Iterate through and conditionally append string values in a Pandas dataframe - r

Question

I've got a dataframe of research participants whose IDs are stored in the following format "0000.000".
Where the first four digits are their family ID number, and the final three digits are their individual index within the family. The majority of individuals have a suffix of ".000", but some have ".001", ".002", etc.
As a result of some inefficiencies, these numbers are stored as floats. I'm trying to import them as strings so that I can use them in a join to another data frame that is formatted correctly.
Those IDs that end in .000 are imported as "0000", rather than "0000.000". All others are imported correctly.
I'm trying to iterate through the IDs and append ".000" to those that are missing the suffix.
If I were using R, I could do it like this.
df %>% mutate(StudyID = ifelse(length(StudyID)<5,
paste(StudyID,".000",sep=""),
StudyID)
I've found a Python solution (below), but it's pretty janky.
row = 0
for i in df["StudyID"]:
if len(i)<5:
df.iloc[row,3] = i + ".000"
else: df.iloc[row,3] = i
index += 1
I think it'd be ideal to do it as a list comprehension, but I haven't been able to find a solution that lets me iterate through the column, changing a single value at a time.
For example, this solution iterates and checks the logic properly, but it replaces every single value that evaluates True during each iteration. I only want the value currently being evaluated to change.
[i + ".000" if len(i)<5 else i for i in df["StudyID"]]
Is this possible?

Answer 1

As you said, your code is doing the trick. One other way of doing what you want that i could think of is the following :
# Start by creating a mask that gives you the index you want to change
mask = [len(i)<5 for i in df.StudyID]
# Change the value of the dataframe on the mask
df.StudyID.iloc[mask] += ".000"

Answer 2

I think by length(StudyID), you meant nchar(StudyID), as #akrun pointed out.
You can do it in the dplyr way in python using datar:
>>> from datar.all import f, tibble, mutate, nchar, if_else, paste
>>>
>>> df = tibble(
... StudyID = ["0000", "0001", "0000.000", "0001.001"]
... )
>>> df
StudyID
<object>
0 0000
1 0001
2 0000.000
3 0001.001
>>>
>>> df >> mutate(StudyID=if_else(
... nchar(f.StudyID) < 5,
... paste(f.StudyID, ".000", sep=""),
... f.StudyID
... ))
StudyID
<object>
0 0000.000
1 0001.000
2 0000.000
3 0001.001
Disclaimer: I am the author of the datar package.

Answer 3

Ultimately, I needed to do this for a few different dataframes so I ended up defining a function to solve the problem so that I could apply it to each one.
I think the list comprehension idea was going to become too complex and potentially too difficult to understand when reviewing so I stuck with a plain old for-loop.
def create_multi_index(data, col_to_split, sep = "."):
"""
This function loops through the original ID column and splits it into
multiple parts (multi-IDs) on the defined separator.
By default, the function assumes the unique ID is formatted like a decimal number
The new multi-IDs are appended into a new list.
If the original ID was formatted like an integer, rather than a decimal
the function assumes the latter half of the ID to be ".000"
"""
# Take a copy of the dataframe to modify
new_df = data
# generate two new lists to store the new multi-index
Family_ID = []
Family_Index = []
# iterate through the IDs, split and allocate the pieces to the appropriate list
for i in new_df[col_to_split]:
i = i.split(sep)
Family_ID.append(i[0])
if len(i)==1:
Family_Index.append("000")
else:
Family_Index.append(i[1])
# Modify and return the dataframe including the new multi-index
return new_df.assign(Family_ID = Family_ID,
Family_Index = Family_Index)
This returns a duplicate dataframe with a new column for each part of the multi-id.
When joining dataframes with this form of ID, as long as both dataframes have the multi index in the same format, these columns can be used with pd.merge as follows:
pd.merge(df1, df2, how= "inner", on = ["Family_ID","Family_Index"])