Basically this:
hash = "355879ACB6"
hash = hash[:4] + '-' + hash[4:]
print (hash)
3558-79ACB6
I got this part above from another stackoverflow post here
but for a DataFrame.
I am only able to successfully add strings before and after, like this:
data ['col1'] = data['col1'] + 'teststring'
If I try the solution from the link above [:amountofcharacterstocutafter] to add values at a certain position, which would be something like:
test = data[:2] + 'zz'
print (test)
It does not seem to be applicable, as the [:2] operator works different for dataframes as it does for strings. It cuts the ouput after the first 2 rows.
Goal:
I want to add a ' - ' at a certain position. Let's say the input row value is 'TTTT1234', output should be 'TTTT-1234'. For every row.
You can perform the operation you presented on a list but you have a column in a dataframe so its (a bit) different.
So while you can do this:
hash = "355879ACB6"
hash = hash[:4] + '-' + hash[4:]
in order to do this on a dataframe you can do it in at least 2 ways:
consider this dummy df:
LOCATION Hash
0 USA 355879ACB6
1 USA 455879ACB6
2 USA 388879ACB6
3 USA 800879ACB6
4 JAPAN 355870BCB6
5 JAPAN 355079ACB6
A. vectorization: the most efficient way
df['new_hash']=df['Hash'].str[:4]+'-'+df['Hash'].str[4:]
LOCATION Hash new_hash
0 USA 355879ACB6 3558-79ACB6
1 USA 455879ACB6 4558-79ACB6
2 USA 388879ACB6 3888-79ACB6
3 USA 800879ACB6 8008-79ACB6
4 JAPAN 355870BCB6 3558-70BCB6
5 JAPAN 355079ACB6 3550-79ACB6
B. apply lambda: intuitive to implement but less attractive in terms of performance
df['new_hash'] = df.apply(lambda x: x['Hash'][:4]+'-'+x['Hash'][4:], axis=1)
Use pd.Series.str. For example:
import pandas as pd
df = pd.DataFrame({
"c": ["TTTT1234"]
})
df["c"].str[:4] + "-" + df["c"].str[4:] # It will output 'TTTT-1234'
pd.Series.str gives vectorized string functions.
Related
I would like to go through every row (entry) in my df and remove every entry that has the value of " " (which yes is an empty string).
So if my data set is:
Name Gender Age
Jack 5
Anna F 6
Carl M 7
Jake M 7
Therefore Jack would be removed from the dataset.
On another note, I would also like to remove entries that has the value "Unspecified" and "Undetermined" as well.
Eg:
Name Gender Age Address
Jack 5 *address*
Anna F 6 *address*
Carl M 7 Undetermined
Jake M 7 Unspecified
Now,
Jack will be removed due to empty field.
Carl will be removed due to the value Undetermined present in a column.
Jake will be removed due to the value Unspecified present in a column.
For now, this has been my approach but I keep getting a TypeError.
list = []
for i in df.columns:
if df[i] == "":
# everytime there is an empty string, add 1 to list
list.append(1)
# count list to see how many entries there are with empty string
len(list)
Please help me with this. I would prefer a for loop being used due to there being about 22 columns and 9000+ rows in my actual dataset.
Note - I do understand that there are other questions asked like this, its just that none of them apply to my situation, meaning that most of them are only useful for a few columns and I do not wish to hardcode all 22 columns.
Edit - Thank you for all your feedbacks, you all have been incredibly helpful.
To delete a row based on a condition use the following:
df = df.drop(df[condition].index)
For example:
df = df.drop(df[Age==5].index) , will drop the row where the Age is 5.
I've come across a post regarding the same dating back to 2017, it should help you understand it more clearer.
Regarding question 2, here's how to remove rows with the specified values in a given column:
df = df[~df["Address"].isin(("Undetermined", "Unspecified"))]
Let's assume we have a Pandas DataFrame object df.
To remove every row given your conditions, simply do:
df = df[df.Gender == " " or df.df.Age == " " or df.Address in [" ", "Undetermined", "Unspecified"]]
If the unspecified fields are NaN, you can also do:
df = df.dropna(how="any", axis = 0)
Answer from #ThatCSFresher or #Bence will help you out in removing rows based on single column... Which is great!
