有条件地替换数据帧列中的字符串的一部分

Question

如果我在列中没有特定的分隔符/字符串，我想替换一个字符串。如果该行中已存在分隔符，则我不想触摸该行。我有大约350万条记录。

以下是样本集。我想替换是：

One:1
Two:2 
Three is 3 
Four is IV:4

输出应该是这样的

One:1
Two:2
Three:3
Four is IV:4

Answer 1

选项1 适用于update

df.update(
    df.myValues.loc[
        lambda x: ~x.str.contains(':')
    ].str.replace('\s+is\s+', ':'))

       myValues
0         One:1
1         Two:2
2       Three:3
3  Four is IV:4

---

选项2 内联和使用map

f = lambda x: x if ':' in x else x.replace(' is ', ':')
df.assign(myValues=list(map(f, v)))

       myValues
0         One:1
1         Two:2
2       Three:3
3  Four is IV:4

Answer 2

除了.contains()之外，您还可以使用简单的字符串操作：

df = pd.DataFrame(["One:1", "Two:2", "Three is 3", "Four is IV:4"], columns=["myValues"])
target = [":" not in e for e in df.myValues]
df.myValues[target] = df.myValues[target].str.replace(" is ",":")

结果：

       myValues
0         One:1
1         Two:2
2       Three:3
3  Four is IV:4

Answer 3

首先，过滤掉所有包含:的字符串。然后，对于剩下的所有行，将“is”替换为“：”。 （在你的例子中，“is”周围的空格也被删除。因此，我将“is”替换为“：”。）

df = pd.DataFrame(["One:1", "Two:2", "Three is 3", "Four is IV:4"], columns=["myValues"])
for idx, v in df[~df.myValues.str.contains(":")].iterrows():
    df.loc[idx].myValues = df.iloc[idx].myValues.replace(" is ", ":")

参考

• Series.str.contains
• How can I obtain the element-wise logical NOT of a pandas Series?

Answer 4

尝试没有循环和使用loc的单行

df = pd.DataFrame(["One:1", "Two:2", "Three is 3", "Four is IV:4", "Five is V"], columns=["myValues"])


df.loc[~df['myValues'].str.contains(':'), 'myValues'] = df.loc[~df['myValues'].str.contains(':'), 'myValues'].str.replace('is', ':')

打印（DF）

    myValues
0   One:1
1   Two:2
2   Three : 3
3   Four is IV:4
4   Five : V