I have put it in simple way so that it's easier for you to understand:
df = dataframe
EXAMPLE:
df1 = pd.DataFrame({
"X": [],
"Y": [],
"Z": []
})
df2 = pd.DataFrame({
"A": ['', '', 'A1'],
"C": ['', '', 'C1'],
"D": ['D1', 'Other', 'D3'],
"F": ['', '', ''],
"G": ['G1', '', 'G3'],
"H": ['H1', 'H2', 'H3']
})
Requirement:
1st step:
We needed to track a value for X column on df1 from columns A, C, D respectively. It would stop searching once it finds any value and would select it.
2nd step:
If the selected value is "Other" then X column of df1 would map columns F, G, and H respectively until it finds any value.
Result:
X
0 D1
1 H2
2 A1
Thank you so much in advance
Try this:
def first_non_empty(df, cols):
"""Return the first non-empty, non-null value among the specified columns per row"""
return df[cols].replace('', pd.NA).bfill(axis=1).iloc[:, 0]
col_x = first_non_empty(df2, ['A','C','D'])
col_x = col_x.mask(col_x == 'Other', first_non_empty(df2, ['F','G','H']))
df1['X'] = col_x
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)