However, I think there are multiple condition in your query needed to check across multiple columns at once in a loop. So, probably apply-lambda can do the job; Try the following code;
df = pd.DataFrame({"Name":["Jack","Anna","Carl","Jake"],
"Gender":["","F","M","M"],
"Age":[5,6,7,7],
"Address":["address","address","Undetermined","Unspecified"]})
df["Noise_Tag"] = df.apply(lambda x: "Noise" if ("" in list(x)) or ("Undetermined" in list(x)) or ("Unspecified" in list(x)) else "No Noise",axis=1)
df1 = df[df["Noise_Tag"] == "No Noise"]
del df1["Noise_Tag"]
# Output of df;
Name Gender Age Address Noise_Tag
0 Jack 5 address Noise
1 Anna F 6 address No Noise
2 Carl M 7 Undetermined Noise
3 Jake M 7 Unspecified Noise
# Output of df1;
Name Gender Age Address
1 Anna F 6 address
Well, OP actually wants to delete any column with "empty" string.
df = df[~(df=="").any(axis=1)] # deletes all rows that have empty string in any column.
If you want to delete specifically for address column, then you can just delete using
df = df[~df["Address"].isin(("Undetermined", "Unspecified"))]
Or if any column with Undetermined or Unspecified, try similar as the first solution in my post, just by replacing the empty string with Undertermined or Unspecified.
df = df[~((df=="Undetermined") | (df=="Unspecified")).any(axis=1)]
You can build masks and then filter the df according to it:
m1 = df.eq('').any(axis=1)
# m1 is True if any cell in a row has an empty string
m2 = df['Address'].isin(['Undetermined', 'Unspecified'])
# m2 is True if a row has one of the values in the list in column 'Address'
out = df[~m1 & ~m2] # invert both condition and get the desired output
print(out)
Output:
Name Gender Age Address
1 Anna F 6 *address*
Used Input:
df = pd.DataFrame({'Name': ['Jack', 'Anna', 'Carl', 'Jake'],
'Gender': ['', 'F', 'M', 'M'],
'Age': [5, 6, 7, 7],
'Address': ['*address*', '*address*', 'Undetermined', 'Unspecified']}
)
using lambda fun
Code:
df[df.apply(lambda x: False if (x.Address in ['Undetermined', 'Unspecified'] or '' in list(x)) else True, axis=1)]
Output:
Name Gender Age Address
1 Anna F 6 *add
I have a subset of data (single column) we'll call ID:
ID
0 07-1401469
1 07-89556629
2 07-12187595
3 07-381962
4 07-99999085
The current format is (usually) YY-[up to 8-character ID].
The desired output format is a more uniformed YYYY-xxxxxxxx:
ID
0 2007-01401469
1 2007-89556629
2 2007-12187595
3 2007-00381962
4 2007-99999085
Knowing that I've done padding in the past, the thought process was to combine
df['id'].str.split('-').str[0].apply(lambda x: '{0:20>4}'.format(x))
df['id'].str.split('-').str[1].apply(lambda x: '{0:0>8}'.format(x))
However I ran into a few problems:
The '20' in '{0:20>4}' must be a singular value and not a string
Trying to do something like the below just results in df['id'] taking the properties of the last lambda & trying any other way to combine multiple apply/lambdas just didn't work. I started going down the pad left/right route but that seemed to be taking be backwards.
df['id'] = (df['id'].str.split('-').str[0].apply(lambda x: '{0:X>4}'.format(x)).str[1].apply(lambda x: '{0:0>8}'.format(x)))
The current solution I have (but HATE because its long, messy, and just not clean IMO) is:
df['idyear'] = df['id'].str.split('-').str[0].apply(lambda x: '{:X>4}'.format(x)) # Split on '-' and pad with X
df['idyear'] = df['idyear'].str.replace('XX', '20') # Replace XX with 20 to conform to YYYY
df['idnum'] = df['id'].str.split('-').str[1].apply(lambda x: '{0:0>8}'.format(x)) # Pad 0s up to 8 digits
df['id'] = df['idyear'].map(str) + "-" + df['idnum'] # Merge idyear and idnum to remake id
del df['idnum'] # delete extra
del df['idyear'] # delete extra
Which does work
ID
0 2007-01401469
1 2007-89556629
2 2007-12187595
3 2007-00381962
4 2007-99999085
But my questions are
Is there a way to run multiple apply() functions in a single line so I'm not making temp variables
Is there a better way than replacing 'XX' for '20'
I feel like this entire code block can be compress to 1 or 2 lines I just don't know how. Everything I've seen on SO and Pandas documentation on highlights/relates to singular manipulation so far.
