Change Letters in A String One at a Time (Pandas,Python3)

Question

I have a list of words in Pandas (DF)

Words
Shirt
Blouse
Sweater

What I'm trying to do is swap out certain letters in those words with letters from my dictionary one letter at a time.

so for example:

mydict = {"e":"q,w",
          "a":"z"}

would create a new list that first replaces all the "e" in a list one at a time, and then iterates through again replacing all the "a" one at a time:

Words
Shirt
Blouse
Sweater
Blousq
Blousw
Swqater
Swwater
Sweatqr
Sweatwr
Swezter

I've been looking around at solutions here: Mass string replace in python?

and have tried the following code but it changes all instances "e" instead of doing so one at a time -- any help?:

mydict = {"e":"q,w"}
s = DF
for k, v in mydict.items():
    for j in v:
          s['Words'] = s["Words"].str.replace(k, j)
DF["Words"] = s

this doesn't seem to work either:

s = DF.replace({"Words": {"e": "q","w"}})

Answer 1

Because you need to replace letters one at a time, this doesn't sound like a good problem to solve with pandas, since pandas is about doing everything at once (vectorized operations). I would dump out your DataFrame into a plain old list and use list operations:

words = DF.to_dict()["Words"].values()

for find, replace in reversed(sorted(mydict.items())):
    for word in words:
        occurences = word.count(find)
        if not occurences:
            print word
            continue
        start_index = 0
        for i in range(occurences):
            for replace_char in replace.split(","):
                modified_word = list(word)
                index = modified_word.index(find, start_index)
                modified_word[index] = replace_char
                modified_word = "".join(modified_word)
                print modified_word
            start_index = index + 1

Which gives:

Words
Shirt
Blousq
Blousw
Swqater
Swwater
Sweatqr
Sweatwr
Words
Shirt
Blouse
Swezter

Instead of printing the words, you can append them to a list and re-create a DataFrame if that's what you want to end up with.

Answer 2

This answer is very similar to Brian's answer, but a little bit sanitized and the output has no duplicates:

words = ["Words", "Shirt", "Blouse", "Sweater"]
md = {"e": "q,w", "a": "z"}
md = {k: v.split(',') for k, v in md.items()}

newwords = []

for word in words:
    newwords.append(word)
    for c in md:
        occ = word.count(c)
        pos = 0
        for _ in range(occ):
            pos = word.find(c, pos)
            for r in md[c]:
                tmp = word[:pos] + r + word[pos+1:]
                newwords.append(tmp)
            pos += 1

Content of newwords:

['Words', 'Shirt', 'Blouse', 'Blousq', 'Blousw', 'Sweater', 'Swqater', 'Swwater', 'Sweatqr', 'Sweatwr', 'Swezter']

Prettyprint:

Words
Shirt
Blouse
Blousq
Blousw
Sweater
Swqater
Swwater
Sweatqr
Sweatwr
Swezter

Any errors are a result of the current time. ;)

---

Update (explanation)

tl;dr

The main idea is to find the occurences of the character in the word one after another. For each occurence we are then replacing it with the replacing-char (again one after another). The replaced word get's added to the output-list.

I will try to explain everything step by step:

words = ["Words", "Shirt", "Blouse", "Sweater"]
md = {"e": "q,w", "a": "z"}

Well. Your basic input. :)

md = {k: v.split(',') for k, v in md.items()}

A simpler way to deal with replacing-dictionary. md now looks like {"e": ["q", "w"], "a": ["z"]}. Now we don't have to handle "q,w" and "z" differently but the step for replacing is just the same and ignores the fact, that "a" only got one replace-char.

newwords = []

The new list to store the output in.

for word in words:
    newwords.append(word)

We have to do those actions for each word (I assume, the reason is clear). We also append the world directly to our just created output-list (newwords).

    for c in md:

c as short for character. So for each character we want to replace (all keys of md), we do the following stuff.

        occ = word.count(c)

occ for occurrences (yeah. count would fit as well :P). word.count(c) returns the number of occurences of the character/string c in word. So "Sweater".count("o") => 0 and "Sweater".count("e") => 2. We use this here to know, how often we have to take a look at word to get all those occurences of c.

        pos = 0

Our startposition to look for c in word. Comes into use in the next loop.

        for _ in range(occ):

For each occurence. As a continual number has no value for us here, we "discard" it by naming it _. At this point where c is in word. Yet.

            pos = word.find(c, pos)

Oh. Look. We found c. :) word.find(c, pos) returns the index of the first occurence of c in word, starting at pos. At the beginning, this means from the start of the string => the first occurence of c. But with this call we already update pos. This plus the last line (pos += 1) moves our search-window for the next round to start just behind the previous occurence of c.

            for r in md[c]:

Now you see, why we updated mc previously: we can easily iterate over it now (a md[c].split(',') on the old md would do the job as well). So we are doing the replacement now for each of the replacement-characters.

                tmp = word[:pos] + r + word[pos+1:]

The actual replacement. We store it in tmp (for debug-reasons). word[:pos] gives us word up to the (current) occurence of c (exclusive c). r is the replacement. word[pos+1:] adds the remaining word (again without c).

                newwords.append(tmp)

Our so created new word tmp now goes into our output-list (newwords).

            pos += 1

The already mentioned adjustment of pos to "jump over c".

---

Additional question from OP: Is there an easy way to dictate how many letters in the string I want to replace [(meaning e.g. multiple at a time)]?

Surely. But I have currently only a vague idea on how to achieve this. I am going to look at it, when I got my sleep. ;)

words = ["Words", "Shirt", "Blouse", "Sweater", "multipleeee"]
md = {"e": "q,w", "a": "z"}
md = {k: v.split(',') for k, v in md.items()}
num = 2     # this is the number of replaces at a time.

newwords = []

for word in words:
    newwords.append(word)
    for char in md:
        for r in md[char]:
            pos = multiples = 0
            current_word = word
            while current_word.find(char, pos) != -1:
                pos = current_word.find(char, pos)
                current_word = current_word[:pos] + r + current_word[pos+1:]
                pos += 1
                multiples += 1
                if multiples == num:
                    newwords.append(current_word)
                    multiples = 0
                    current_word = word

Content of newwords:

['Words', 'Shirt', 'Blouse', 'Sweater', 'Swqatqr', 'Swwatwr', 'multipleeee', 'multiplqqee', 'multipleeqq', 'multiplwwee', 'multipleeww']

Prettyprint:

Words
Shirt
Blouse
Sweater
Swqatqr
Swwatwr
multipleeee
multiplqqee
multipleeqq
multiplwwee
multipleeww

I added multipleeee to demonstrate, how the replacement works: For num = 2 it means the first two occurences are replaced, after them, the next two. So there is no intersection of the replaced parts. If you would want to have something like ['multiplqqee', 'multipleqqe', 'multipleeqq'], you would have to store the position of the "first" occurence of char. You can then restore pos to that position in the if multiples == num:-block.

If you got further questions, feel free to ask. :)

Answer 3

If you are looping, you need to update s at each cycle of the loop. You also need to loop over v.

mydict = {"e":"q,w"}
s=deduped
for k, v in mydict.items():
     for j in v:
          s = s.replace(k, j)

Then reassign it to your dataframe:

df["Words"] = s

If you can write this as a function that takes in a 1d array (list, numpy array etc...), you can use df.apply to apply it to any column, using df.apply().