One option is to split; then use str.zfill to pad '0's. Also prepend '20's before splitting, since you seem to need it anyway:
tmp = df['ID'].radd('20').str.split('-')
df['ID'] = tmp.str[0] + '-'+ tmp.str[1].str.zfill(8)
Output:
ID
0 2007-01401469
1 2007-89556629
2 2007-12187595
3 2007-00381962
4 2007-99999085
I'd do it in two steps, using .str.replace:
df["ID"] = df["ID"].str.replace(r"^(\d{2})-", r"20\1-", regex=True)
df["ID"] = df["ID"].str.replace(r"-(\d+)", lambda g: f"-{g[1]:0>8}", regex=True)
print(df)
Prints:
ID
0 2007-01401469
1 2007-89556629
2 2007-12187595
3 2007-00381962
4 2007-99999085
I am working on a python script that automates some phone calls for me. I have a tool to test with that I can interact with REST API. I need to select a specific carrier based on which country code is entered. So let's say my user enters 12145221414 in my excel document, I want to choose AT&T as the carrier. How would I accept input from the first column of the table and then output what's in the 2nd column?
Obviously this can get a little tricky, since I would need to match up to 3-4 digits on the front of a phone number. My plan is to write a function that then takes the initial number and then plugs the carrier that needs to be used for that country.
Any idea how I could extract this data from the table? How would I make it so that if you entered Barbados (1246), then Lime is selected instead of AT&T?
Here's my code thus far and tables. I'm not sure how I can read one table and then pull data from that table to use for my matching function.
testlist.xlsx
| Number |
|:------------|
|8155555555|
|12465555555|
|12135555555|
|96655555555|
|525555555555|
carriers.xlsx
| countryCode | Carrier |
|:------------|:--------|
|1246|LIME|
|1|AT&T|
|81|Softbank|
|52|Telmex|
|966|Zain|
import pandas as pd
import os
FILE_PATH = "C:/temp/testlist.xlsx"
xl_1 = pd.ExcelFile(FILE_PATH)
num_df = xl_1.parse('Numbers')
FILE_PATH = "C:/temp/carriers.xlsx"
xl_2 = pd.ExcelFile(FILE_PATH)
car_df = xl_2.parse('Carriers')
for index, row in num_df.iterrows():
Any idea how I could extract this data from the table? How would I
make it so that if you entered Barbados (1246), then Lime is selected
instead of AT&T?
carriers.xlsx
countryCode
Carrier
1246
LIME
1
AT&T
81
Softbank
52
Telmex
966
Zain
script.py
import pandas as pd
FILE_PATH = "./carriers.xlsx"
df = pd.read_excel(FILE_PATH)
rows_list = df.to_dict('records')
code_carrier_map = {}
for row in rows_list:
code_carrier_map[row["countryCode"]] = row["Carrier"]
print(type(code_carrier_map), code_carrier_map)
print(f"{code_carrier_map.get(1)=}")
print(f"{code_carrier_map.get(1246)=}")
print(f"{code_carrier_map.get(52)=}")
print(f"{code_carrier_map.get(81)=}")
print(f"{code_carrier_map.get(966)=}")
Output
$ python3 script.py
<class 'dict'> {1246: 'LIME', 1: 'AT&T', 81: 'Softbank', 52: 'Telmex', 966: 'Zain'}
code_carrier_map.get(1)='AT&T'
code_carrier_map.get(1246)='LIME'
code_carrier_map.get(52)='Telmex'
code_carrier_map.get(81)='Softbank'
code_carrier_map.get(966)='Zain'
Then if you want to parse phone numbers, don't reinvent the wheel, just use this phonenumbers library.
Code
import phonenumbers
num = "+12145221414"
phone_number = phonenumbers.parse(num)
print(f"{num=}")
print(f"{phone_number.country_code=}")
print(f"{code_carrier_map.get(phone_number.country_code)=}")
Output
num='+12145221414'
phone_number.country_code=1
code_carrier_map.get(phone_number.country_code)='AT&T'
Let's assume the following input:
>>> df1
Number
0 8155555555
1 12465555555
2 12135555555
3 96655555555
4 525555555555
>>> df2
countryCode Carrier
0 1246 LIME
1 1 AT&T
2 81 Softbank
3 52 Telmex
4 966 Zain
First we need to rework a bit df2 to sort the countryCode in descending order, make it as string and set it to index.
The trick for later is to sort countryCode in descending order. This will ensure that a longer country codes, such as "1246" is matched before a shorter one like "1".
>>> df2 = df2.sort_values(by='countryCode', ascending=False).astype(str).set_index('countryCode')
>>> df2
Carrier
countryCode
1246 LIME
966 Zain
81 Softbank
52 Telmex
1 AT&T
Finally, we use a regex (here '1246|966|81|52|1' using '|'.join(df2.index)) made from the country codes in descending order to extract the longest code, and we map it to the carrier:
(df1.astype(str)['Number']
.str.extract('^(%s)'%'|'.join(df2.index))[0]
.map(df2['Carrier'])
)
output:
0 Softbank
1 LIME
2 AT&T
3 Zain
4 Telmex
Name: 0, dtype: object
NB. to add it to the initial dataframe:
df1['carrier'] = (df1.astype(str)['Number']
.str.extract('^(%s)'%'|'.join(df2.index))[0]
.map(df2['Carrier'])
).to_clipboard(0)
output:
Number carrier
0 8155555555 Softbank
1 12465555555 LIME
2 12135555555 AT&T
3 96655555555 Zain
4 525555555555 Telmex
If I understand it correctly, you just want to get the first characters from the input column (Number) and then match this with the second dataframe from carriers.xlsx.
Extract first characters of a Number column. Hint: The nbr_of_chars variable should be based on the maximum character length of the column countryCode in the carriers.xlsx
nbr_of_chars = 4
df.loc[df['Number'].notnull(), 'FirstCharsColumn'] = df['Number'].str[:nbr_of_chars]
Then the matching should be fairly easy with dataframe joins.
I can think only of an inefficient solution.
First, sort the data frame of carriers in the reverse alphabetical order of country codes. That way, longer prefixes will be closer to the beginning.
codes = xl_2.sort_values('countryCode', ascending=False)
Next, define a function that matches a number with each country code in the second data frame and finds the index of the first match, if any (remember, that match is the longest).
def cc2carrier(num):
matches = codes['countryCode'].apply(lambda x: num.startswith(x))
if not matches.any(): #Not found
return np.nan
return codes.loc[matches.idxmax()]['Carrier']
Now, apply the function to the numbers dataframe:
xl_1['Number'].apply(cc2carrier)
#1 Softbank
#2 LIME
#3 AT&T
#4 Zain
#5 Telmex
#Name: Number, dtype: object
first of all, I have no background in computer language and I am learning Python.
I'm trying to group some data in a data frame.
[dataframe "cafe_df_merged"]
Actually, I want to create a new data frame shows the 'city_number', 'city' (which is a name), and also the number of cafes in the same city. So, it should have 3 columns; 'city_number', 'city' and 'number_of_cafe'
However, I have tried to use the group by but the result did not come out as I expected.
city_directory = cafe_df_merged[['city_number', 'city']]
city_directory = city_directory.groupby('city').count()
city_directory
[the result]
How should I do this? Please help, thanks.
There are likely other ways of doing this as well, but something like this should work:
import pandas as pd
import numpy as np
# Create a reproducible example
places = [[['starbucks', 'new_york', '1234']]*5, [['bean_dream', 'boston', '3456']]*4, \
[['coffee_today', 'jersey', '7643']]*3, [['coffee_today', 'DC', '8902']]*3, \
[['starbucks', 'nowwhere', '2674']]*2]
places = [p for sub in places for p in sub]
# a dataframe containing all information
city_directory = pd.DataFrame(places, columns=['shop','city', 'id'])
# make a new dataframe with just cities and ids
# drop duplicate rows
city_info = city_directory.loc[:, ['city','id']].drop_duplicates()
# get the cafe counts (number of cafes)
cafe_count = city_directory.groupby('city').count().iloc[:,0]
# add the cafe counts to the dataframe
city_info['cafe_count'] = cafe_count[city_info['city']].to_numpy()
# reset the index
city_info = city_info.reset_index(drop=True)
city_info now yields the following:
city id cafe_count
0 new_york 1234 5
1 boston 3456 4
2 jersey 7643 3
3 DC 8902 3
4 nowwhere 2674 2
And part of the example dataframe, city_directory.tail(), looks like this:
shop city id
12 coffee_today DC 8902
13 coffee_today DC 8902
14 coffee_today DC 8902
15 starbucks nowwhere 2674
16 starbucks nowwhere 2674
Opinion: As a side note, it might be easier to get comfortable with regular Python first before diving deep into the world of pandas and numpy. Otherwise, it might be a bit overwhelming.
I have a CSV file that looks like below, this is same like my last question but this is by using Pandas.
Group Sam Dan Bori Son John Mave
A 0.00258844 0.983322 1.61479 1.2785 1.96963 10.6945
B 0.0026034 0.983305 1.61198 1.26239 1.9742 10.6838
C 0.0026174 0.983294 1.60913 1.24543 1.97877 10.6729
D 0.00263062 0.983289 1.60624 1.22758 1.98334 10.6618
E 0.00264304 0.98329 1.60332 1.20885 1.98791 10.6505
I have a function like below
def getnewno(value):
value = value + 30
if value > 40 :
value = value - 20
else:
value = value
return value
I want to send all these values to the getnewno function and get a newvalue and update the CSV file. How can this be accomplished in Pandas.
Expected output:
Group Sam Dan Bori Son John Mave
A 30.00258844 30.983322 31.61479 31.2785 31.96963 20.6945
B 30.0026034 30.983305 31.61198 31.26239 31.9742 20.6838
C 30.0026174 30.983294 31.60913 31.24543 31.97877 20.6729
D 30.00263062 30.983289 31.60624 31.22758 31.98334 20.6618
E 30.00264304 30.98329 31.60332 31.20885 31.98791 20.6505
The following should give you what you desire.
Applying a function
Your function can be simplified and here expressed as a lambda function.
It's then a matter of applying your function to all of the columns. There are a number of ways to do so. The first idea that comes to mind is to loop over df.columns. However, we can do better than this by using the applymap or transform methods:
import pandas as pd
# Read in the data from file
df = pd.read_csv('data.csv',
sep='\s+',
index_col=0)
# Simplified function with which to transform data
getnewno = lambda value: value + 10 if value > 10 else value + 30
# Looping over columns
#for col in df.columns:
# df[col] = df[col].apply(getnewno)
# Apply to all columns without loop
df = df.applymap(getnewno)
# Write out updated data
df.to_csv('data_updated.csv')
Using broadcasting
You can achieve your result using broadcasting and a little boolean logic. This avoids looping over any columns, and should ultimately prove faster and less memory intensive (although if your dataset is small any speed-up would be negligible):
import pandas as pd
df = pd.read_csv('data.csv',
sep='\s+',
index_col=0)
df += 30
make_smaller = df > 40
df[make_smaller] -= 20
First of all, your getnewno function looks too complicated... it can be simplified to e.g.:
def getnewno(value):
if value + 30 > 40:
return value - 20
else:
return value
you can even change value + 30 > 40 to value > 10.
Or even a oneliner if you want:
getnewno = lambda value: value-20 if value > 10 else value
Having the function you can apply it to specific values/columns. For example, if want you to create a column Mark_updated basing on Mark column, it should look like this (I assume your pandas DataFrame is called df):
df['Mark_updated'] = df['Mark'].apply(getnewno)
Use the mask function to do an if-else solution, before writing the data to csv
res = (df
.select_dtypes('number')
.add(30)
#the if-else comes in here
#if any entry in the dataframe is greater than 40, subtract 20 from it
#else leave as is
.mask(lambda x: x>40, lambda x: x.sub(20))
)
#insert the group column back
res.insert(0,'Group',df.Group.array)
write to csv
res.to_csv(filename)
Group Sam Dan Bori Son John Mave
0 A 30.002588 30.983322 31.61479 31.27850 31.96963 20.6945
1 B 30.002603 30.983305 31.61198 31.26239 31.97420 20.6838
2 C 30.002617 30.983294 31.60913 31.24543 31.97877 20.6729
3 D 30.002631 30.983289 31.60624 31.22758 31.98334 20.6618
4 E 30.002643 30.983290 31.60332 31.20885 31.98791 20.6505